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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