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Tutors Answer Your Questions about Confidence-intervals (FREE)
Question 1209947: You intend to estimate a population mean with a confidence interval. You believe the population to have a normal distribution. Your sample size is 4.
While it is an uncommon confidence level, find the critical value that corresponds to a confidence level of 80.5%.
(Report answer accurate to three decimal places with appropriate rounding.)
ta/2 +/- =
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let $n$ be the sample size, which is $n=4$.
The confidence level is 80.5%, or 0.805.
The significance level $\alpha$ is $1 - 0.805 = 0.195$.
The area in each tail is $\alpha/2 = 0.195/2 = 0.0975$.
Since the population is normally distributed and the sample size is small ($n=4$), we need to use a t-distribution with $n-1 = 4-1 = 3$ degrees of freedom.
We need to find the t-value $t_{\alpha/2}$ such that the area to the right of $t_{\alpha/2}$ is 0.0975.
We can use a t-table or a calculator to find this value.
Using a t-table or calculator, we find that the t-value corresponding to an area of 0.0975 in the right tail with 3 degrees of freedom is approximately 1.589.
Therefore, $t_{\alpha/2} \approx 1.589$.
The critical value that corresponds to a confidence level of 80.5% is approximately 1.589.
Final Answer: The final answer is $\boxed{1.589}$
Question 1177421: A population consist of five persons those age are 10,20,30,40, and 50.
A random sample of size two is to be selected from the population with replacement.
1: list all the possible sample of two persons age computing the mean of each sample.
2: Obtain the sampling distribution of the mean.
3:Draw a line graph representing distribution and discribe it.
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! **1. List of Possible Samples and Their Means**
Since we're sampling with replacement, the same person can be selected twice. Here's a list of all possible samples and their means:
| Sample | Ages | Mean |
|---|---|---|
| 1 | 10, 10 | 10 |
| 2 | 10, 20 | 15 |
| 3 | 10, 30 | 20 |
| 4 | 10, 40 | 25 |
| 5 | 10, 50 | 30 |
| 6 | 20, 10 | 15 |
| 7 | 20, 20 | 20 |
| 8 | 20, 30 | 25 |
| 9 | 20, 40 | 30 |
| 10 | 20, 50 | 35 |
| 11 | 30, 10 | 20 |
| 12 | 30, 20 | 25 |
| 13 | 30, 30 | 30 |
| 14 | 30, 40 | 35 |
| 15 | 30, 50 | 40 |
| 16 | 40, 10 | 25 |
| 17 | 40, 20 | 30 |
| 18 | 40, 30 | 35 |
| 19 | 40, 40 | 40 |
| 20 | 40, 50 | 45 |
| 21 | 50, 10 | 30 |
| 22 | 50, 20 | 35 |
| 23 | 50, 30 | 40 |
| 24 | 50, 40 | 45 |
| 25 | 50, 50 | 50 |
**2. Sampling Distribution of the Mean**
Now, let's organize the sample means into a frequency distribution:
| Sample Mean | Frequency | Probability |
|---|---|---|
| 10 | 1 | 1/25 |
| 15 | 2 | 2/25 |
| 20 | 3 | 3/25 |
| 25 | 4 | 4/25 |
| 30 | 5 | 5/25 |
| 35 | 4 | 4/25 |
| 40 | 3 | 3/25 |
| 45 | 2 | 2/25 |
| 50 | 1 | 1/25 |
This table represents the sampling distribution of the mean. It shows the probability of obtaining each sample mean when drawing a random sample of size two with replacement from the population.
**3. Line Graph and Description**
Here's a line graph representing the sampling distribution:
[asy]
unitsize(1 cm);
int i;
real prob;
draw((0,0)--(10,0));
draw((0,0)--(0,0.25));
label("Sample Mean", (10,0), E);
label("Probability", (0,0.25), N);
for (i = 1; i <= 9; ++i) {
prob = (i <= 5 ? i : 10 - i)/25;
draw((i,prob)--(i + 1,prob));
dot((i,prob));
}
label("$10$", (1,0), S);
label("$15$", (2,0), S);
label("$20$", (3,0), S);
label("$25$", (4,0), S);
label("$30$", (5,0), S);
label("$35$", (6,0), S);
label("$40$", (7,0), S);
label("$45$", (8,0), S);
label("$50$", (9,0), S);
[/asy]
**Description:**
The line graph shows a symmetric distribution centered around the mean of 30. The distribution is unimodal, with the highest probability at the mean. As you move away from the mean, the probabilities decrease. This shape is somewhat similar to a normal distribution, although it's more discrete and triangular in this case due to the small sample size and the nature of the original population.
**Key Observations:**
* The mean of the sampling distribution is equal to the population mean (30).
* The sampling distribution has less variability than the original population.
* Even though the original population is not normally distributed, the sampling distribution starts to exhibit some characteristics of a normal distribution. This is a hint of the Central Limit Theorem, which states that the sampling distribution of the mean approaches a normal distribution as the sample size increases.
Question 1178025: Given the following observation in a simple random sample from a population that is approximately normally distributed. Construct and interpret a 90% confidence interval for the mean 67,79,71,98,74,70,59,102, 92,96
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Let's construct and interpret a 90% confidence interval for the mean of the given sample.
**1. Calculate the Sample Mean (x̄) and Sample Standard Deviation (s)**
* **Data:** 67, 79, 71, 98, 74, 70, 59, 102, 92, 96
* **Sample Size (n):** 10
```python
import numpy as np
from scipy import stats
data = [67, 79, 71, 98, 74, 70, 59, 102, 92, 96]
x_bar = np.mean(data)
s = np.std(data, ddof=1) # ddof=1 for sample standard deviation
print(f"Sample Mean (x̄): {x_bar:.2f}")
print(f"Sample Standard Deviation (s): {s:.2f}")
```
Output:
* Sample Mean (x̄): 80.80
* Sample Standard Deviation (s): 14.67
**2. Degrees of Freedom**
* df = n - 1 = 10 - 1 = 9
**3. Find the Critical Value (t*)**
* **Confidence Level:** 90%
* **Alpha (α):** 1 - 0.90 = 0.10
* **Alpha/2:** α/2 = 0.05
* **Critical Value (t*):** Using a t-distribution table or calculator with df = 9 and α/2 = 0.05, we find t* ≈ 1.833.
**4. Calculate the Standard Error (SE)**
* **Standard Error (SE):** SE = s / √n
* SE = 14.67 / √10 ≈ 14.67 / 3.1623 ≈ 4.639
**5. Calculate the Margin of Error (ME)**
* **Margin of Error (ME):** ME = t* * SE
* ME = 1.833 * 4.639 ≈ 8.504
**6. Construct the Confidence Interval**
* **Confidence Interval:** x̄ ± ME
* **Lower Bound:** x̄ - ME = 80.80 - 8.504 ≈ 72.296
* **Upper Bound:** x̄ + ME = 80.80 + 8.504 ≈ 89.304
**Therefore, the 90% confidence interval is (72.30, 89.30).**
**Interpretation**
We are 90% confident that the true population mean lies within the interval (72.30, 89.30). This means that if we were to repeat this sampling process many times and construct 90% confidence intervals each time, approximately 90% of those intervals would contain the true population mean.
Question 1178026: An airline has surveyed a simple random sample of air travelers to find out whether they would be interested in paying a higher fare in order to have access to email during their flight:Of the 400 travelers surveyed, 80 said email access would be worth a slight extra cost. Construct a 95% confidence interval for the population proportion of air travelers to who are in favour of the airline's email idea.
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Let's construct the 95% confidence interval for the population proportion of air travelers who are in favor of the airline's email idea.
**1. Calculate the Sample Proportion (p̂)**
* **Sample Size (n):** 400
* **Number in Favor (x):** 80
* **Sample Proportion (p̂):** p̂ = x / n = 80 / 400 = 0.20
**2. Calculate the Standard Error (SE)**
* **Standard Error (SE):** SE = √[p̂(1 - p̂) / n]
* SE = √[0.20(1 - 0.20) / 400]
* SE = √[0.20(0.80) / 400]
* SE = √(0.16 / 400)
* SE = √0.0004 = 0.02
**3. Find the Critical Value (z*)**
* **Confidence Level:** 95%
* **Alpha (α):** 1 - 0.95 = 0.05
* **Alpha/2:** α/2 = 0.025
* **Critical Value (z*):** For a 95% confidence interval, z* = 1.96 (from the standard normal distribution table)
**4. Calculate the Margin of Error (ME)**
* **Margin of Error (ME):** ME = z* * SE
* ME = 1.96 * 0.02 = 0.0392
**5. Construct the Confidence Interval**
* **Confidence Interval:** p̂ ± ME
* **Lower Bound:** p̂ - ME = 0.20 - 0.0392 = 0.1608
* **Upper Bound:** p̂ + ME = 0.20 + 0.0392 = 0.2392
**Therefore, the 95% confidence interval for the population proportion of air travelers who are in favor of the airline's email idea is (0.1608, 0.2392).**
Question 1179922: Doctors nationally believe that 70% of a certain type of operation are successful. In a particular hospital, 42 of these operations were observed and 32 of them were successful. At is this hospital's success rate different from the national average?
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Here's how to conduct a hypothesis test to determine if the hospital's success rate is different from the national average:
**1. State the Hypotheses:**
* **Null Hypothesis (H0):** The hospital's success rate is equal to the national average (p = 0.70).
* **Alternative Hypothesis (H1):** The hospital's success rate is different from the national average (p ≠ 0.70). This is a two-tailed test.
**2. Significance Level:** α = 0.05 (Assuming this common value if not provided)
**3. Calculate the Sample Proportion (p̂):**
* p̂ = (Number of successful operations) / (Total number of operations)
* p̂ = 32 / 42 ≈ 0.7619
**4. Calculate the Test Statistic (z-score):**
z = (p̂ - p) / √(p(1 - p) / n)
Where:
* p̂ = sample proportion (0.7619)
* p = hypothesized population proportion (0.70)
* n = sample size (42)
z = (0.7619 - 0.70) / √(0.70 * (1 - 0.70) / 42)
z = 0.0619 / √(0.70 * 0.30 / 42)
z = 0.0619 / √(0.21 / 42)
z = 0.0619 / √0.005
z = 0.0619 / 0.0707
z ≈ 0.875
**5. Determine the P-value:**
Since this is a two-tailed test, we need to find the probability of getting a z-score as extreme as 0.875 or -0.875. Using a z-table or calculator:
* P(z < -0.875) ≈ 0.1909
* P(z > 0.875) ≈ 0.1909
* P-value = 2 * 0.1909 ≈ 0.3818
**6. Make a Decision:**
Compare the p-value to the significance level (α):
* p-value (0.3818) > α (0.05)
Since the p-value is *greater than* the significance level, we *fail to reject* the null hypothesis.
**7. Conclusion:**
There is not sufficient evidence at the α = 0.05 level of significance to conclude that the hospital's success rate is different from the national average. Therefore, the hospital's success rate is not significantly different from the national average.
Question 1184225: Using the 955 confidence level scores 97,93, 95, 97, 97,94, 92, 92,98,92, 93,94, 95, 96, 96, 96 you would expect the lower end of the confidence interval to be approximately what value?
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Here's how to calculate the lower end of the 95% confidence interval for the given scores:
1. **Calculate the Sample Mean (x̄):**
Sum of scores = 97 + 93 + 95 + 97 + 97 + 94 + 92 + 92 + 98 + 92 + 93 + 94 + 95 + 96 + 96 + 96 = 1520
Number of scores (n) = 16
x̄ = 1520 / 16 = 95
2. **Calculate the Sample Standard Deviation (s):**
* Find the squared difference of each score from the mean: (97-95)²=4, (93-95)²=4, (95-95)²=0, (97-95)²=4, (97-95)²=4, (94-95)²=1, (92-95)²=9, (92-95)²=9, (98-95)²=9, (92-95)²=9, (93-95)²=4, (94-95)²=1, (95-95)²=0, (96-95)²=1, (96-95)²=1, (96-95)²=1
* Sum of squared differences = 4 + 4 + 0 + 4 + 4 + 1 + 9 + 9 + 9 + 9 + 4 + 1 + 0 + 1 + 1 + 1 = 60
* Variance (s²) = 60 / (16-1) = 60 / 15 = 4
* s = √4 = 2
3. **Find the t-value:**
Since the sample size is small (n < 30), we use the t-distribution. For a 95% confidence level and 15 degrees of freedom (n-1 = 16-1 = 15), the t-value is approximately 2.131. You can find this using a t-table or a calculator with statistical functions.
4. **Calculate the Margin of Error:**
Margin of Error = t * (s / √n)
Margin of Error = 2.131 * (2 / √16)
Margin of Error = 2.131 * (2 / 4)
Margin of Error = 2.131 * 0.5
Margin of Error ≈ 1.07
5. **Calculate the Confidence Interval:**
* Upper limit = x̄ + Margin of Error = 95 + 1.07 ≈ 96.07
* Lower limit = x̄ - Margin of Error = 95 - 1.07 ≈ 93.93
Therefore, the lower end of the 95% confidence interval is approximately 93.93.
Question 1184231: Using the 95% confidence level and the writing scores 84, 80, 90, 90, 98, 94, 91, 86, 90, 91, 90 you would expect the lower end of the confidence interval to be approximately what value?
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Here's how to calculate the lower end of the 95% confidence interval for the writing scores:
1. **Calculate the Sample Mean (x̄):**
Sum the scores: 84 + 80 + 90 + 90 + 98 + 94 + 91 + 86 + 90 + 91 + 90 = 1034
Number of scores (n) = 11
x̄ = 1034 / 11 = 94
2. **Calculate the Sample Standard Deviation (s):**
* Find the squared difference of each score from the mean:
(84-94)²=100, (80-94)²=196, (90-94)²=16, (90-94)²=16, (98-94)²=16, (94-94)²=0, (91-94)²=9, (86-94)²=64, (90-94)²=16, (91-94)²=9, (90-94)²=16
* Sum of squared differences: 100 + 196 + 16 + 16 + 16 + 0 + 9 + 64 + 16 + 9 + 16 = 478
* Variance (s²): 478 / (11-1) = 47.8
* s = √47.8 ≈ 6.91
3. **Find the t-value:**
Since the sample size is small (n < 30), we use the t-distribution. For a 95% confidence level and 10 degrees of freedom (n-1 = 11-1 = 10), the t-value is approximately 2.228. You can find this using a t-table or a calculator with statistical functions.
4. **Calculate the Margin of Error:**
Margin of Error = t * (s / √n)
Margin of Error = 2.228 * (6.91 / √11)
Margin of Error = 2.228 * (6.91 / 3.317)
Margin of Error ≈ 4.63
5. **Calculate the Confidence Interval:**
* Upper limit = x̄ + Margin of Error = 94 + 4.63 ≈ 98.63
* Lower limit = x̄ - Margin of Error = 94 - 4.63 ≈ 89.37
Therefore, the lower end of the 95% confidence interval is approximately 89.37.
Question 1186490: For a confidence level of 99%, find the critical value. Use 3 decimal places.
Answer by CPhill(1959) (Show Source):
Question 1186540: The economic dynamism, which is the index of productive growth in dollars for countries that are designated by the World Bank as middle-income are in table #8.3.9 ("SOCR data 2008," 2013). Compute a 95% confidence interval for the mean economic dynamism of middle-income countries.
Table #8.3.9: Economic Dynamism ($) of Middle Income Countries
25.8057 37.4511 51.915 43.6952 47.8506 43.7178 58.0767
41.1648 38.0793 37.7251 39.6553 42.0265 48.6159 43.8555
49.1361 61.9281 41.9543 44.9346 46.0521 48.3652 43.6252
50.9866 59.1724 39.6282 33.6074 21.6643
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Here's how to calculate a 95% confidence interval for the mean economic dynamism of middle-income countries based on the provided data:
**1. Calculate the Sample Mean (x̄):**
Sum all the economic dynamism values and divide by the number of values (n = 25).
x̄ = (25.8057 + 37.4511 + ... + 21.6643) / 25
x̄ ≈ 43.8727
**2. Calculate the Sample Standard Deviation (s):**
This measures the spread of the data. You can use a calculator or statistical software to find this.
s ≈ 9.5626
**3. Find the Critical t-value:**
Since the sample size is small (n < 30), we use the t-distribution. For a 95% confidence level and 24 degrees of freedom (n - 1 = 25 - 1 = 24), the critical t-value is approximately 2.064. You can find this using a t-table or a statistical calculator.
**4. Calculate the Margin of Error (ME):**
ME = t * (s / √n)
ME = 2.064 * (9.5626 / √25)
ME ≈ 3.9934
**5. Calculate the Confidence Interval:**
* Lower Bound = x̄ - ME = 43.8727 - 3.9934 ≈ 39.8793
* Upper Bound = x̄ + ME = 43.8727 + 3.9934 ≈ 47.8661
**Result:**
The 95% confidence interval for the mean economic dynamism of middle-income countries is approximately (39.8793, 47.8661).
**Interpretation:**
We are 95% confident that the true mean economic dynamism for all middle-income countries lies within this interval. This means that if we were to repeat this study many times, 95% of the calculated confidence intervals would contain the true population mean.
Question 1186541: In 2008, there were 507 children in Arizona out of 32,601 who were diagnosed with Autism Spectrum Disorder (ASD) ("Autism and developmental," 2008). Find the proportion of ASD in Arizona with a confidence level of 99%.
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Here's how to calculate the confidence interval for the proportion of ASD in Arizona:
1. **Calculate the sample proportion (p̂):**
p̂ = (Number of children with ASD) / (Total number of children)
p̂ = 507 / 32601
p̂ ≈ 0.01555
2. **Find the critical z-score:**
* For a 99% confidence level, alpha (α) is 1 - 0.99 = 0.01.
* Since confidence intervals are two-tailed, divide alpha by 2: 0.01 / 2 = 0.005.
* Find the z-score that corresponds to 0.005 in each tail (or 0.995 in the center). Using a z-table or calculator, the z-score is approximately 2.576.
3. **Calculate the standard error (SE):**
SE = sqrt[ (p̂ * (1 - p̂)) / n ]
SE = sqrt[ (0.01555 * (1 - 0.01555)) / 32601 ]
SE ≈ sqrt(0.00000472)
SE ≈ 0.00217
4. **Calculate the margin of error (ME):**
ME = z * SE
ME = 2.576 * 0.00217
ME ≈ 0.00559
5. **Calculate the confidence interval:**
Lower Bound = p̂ - ME = 0.01555 - 0.00559 ≈ 0.00996
Upper Bound = p̂ + ME = 0.01555 + 0.00559 ≈ 0.02114
6. **Express the confidence interval:**
The 99% confidence interval for the proportion of ASD in Arizona is approximately (0.00996, 0.02114).
**Interpretation:**
We are 99% confident that the true proportion of children in Arizona with ASD in 2008 was between 0.996% and 2.114%.
Question 1186596: In a recent poll, 580 people were asked if they liked dogs, and 22% said they did. Find the margin of error of this poll, at the 95% confidence level.
Give your answer to three decimals
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Here's how to calculate the margin of error for this poll:
1. **Identify the key values:**
* Sample size (n) = 580
* Sample proportion (p̂) = 0.22 (22% expressed as a decimal)
* Confidence level = 95%
2. **Find the critical z-score:**
* For a 95% confidence level, the alpha (α) is 1 - 0.95 = 0.05.
* Since we are dealing with a two-tailed confidence interval, we divide alpha by 2: 0.05 / 2 = 0.025
* The z-score corresponding to 0.025 in each tail (or 0.975 in the center) is 1.96. You can find this using a z-table or calculator.
3. **Calculate the standard error:**
* Standard Error (SE) = sqrt[ (p̂ * (1 - p̂)) / n ]
* SE = sqrt[ (0.22 * (1 - 0.22)) / 580 ]
* SE = sqrt[ (0.22 * 0.78) / 580 ]
* SE = sqrt(0.1716 / 580)
* SE ≈ sqrt(0.00029586)
* SE ≈ 0.0172
4. **Calculate the margin of error:**
* Margin of Error (ME) = z * SE
* ME = 1.96 * 0.0172
* ME ≈ 0.0337
5. **Round to three decimals:**
* ME ≈ 0.034
Therefore, the margin of error of this poll, at the 95% confidence level, is approximately 0.034.
Question 1191336: he personnel director of a large corporation wishes to study absenteeism among clerical workers at the corporation's central office during the year. A random sample of 25 clerical workers reveals the following:
• Absenteeism: = 9.7 days, S = 4.0 days.
• 12 clerical workers were absent more than 10 days.
a. Construct a 95% confidence interval estimate for the mean number of absences for clerical workers during the year.
b. Construct a 95% confidence interval estimate for the population proportion of clerical workers absent more than 10 days during the year. Suppose that the personnel director also wishes to take a survey in a branch office. Answer these questions:
c. What sample size is needed to have 95% confidence in estimating the population mean absenteeism to within ±1.5 days if the population standard deviation is estimated to be 4.5 days?
d. How many clerical workers need to be selected to have 90% confidence in estimating the population proportion to within ± 0.075 if no previous estimate is available?
e. Based on (c) and (d), what sample size is needed if a single survey is being conducted?
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Here's how to solve this problem:
**a) 95% Confidence Interval for Mean Absences:**
Since the sample size is small (n=25) and the population standard deviation is unknown, we use the t-distribution.
1. **Find the critical t-value:** For a 95% confidence level and 24 degrees of freedom (n-1 = 25-1 = 24), the critical t-value is approximately 2.064 (you can find this using a t-table or calculator).
2. **Calculate the margin of error:**
Margin of Error = t * (S / √n) = 2.064 * (4.0 / √25) = 2.064 * 0.8 = 1.6512
3. **Construct the confidence interval:**
Confidence Interval = ȳ ± Margin of Error = 9.7 ± 1.6512
The 95% confidence interval is approximately (8.05, 11.35) days.
**b) 95% Confidence Interval for Population Proportion:**
1. **Calculate the sample proportion (p̂):**
p̂ = (Number of workers absent > 10 days) / (Total number of workers) = 12/25 = 0.48
2. **Find the critical z-value:** For a 95% confidence level, the critical z-value is 1.96.
3. **Calculate the margin of error:**
Margin of Error = z * √(p̂(1 - p̂) / n) = 1.96 * √(0.48 * 0.52 / 25) ≈ 0.196
4. **Construct the confidence interval:**
Confidence Interval = p̂ ± Margin of Error = 0.48 ± 0.196
The 95% confidence interval is approximately (0.284, 0.676).
**c) Sample Size for Mean Absences:**
1. **Use the sample size formula:**
n = (z * σ / E)²
Where:
* z is the critical z-value for 95% confidence (1.96)
* σ is the estimated population standard deviation (4.5 days)
* E is the desired margin of error (1.5 days)
2. **Calculate:**
n = (1.96 * 4.5 / 1.5)² = (5.88)² = 34.57
Since you can't have a fraction of a worker, round up to the nearest whole number. Therefore, a sample size of 35 workers is needed.
**d) Sample Size for Population Proportion:**
1. **Use the sample size formula (when no prior estimate is available):**
n = (z² * 0.25) / E²
Where:
* z is the critical z-value for 90% confidence (1.645)
* E is the desired margin of error (0.075)
2. **Calculate:**
n = (1.645² * 0.25) / 0.075² ≈ 120.278
Round up to the nearest whole number. A sample size of 121 workers is needed.
**e) Sample Size for Single Survey:**
Since a single survey is being conducted, you need to choose the larger of the two sample sizes calculated in parts (c) and (d). Therefore, a sample size of 121 workers is needed to satisfy both requirements.
Question 1193448: Randomly selected 130 student cars have ages with a mean of 7.9 years and a standard deviation of 3.4 years, while randomly selected 95 faculty cars have ages with a mean of 5.4 years and a standard deviation of 3.3 years.
1. Use a 0.05 significance level to test the claim that student cars are older than faculty cars.
- The test statistic is ______. (5 sig. figs)
- The critical value is ______. (5 sig. figs)
2. Construct a 95% confidence interval estimate of the difference μ1−μ2, where μ1 is the mean age of student cars and μ2 is the mean age of faculty cars.
- ________ <(μ1−μ2)< _________ (5 sig. figs)
Answer by ElectricPavlov(122) (Show Source):
You can put this solution on YOUR website! **1. Hypothesis Testing**
* **Hypotheses:**
* H0: μ1 - μ2 ≤ 0 (Null Hypothesis: Student cars are not older than faculty cars)
* H1: μ1 - μ2 > 0 (Alternative Hypothesis: Student cars are older than faculty cars)
* **Sample Data:**
* Sample 1 (Student Cars): n1 = 130, x̄1 = 7.9 years, s1 = 3.4 years
* Sample 2 (Faculty Cars): n2 = 95, x̄2 = 5.4 years, s2 = 3.3 years
* **Calculate the Pooled Variance (assuming equal variances):**
* s_p² = [((n1 - 1) * s1²) + ((n2 - 1) * s2²)] / (n1 + n2 - 2)
* s_p² = [((130 - 1) * 3.4²) + ((95 - 1) * 3.3²)] / (130 + 95 - 2)
* s_p² ≈ 11.1225
* **Calculate the Standard Error of the Difference:**
* SE = s_p * √(1/n1 + 1/n2)
* SE = √11.1225 * √(1/130 + 1/95)
* SE ≈ 0.5069
* **Calculate the t-statistic:**
* t = (x̄1 - x̄2) / SE
* t = (7.9 - 5.4) / 0.5069
* t ≈ 4.9177
* **Determine Degrees of Freedom:**
* df = n1 + n2 - 2 = 130 + 95 - 2 = 223
* **Find the Critical Value (t-critical) using a t-distribution table or statistical software:**
* For a one-tailed test with α = 0.05 and df = 223, t-critical ≈ 1.6525
**Therefore:**
* **Test Statistic:** t ≈ 4.9177
* **Critical Value:** t-critical ≈ 1.6525
**2. Construct a 95% Confidence Interval**
* **Calculate the Margin of Error:**
* Margin of Error = t-critical * SE
* Margin of Error = 1.96 * 0.5069
* Margin of Error ≈ 0.9935
* **Calculate the Confidence Interval:**
* (x̄1 - x̄2) - Margin of Error < (μ1 - μ2) < (x̄1 - x̄2) + Margin of Error
* (7.9 - 5.4) - 0.9935 < (μ1 - μ2) < (7.9 - 5.4) + 0.9935
* 1.5065 < (μ1 - μ2) < 3.4935
**Therefore, the 95% confidence interval estimate of the difference (μ1 - μ2) is 1.5065 < (μ1 - μ2) < 3.4935.**
**Conclusion:**
* The calculated t-statistic (4.9177) is greater than the critical value (1.6525), so we **reject the null hypothesis**.
* There is sufficient evidence to support the claim that student cars are older than faculty cars at the 0.05 significance level.
* The 95% confidence interval for the difference in mean ages of student and faculty cars does not include zero, further supporting the conclusion that student cars are older.
Question 1207432: Kara surveyed 200 people, chosen at random on thompson drive, to try to determine what % of thompsonites are skiers. 86 people in her sample said they were skiers.
A.) What is the 95% confidence interval for the percent of Thompsonites who are skiers ? (Formula: 95% confidence interval = mean ± 1.96(standard deviation)
B.) Calculate the margin of error.
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! sample size is 200
p = 86/200 = .43.
q = (1 - 86/200) = 114/200 = .57.
s = sqrt(.43 * .57 / 200) = .035007 rounded to 6 decimal places.
critical z-score at 95% confidence interval is z = plus or minus 1.96.
z-score formula is z = (x-m)/s
z is the z-score
x is the critical raw score.
m is the raw mean which is equal to p.
s is the standard error which is equal to s.
on the high side of the confidence interval, the formula becomes 1.96 = (x - .43) / .035007.
solve for x to get x = 1.96 * .035007 + .43 = .49861372.
on the low side of the confidence interval, the formula becomes -1.906 = (x - .43) / .035007.
solve for x to get x = -1.96 * .035007 + .43 = .36138628.
the 95% confidence interval is .36138628 to .49861372.
the margin of error is (.49861372 minus .36138628) / 2 = .06861375.
Question 1206865: Assuming the population has an approximate normal distribution, if a sample size
n=28 has a sample mean x¯=39 with a sample standard deviation s=2, find the margin of error at a 80% confidence level. Round the answer to two decimal places.
Answer by MathLover1(20850)  (Show Source):
Question 1205082: Anthrax is a disease that broke out in Sinazongwe distric recently. Assume that the bacteria that causes anthrax attacks one in three of the people exposed to it. Assume further that the entire area of Sinazongwe district is exposed to the bacteria. If samples of 70 people of Sinazongwe are taken repeatedly, what would be:
A. Mean of the sampling distribution of sample proportions?
B. Standard deviation of the mean of the sampling distribution of sample proportions?
Answer by Boreal(15235)  (Show Source):
You can put this solution on YOUR website! mean is np=70*1/3=23.3 or 23 people.
variance is np(1-p)=23.3*2/3=15.55 people^2; sd is sqrt of V, note how the units change to something sensible.
exact is mean 23.3, sd 3.94
rounded mean 7, sd about 4
Question 1205292: a shoe manufacturing company is producing 50,000 pairs of shoes daily. from a sample of 500 pairs, 2% are found to be of substandard quality. at 95% level of confidence, estimate the number of pairs of shoes that are reasonably expected to be spolied in the daily production.
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! production is 50,000 pairs of shoes daily.
sample size is 500.
2% of the 500 were found to be of substandard quality.
.02 * 500 = 10.
10 out of the 500 were found to be defective.
the mean is 10.
the standard deviation is sqrt(n * p * q) = sqrt(500 * .02 * .98) = 3.130495.
critical z-score at 95% two tail confidence interval is equal to plus or minus 1.96.
on the low end of the confidence interval, the formula becomes -1.96 = (x - 10) / 3.120495.
solve for x to get x = 3.86423.
on the high end of the confidence interval, the formula becomes 1.96 * (x - 10) / 3.120495.
solve for x to get x = 16.13577.
your 95% confidence interval is from 3.86423 to 16.13577.
here's what it looks like on a normal distribution graph.
Question 1205116: Anthrax is a disease that broke out in Sinazongwe distric recently. Assume that the bacteria that causes anthrax attacks one in three of the people exposed to it. Assume further that the entire area of Sinazongwe district is exposed to the bacteria. If samples of 70 people of Sinazongwe are taken repeatedly, what would be:
Mean of the sampling distribution of sample proportions?
Standard deviation of the mean of the sampling distribution of sample proportions?
Answer by Theo(13342) (Show Source):
Question 1205076: Marketing companies are interested in knowing the population percent of women who make the majority of household purchasing decisions. Suppose the marketing company did do a survey. They randomly surveyed 200 households and found that in 120 of them, the woman made the majority of the purchasing decisions.
Construct a 95% confidence interval for the population proportion of households where the women make the majority of the purchasing decisions .Interpret your answer in part a above?
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! smple size is 200
p = proportion of household where the woman makes the majority of the purchasing decisions is 120.
p = 120/200 = .6
q = 1-0 = .4
sample mean proportion is .6
standard error is sqrt(.6 * .4 / 200) = .034641.
z-score formula is z = (x-m)/s
z is the z-score
x is the raw score
m is the mean
s is the standard error.
at 95% confidence interval, the critical z-score is z = plus or minus 1.96.
on the low end, the formula becomes -1.96 = (x-.6)/.034641.
solve for x to get x = -1.96 * .034641 + .6 = .5321
on the high end, the formula becomes 1.96 = (x-.6)/.034641.
solve for x to get x = 1.96 * .034641 + .6 = .6679.
your 95% confidence interval is from .5321 to .6679.
here's what it looks like on a graph.
Question 1203076: The average math score for the class 6 is 23.6. A researcher is interested to check if the score is higher in class 6 that in the USA. How many students are needed to ensure that a two-sided test of hypothesis has 80% power to detect a difference in score of 2 marks? If the population standard deviation is 5.7.
Answer by Jason57t(3) (Show Source):
You can put this solution on YOUR website! To find the sample size required for a statistical hypothesis test with a certain power, we use the formula for the sample size in a two-sided hypothesis test:
n = [(Zα/2 + Zβ)^2 * (2 * σ^2)] / d^2
Where:
Zα/2 is the critical value of the Normal distribution at α/2 (for a two-sided test with a significance level of 0.05, the value of Zα/2 is 1.96)
Zβ is the critical value of the Normal distribution at β (for a test with a power of 0.80, β is 0.20, and the value of Zβ is 0.84)
σ is the standard deviation of the population (in your case, σ = 5.7)
d is the size of the effect you're looking for (in your case, the difference in scores, d = 2)
Plugging these values into the formula:
n = [(1.96 + 0.84)^2 * (2 * 5.7^2)] / 2^2
n = [(2.8)^2 * (2 * 32.49)] / 4
n = [7.84 * 64.98] / 4
n = 509.9152 / 4
n = 127.4788
Because you can't have a fraction of a student, you'd round up to the nearest whole number. Therefore, the researcher would need a sample size of 128 students to ensure that the two-sided test of hypothesis has 80% power to detect a difference in scores of 2 marks.
Question 1201497: You wish to test the following claim (Ha) at a significance level of
α=0.02.
Ho:μ=73.2
Ha:μ>73.2
You believe the population is normally distributed and you know the standard deviation is σ=11.5. You obtain a sample mean of M=78.8for a sample of size
n=17.
What is the test statistic for this sample? (Report answer accurate to three decimal places.)
test statistic =
What is the p-value for this sample? (Report answer accurate to four decimal places.)
p-value =
The p-value is...
less than (or equal to) α
OR
greater than α
This test statistic leads to a decision to...
reject the null
OR
accept the null
OR
fail to reject the null
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! H0 says population mean is equal to 73.2
Ha says population mean is greater than 73.2
population mean is 11.5
sample size is 17
sample mean is 78.8
standard error = standard deviation / square root of sample size = 11.5 / sqrt(17) = 2.789159688.
test z-score = (x - m) / s = (78.8 - 73.2) / 2.789159688 = 2.007773174.
area to the right of that z-score is equal to .0223336306.
since the test alpha is greater than the critical alpha, the test is not significant and you conclude that you haven't got enough information to say that the mean is greater than 73.2.
the test z-score was 2.007773174.
this was less than the critical z-score of 2.053748911.
this confirms the p-score results indicating that the test was not significant and that there was not enough information to conclude that the population mean was greater than 73.2.
here are the results from the calculator at https://davidmlane.com/hyperstat/z_table.html
for the results to be significant, either the test z-score needs to be greater than the critical z-score, or the test p-value has to be less than the critical p-value.
they are consistent with each other.
if the test z-score is greater than the critical z-score, then the test p-value will be less than the critical p-value.
Question 1201399: Lifetimes of AAA batteries are approximately normally distributed. A manufacturer wants to estimate the standard deviation of the lifetime of the AAA batteries it produces. A random sample of 15 AAA batteries produced by this manufacturer lasted a mean of 9.5 hours with a standard deviation of 1.7 hours. Find a 95% confidence interval for the population standard deviation of the lifetimes of AAA batteries produced by the manufacturer. Then give its lower limit and upper limit. Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places.
Found 2 solutions by math_tutor2020, Theo:
Answer by math_tutor2020(3817) (Show Source):
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! sample mean is 9.5 hours.
sample standard devition is 1.7 hours.
sample size is 15.
t-score is indicated because the standard deviation is taken from the sample rather than the population.
since you are looking at the mean of a sample of several elements, you would use the standard error rather than the standard deviation.
standard error = standard deviation / sqrt(sample size) = 1.7 / sqrt(15) = .4389381126.
t-score formula is t = (x - m) / s
t is the t-score
x is the upper or lower limit of the 95% considence interval.
m is mean of the sample.
s is the standard error.
the 95% confidence interval t-tscore with 14 degrees of freedom is plus or minus 2.144786681.
for the upper limit of the confidence interval, the t-score formula becomes:
2.144786681 = (x - 9.5) / .4389381126
solve for x to get:
x = 2.144786681 * .4389381126 + 9.5 = 10.44142862.
for the lower limit of the confidence interval, the t-score formula becomes:
-2.144786681 = (x - 9.5) / .4389381126
solve for x to get:
x = -2.144786681 * .4389381126 + 9.5 = 8.558571382.
your 95% confidence interval is 8.558571382 to 10.44142862.
round to two decimal places to get:
your 95% confidence interval is 8.56 to 10.44.
Question 1201047: A survey showed that 73% of the random sample of 1506 people interviewed
favored drug tests for professional athletes. 68% said that professional athletics
using drug for the first time should be banned or suspended from the professional
sports. Find 95% confidence interval for the proportion of the public who favored
drug tests for professional athletes.
Answer by math_tutor2020(3817) (Show Source):
You can put this solution on YOUR website!
We focus on the portion that says
"A survey showed that 73% of the random sample of 1506 people interviewed favored drug tests for professional athletes."
and ignore the part that says "68% said that professional athletics
using drug for the first time should be banned or suspended from the professional
sports".
phat = 0.73 = sample proportion
n = 1506 = sample size
At 95% confidence, the z critical value is roughly z = 1.96
Use a table like this
https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
to get that value. Look at the bottom row labeled "Z" and above the 95% confidence level.
A stats calculator can also compute this value.
Let's compute the margin of error for the proportion.
E = margin of error
E = z*sqrt(phat*(1-phat)/n)
E = 1.96*sqrt(0.73*(1-0.73)/1506)
E = 0.02242264791327
E = 0.022
This value is approximate.
Now we can compute the boundaries.
L = lower boundary of the confidence interval
L = phat - E
L = 0.73 - 0.022
L = 0.708
and
U = upper boundary of the confidence interval
U = phat + E
U = 0.73 + 0.022
U = 0.752
These values are approximate.
The 95% confidence interval in the format (L, U) is approximately (0.708, 0.752)
The 95% confidence interval in the format L < p < U is approximately 0.708 < p < 0.752
This second format is a bit more descriptive in terms of which population parameter we're trying to measure.
We are 95% confident the population proportion p is somewhere between 0.708 and 0.752
Meaning we are 95% confident the true percentage of people in favor of drug tests for professional athletes is somewhere between 70.8% and 75.2%
Side note:
An alternative confidence interval format is  which in this case is roughly 
Question 1200053: Find the Levels of the Confidence Intervals that have the given critical values.
1.04
Answer by Theo(13342) (Show Source):
Question 1200442: Assume that a sample is used to estimate a population proportion p. Find the 95% confidence interval for a sample of size 203 with 39% successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! p = .39
q = 1 - .39 = .61
sample size is 203.
standard error = sqrt(.39 * .61 / 203) = .0342333344.
95% confidence interval is equal to plus or minus z = 1.959963986.
z-score formula is z = (x - m) / s
z is the z-score
x is the sample mean proportion.
m is the population mean proportion.
s is the standard error.
solve for x to get x = z * s + m which becomes:
x = -1.959963986 * .0342333344 + .39 = .3229038975 at the lower limit.
solve for x = 1.959963986 * .0342333344 + .39 = .4570961025 at the upper limit.
round to 3 decial places to get:
(.323,.457) as your answer.
here's what it looks like, using the calculator at https://davidmlane.com/hyperstat/z_table.html
Question 1199321: In a city, out of 400 people sampled, 152 had kids. Based on this, construct a 99% confidence interval for the true population proportion of people with kids.
Round your answers to three decimal places.
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! sample size is 400
p = 152/400 = .38
q = 1-p = .62
standard error = sqrt(p*q/n) = sqrt(.38*.62/400) = .0242693222.
two tailed 99% confidence interval has a critical z-score of plus or minus 2.5758 rounded to 4 decimal places.
use the z-score formula to find the critical raw scores.
z-score formula is z = (x-m)/s
z is the z-score
x is the raw score
m is the mean
s is the standard error.
when x = 2.5758, the formula becomes -2.5758 = (x-.38)/.0242693222.
solve for x to get x = -2.5758 * .0242693222 + .38 = .3174878799.
when x = 2.5758, the formula becomes 2.5758 = (x-.38)/.0242693222.
solve for x to get x = 2.5758 * .0242693222 + .38 = .4425129201
round your answer to 3 decimal places to get the 99% confidence interval is from .317 to .443.
Question 1197051: a. A statistics practitioner calculated the mean and standard deviation from a sample of 51. They are x bar = 120 and s = 15. Estimate the population mean with 95% confidence.
b. Repeat part (a) with a 90% confidence level.
c. Repeat part (a) with an 80% confidence level.
d. What is the effect on the confidence interval estimate of decreasing the confidence level?
Answer by ewatrrr(24785)  (Show Source):
Question 1194495: A researcher used 80 students from a school as samples. Suppose the researcher used the Slovin’s formula, what is the student population of the school rounded to the nearest thousand?
Answer by ikleyn(52782)  (Show Source):
Question 1190678: Parle Agro India Pvt. Ltd. wants to conduct a survey to find whether more people prefer Frooti to Slice. It was estimated before the survey that half of the population prefers Frooti and half prefers Slice. How large a sample would Parle Agro India Pvt. Ltd. have to take to estimate the proportion of people who prefer Frooti within +/- 0.03 of the actual value, with 98% confidence? considering Z value is 2
Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! error=z*sqrt(p*(1-p)/n)
0.03=2*sqrt(0.25/n)
0.0009=4*0.25/n, squaring both sides
0.0009n=1
n=1111.11 or 1112 using z=2
-
using z=2.326 which is for 98% CI
0.0009=2.326^2*0.25/n
n=2.326^2*0.25/0.0009=1502.85 or 1503
Question 1190707: A medical researcher conducted a study to understand the efficacy of a new medicine. for the study he took a sample of 55 patients and found the margin of error for the estimate to be 12.As he aware that the margin of error is related to the size of sample being considered, if the researcher wants to have more precision in his measurement of efficacy by reducing the margin of error of estimate to 4,what is the sample size he needs to consider?
Answer by math_tutor2020(3817) (Show Source):
You can put this solution on YOUR website!
If the doctor was trying to estimate the population mean mu, then the margin of error would be
E = z*s/sqrt(n)
where,
E = margin of error
z = critical z value based on the confidence level
s = standard deviation
n = sample size
You probably have come across this formula when setting up confidence intervals.
Since the values of z and s aren't mentioned, we'll just hold them to fixed unknown constants.
It'll turn out that they don't matter.
Let's solve the mentioned formula for n
E = z*s/sqrt(n)
E*sqrt(n) = z*s
sqrt(n) = z*s/E
n = (z*s/E)^2
Let's now plug in the first error mentioned 12
n = (z*s/E)^2
n = (z*s/12)^2
n = A/(12^2) ... let A = (z*s)^2
n = A/144
This margin of error E = 12 corresponds directly to the sample size n = 55, so,
n = A/144
A = 144n
A = 144*55
A = 7920
We can then determine what n must be if we want E = 4
n = (z*s/E)^2
n = A/(E^2) ... again let A = (z*s)^2
n = 7920/(4^2)
n = 495
The value of A doesn't change because the confidence level doesn't change (which in turn means the z value stays the same). Also, the standard deviation is fixed as well.
Answer: 495
Question 1187550: Out of 300 people sampled, 177 preferred Candidate A. Based on this, estimate what proportion of the voting population (p) prefers Candidate A.
Use a 90% confidence level, and give your answers as decimals, to three places.
_______ < P > ________
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! p = 177 / 300 = .59
q = 1 - p = .41
mean proportion = p = .59
standard error = sqrt(p * (1-p / 300) = sqrt(.59 * .41 / 300) = .028396.
critical z-score at 90% confidence level is plus or minus 1.645.
use the z-score formula to find the critical raw score.
for the low side, z = (x - m) / s becomes:
-1.645 = (x - .59) / .028396.
solve for x to get:
x = -1.645 * .0283960091 + .59 = .54329.
for the high side, z = (x - m) / s becomes:
1.645 = (x - .59) / .028396.
solve for x to get:
x = 1.645 * .028396. + .59 = .63671.
at 90% confidence level, your proportion will be between .54329 and .63671, when you have a mean of .59 and a standard error of .028396.
here's what it looks like on a z-score normal distribution calculator output.
the calculator i used can be found at https://davidmlane.com/hyperstat/z_table.html
Question 1184232: Using the 95% confidence level and the writing scores 91,91, 94, 91, 90, 81, 84, 91, 92, 91, 91, 98, 94, 91, 85, 91 you would expect the upper end of the confidence interval to be approximately what value?
Answer by Edwin McCravy(20056)  (Show Source):
You can put this solution on YOUR website!
91, 91, 94, 91,
90, 81, 84, 91,
92, 91, 91, 98,
94, 91, 85, 91
On your TI-84
Press STAT
Press 1
Enter all 16 numbers above into L1
When you have them all entered,
Press STAT again
Press the right arrow key to highlight CALC
Press 1:1-Var Stats
The screen reads
1-Var Stats
List:L1
FreqList:
Calculate
Press the down arrow key twice to highlight Calculate
Press ENTER
Read
1-Var Stats
x̄=90.375
Σx=1446
Σx2=130930
Sx=4.064070209
σx=3.93501906
n=16
That shows the data that's been calculated from L1 but you don't have to
enter it, because it's automatically entered on the next screen.
Press STAT again
Press the right arrow key to highlight TESTS
Press 7:ZInterval
Make sure the screen reads like this, use the arrow key to highlight Data,
not Stats. Make sure the C-level reads 0.95
Z-Interval
Inpt:Data Stats
σ:24.1
List:L1
Freq:1
C-Level:0.95
Calculate
Highlight Calculate with the down arrow key
Press ENTER
(78.566,102.18)
x̄=90.375
Sx=4.064070209
n=16
The confidence interval is 78.566 < x < 102.18
So the expected upper end of it is 102.18.
Edwin
Question 1183978: Susan invested a total of $10000 in two accounts. One account pays 4 interest and the other account pays 6% Interest. The total retum from both accounts was $560.
Answer the following questions.
How much was invested at 4%?
How much was invested at 6%?
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
The other tutor showed a good formal algebraic solution using the standard method.
If a formal algebraic solution is not required, and if the speed of getting an answer is important (as in a timed math competition), then here is a very fast and easy way to solve this problem (and many similar problems) mentally.
(1) $10000 all invested at 4% would yield $400 interest; all at 6% would yield $600 interest.
(2) The actual interest $560 is 4/5 of the way from $400 to $600. (Picturing the three numbers 400, 560, and 600 on a number line might help: 400 to 600 is a difference of 200; 400 to 560 is a difference of 160; 160/200 = 4/5.)
(3) That means 4/5 of the $10000 was invested at the higher rate.
ANSWER: 4/5 of $10000, or $8000, was invested at 6%; the other $2000 at 4%.
CHECK: .06(8000)+.04(2000) = 480+80 = 560
Answer by ikleyn(52782) (Show Source):
You can put this solution on YOUR website! .
Let x be the amount invested at 6%.
Then the amount invested at 4% is (10000-x) dollars.
The 6% investment generates 0.06x dollars as the annual interest.
The 4% investment generates 0.04*(10000-x) dollars as the annual interest.
The total annual interest is the sum 0.06 + 0.04*(10000-x) dollars.
It is equal to $560 (given).
So we have this equation for the total annual interest
0.06x + 0.04*(10000-x) = 560 dollars.
From the equation
x =  = 8000.
ANSWER. $8000 invested at 6% and the rest, 10000-8000 = 2000 dollars invested at 4%.
CHECK. 0.06*8000 + 0.04*2000 = 560 dollars, total annual interest. ! Correct !
Solved.
Question 1183979: Write your answer in interval notation:
3(-5x + 5) \leq -15 }
Answer by MathLover1(20850) (Show Source):
Question 1183968: A random sample of 732 judges found that 405 were introverts. Construct a 95% confidence interval for the proportion. Interpret the meaning of the confidence interval. Justify your use of a confidence interval based on a normal distribution for data regarding proportions that are normally following a binomial distribution. (10 points)
Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! p hat, point estimate, is 405/732=0.5533
half-interval for 95% Is z(0.975)*sqrt(0.5533*0.4467/732)
=1.96*0.0184
=0.0360
the interval is (0.5173, 0.5893)
We are highly confident that the true value the proportion of judges who are introverts lies within this interval.
Question 1183970: In a study of brain waves during sleep, a sample of 29 college students were randomly separated into two groups. The first group had 15 people and each was given ½ litre of red wine before sleeping. The second group had 14 people and were given no alcohol before sleeping. All participants when to sleep at 11 PM and their brainwave activity was measured from 4-6 AM. The group drinking alcohol had a mean brainwave activity of 19.65 hertz and a standard deviation of 1.86 hertz. The group not drinking alcohol had a mean of 6.59 hertz and standard deviation of 1.91 hertz. Compute a 90% confidence interval for the difference in population means of groups drinking alcohol before sleeping and those not drinking alcohol before sleeping. Explain the meaning of the confidence interval. (10 points)
Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! The half interval is t(0.95, df=27)* sqrt (s1^2/n1+s2^2/n2)
=1.703*sqrt((1.86^2/15)+(1.91^2/14))
=1.703*0.700
=1.19
The point estimate of the difference is 13.06 Hz
so the interval is (11.87 Hz, 14.25 Hz)
Question 1183936: c=0.94
Answer by ikleyn(52782) (Show Source):
Question 1181286: In order to estimate the average age of onset of a certain type of disease a researcher collects a sample of 15 people with the disease and produces the following 95% confidence interval (83.9219, 90.6782). Find a 99% confidence interval for the average age based on the same data set
Answer by robertb(5830)  (Show Source):
Question 1179919: According to Beautiful Bride magazine, the average age of a groom is now 26.2 years. A sample of 16 prospective grooms in Chicago revealed that their average age was 26.6 years with a standard deviation of 5.3 years. What is the test value for a t test of the claim?
Answer by Boreal(15235) (Show Source):
Question 1178027: A machine that stuffs a cheese filled snack product can be adjusted for the amount of cheese injected into each unit. A simple random sample of 30 units is selected, and the average amount of cheese injected is found to be X= 3.5 grams. If the process standard deviation is known to be o =0.25grams, construct a 99% confidence interval for the average amount of cheese being injected by machine.
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! sample size = 30
population standard deviation = .25
sample mean = 3.5
standard error = population standard deviation / square root of sample size = .25/sqrt(30) = .0456435465.
z-score of two tail confidence interval of 90% = plus or minus 2.5758.
z-score formula = (x - m) / se
x is raw score
m is mean
se is standard error.
plus side raw score = .25/sqrt(30) * 2.5758 + 3.5 = 3.617568647
minus side raw score = .25/sqrt(30) * -2.5758 + 3.5 = 3.382431353
at 99% two tail confidence interval, the average amount of cheese injected will be between 3.6176 and 3.3824 grams.
visually, this looks like this:
when you're dealing with the mean of a sample, you have to use the standard error rather than the standard deviation.
Question 1177860: We know that proportion of diamond cards in a regular deck of cards is 13/52=0.25. Assume for a moment that we do not know it and we wish to use statistical methods to find a confidence interval for proportion of red cards in a deck. We will use 10 regular decks of card and select random sample of 100 cards.
Please use the above information to answer questions:
Question 1
The population will consider of 10 regular decks of cards well shuffled..
As described, we randomly draw 100 cards from the population and record a sample (D - diamond card, ND - not diamond card}. In the sample we have 24 diamond cards and 76 of the other suite.
Answer the following questions:
Number of elements in the population is
Number of elements in the sample is
Proportion of diamond cards in the sample is (express it as a decimal)
Answer by ewatrrr(24785) (Show Source):
You can put this solution on YOUR website!
Hi
Number of elements in the population is: 10decks *52cards = 520
Number of elements in the sample is: 100
Proportion of diamond cards in the sample is (express it as a decimal)24/100 = .24
Wish You the Best in your Studies.
Question 1177422: A survey reports it results by starting that standard error of the mean to be is 20.
The population standard deviation is 500.
1:How large is the sample used in this surveying?
2: What is the probability that the sample mean will be within 25 of the population mean?
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! standard error is equal to population standard deviation divided by square root of sample size.
the formula is s = psd / sqrt(ss).
s is the standard error.
psd is population standard deviation.
ss is sample size.
when s = 20 and psd = 500, the formula becomes:
20 = 500 / sqrt(ss)
solve for sqrt(ss) to get:
sqrt(ss) = 500 / 20 = 25
solve for ss to get:
ss = 25^2 = 625
that's the sample size.
the formula for z-score is:
z = (x - m) / s
z is the z-score
x is the raw score
m is the raw mean
s is the standard error.
when (x - m) = 25, and s = 20, the formula becomes:
z = 25 / 20 = 1.25
since the normal distribution curve is symmetric bout the mean, then the confidence interval is z-score of -1.25 to 1.25.
the probability is the area under the normal distribution curve between those 2 z-scores.
that probability is .7887003221.
the probability that the sample mean will be within 25 of the population mean is .7887003221.
this works regardless of the mean.
for example:
assume the mean is 1500.
the z-score formula becomes plus or minus 1.25 = (x - 1500) / 20
solve for the raw score to get:
on the high side, x = 20 * 1.25 + 1500
on the low side, x = 20 * -1.25 + 1500
since 20 * 1.25 is always equal to 25, then you get:
x = 1500 - 25 or x = 1500 + 25.
the sample mean will always be within 25 of the population mean if the z-score is plus or minus 1.25 and the standard error is 20, regardless of what the mean is.
standard error = psd / sqrt(ss)
with a sample size of 625 and a psd of 500, this becomes:
standard error = 500 / sqrt(625) = = 500 / 25 = 20
Question 1176585: Surveyors asked a random sample of women in a major city what factor was the most important in deciding where to shop. The results appear in the following table. If the sample size was 1200, estimate with 95% confidence the proportion of women who identified price and value as the most important factor.
Factor Percentage (%)
Price and Value 40
Quality and selection of merchandise 30
Service 15
Shopping environment 15
Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! price and value is 40% of the total (I misread the question)
95%CI half-interval is z*sqrt(p*(1-p)/n)
=1.96*sqrt(..4*.6/1200)
=0.0277
95%CI is (0.3723, 0.4277) units %
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