SOLUTION: A group of students estimated the length of one minute without reference to a watch or​ clock, and the times​ (seconds) are listed below. Use a 0.10 significance level to test
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Question 1169008: A group of students estimated the length of one minute without reference to a watch or​ clock, and the times​ (seconds) are listed below. Use a 0.10 significance level to test the claim that these times are from a population with a mean equal to 60 seconds. Does it appear that students are reasonably good at estimating one​ minute?
70 84 42 65 41 24 58 63 69 49 61 72 90 91 67
Need to find
Hypothesis test
p-value
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Here's how to conduct the hypothesis test using the given data and significance level:
**1. Define the Hypotheses:**
* **Null Hypothesis ($H_0$):** The population mean of the estimated times is equal to 60 seconds.
$H_0: \mu = 60$
* **Alternative Hypothesis ($H_a$):** The population mean of the estimated times is not equal to 60 seconds. This is a two-tailed test because we are testing if the mean is *different* from 60 seconds.
$H_a: \mu \neq 60$
**2. Set the Significance Level:**
The significance level is given as $\alpha = 0.10$.
**3. Calculate the Sample Statistics:**
First, we need to calculate the sample mean ($\bar{x}$) and the sample standard deviation ($s$) from the given data:
Data: 70, 84, 42, 65, 41, 24, 58, 63, 69, 49, 61, 72, 90, 91, 67
* **Sample Size ($n$) = 15**
* **Calculate the sample mean ($\bar{x}$):**
$\bar{x} = \frac{70 + 84 + 42 + 65 + 41 + 24 + 58 + 63 + 69 + 49 + 61 + 72 + 90 + 91 + 67}{15}$
$\bar{x} = \frac{946}{15} \approx 63.07$ seconds
* **Calculate the sample standard deviation ($s$):**
To do this, we first find the deviations from the mean, square them, sum the squared deviations, divide by $n-1$, and then take the square root.
| Value | Deviation ($x - \bar{x}$) | Squared Deviation ($(x - \bar{x})^2$) |
|-------|--------------------------|--------------------------------------|
| 70 | $70 - 63.07 = 6.93$ | $6.93^2 = 48.0249$ |
| 84 | $84 - 63.07 = 20.93$ | $20.93^2 = 438.0649$ |
| 42 | $42 - 63.07 = -21.07$ | $(-21.07)^2 = 443.9449$ |
| 65 | $65 - 63.07 = 1.93$ | $1.93^2 = 3.7249$ |
| 41 | $41 - 63.07 = -22.07$ | $(-22.07)^2 = 487.0849$ |
| 24 | $24 - 63.07 = -39.07$ | $(-39.07)^2 = 1526.4649$ |
| 58 | $58 - 63.07 = -5.07$ | $(-5.07)^2 = 25.7049$ |
| 63 | $63 - 63.07 = -0.07$ | $(-0.07)^2 = 0.0049$ |
| 69 | $69 - 63.07 = 5.93$ | $5.93^2 = 35.1649$ |
| 49 | $49 - 63.07 = -14.07$ | $(-14.07)^2 = 197.9649$ |
| 61 | $61 - 63.07 = -2.07$ | $(-2.07)^2 = 4.2849$ |
| 72 | $72 - 63.07 = 8.93$ | $8.93^2 = 79.7449$ |
| 90 | $90 - 63.07 = 26.93$ | $26.93^2 = 725.2249$ |
| 91 | $91 - 63.07 = 27.93$ | $27.93^2 = 780.0849$ |
| 67 | $67 - 63.07 = 3.93$ | $3.93^2 = 15.4449$ |
| **Sum** | | **4820.0000** |
$s^2 = \frac{\sum(x - \bar{x})^2}{n-1} = \frac{4820}{15 - 1} = \frac{4820}{14} \approx 344.2857$
$s = \sqrt{344.2857} \approx 18.56$ seconds
**4. Determine the Test Statistic:**
Since the population standard deviation is unknown and the sample size is small ($n < 30$), we will use a t-test. The test statistic is:
$t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}$
where:
* $\bar{x}$ is the sample mean ($\approx 63.07$)
* $\mu_0$ is the hypothesized population mean ($60$)
* $s$ is the sample standard deviation ($\approx 18.56$)
* $n$ is the sample size ($15$)
$t = \frac{63.07 - 60}{18.56 / \sqrt{15}}$
$t = \frac{3.07}{18.56 / 3.873}$
$t = \frac{3.07}{4.792} \approx 0.641$
**5. Determine the P-value:**
For a two-tailed t-test with $n-1 = 15 - 1 = 14$ degrees of freedom, we need to find the p-value associated with a test statistic of $t \approx 0.641$.
Looking at a t-distribution table or using a statistical calculator, the p-value for a two-tailed test with $t = 0.641$ and 14 degrees of freedom is greater than $2 \times 0.25 = 0.50$. More precisely, it's approximately $0.532$.
**Hypothesis Test:**
* **Test Statistic:** $t \approx 0.641$
* **Degrees of Freedom:** $df = 14$
* **Type of Test:** Two-tailed
**P-value:**
* **P-value $\approx 0.532$**
**Conclusion:**
We compare the p-value to the significance level ($\alpha = 0.10$).
Since the p-value ($0.532$) is greater than the significance level ($0.10$), we **fail to reject the null hypothesis**.
**Answer to the question: Does it appear that students are reasonably good at estimating one minute?**
Based on this hypothesis test at a 0.10 significance level, there is **not enough statistical evidence** to conclude that the mean estimated time by students is significantly different from 60 seconds. Therefore, it appears that students are **reasonably good** at estimating one minute, as the data does not provide strong evidence to suggest their estimates are consistently off.
Final Answer: The final answer is $\boxed{p-value \approx 0.532}$
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