Question
How many tissues should the Kimberly Clark Corporation package of Kleenex® contain? Researchers determined that 80 tissues is the mean number of tissues used during a cold. Suppose a random sample of 100 Kleenex users yielded the following data on the number of tissues used during a cold: = 72, S = 20. Suppose the alternative you wanted to test was H1: μ < 60. State the correct rejection region for α = 0.05.
Answers
GPT-4o mini
To determine the rejection region for the hypothesis test, we follow these steps:
1. **Define Hypotheses**:
- Null hypothesis, \( H_0: \mu \geq 60 \)
- Alternative hypothesis, \( H_1: \mu < 60 \)
2. **Determine the significance level (\( \alpha \))**:
- Given \( \alpha = 0.05 \)
3. **Identify the test statistic**:
We will use the t-test for the mean. The test statistic \( t \) is given by:
\[
t = \frac{\bar{x} - \mu_0}{S / \sqrt{n}}
\]
where:
- \( \bar{x} = 72 \) (sample mean),
- \( \mu_0 = 60 \) (hypothesized mean under \( H_0 \)),
- \( S = 20 \) (sample standard deviation),
- \( n = 100 \) (sample size).
4. **Calculate the test statistic**:
\[
t = \frac{72 - 60}{20 / \sqrt{100}} = \frac{12}{2} = 6
\]
5. **Determine the critical value**:
We need the critical value for a one-tailed t-test with \( \alpha = 0.05 \) and \( n - 1 = 99 \) degrees of freedom.
Using a t-distribution table or calculator for \( \alpha = 0.05 \) in the left tail and \( 99 \) degrees of freedom (df), the critical value \( t_{\text{critical}} \) is approximately \(-1.660\).
6. **State the rejection region**:
The rejection region for this one-tailed test is:
\[
t < -1.660
\]
7. **Conclusion**:
If the calculated t-statistic is less than \(-1.660\), we reject the null hypothesis \( H_0 \).
Given the calculated \( t = 6 \) is greater than \(-1.660\), we do not reject the null hypothesis in this case. Thus, the rejection region for \( \alpha = 0.05 \) is \( t < -1.660 \).
1. **Define Hypotheses**:
- Null hypothesis, \( H_0: \mu \geq 60 \)
- Alternative hypothesis, \( H_1: \mu < 60 \)
2. **Determine the significance level (\( \alpha \))**:
- Given \( \alpha = 0.05 \)
3. **Identify the test statistic**:
We will use the t-test for the mean. The test statistic \( t \) is given by:
\[
t = \frac{\bar{x} - \mu_0}{S / \sqrt{n}}
\]
where:
- \( \bar{x} = 72 \) (sample mean),
- \( \mu_0 = 60 \) (hypothesized mean under \( H_0 \)),
- \( S = 20 \) (sample standard deviation),
- \( n = 100 \) (sample size).
4. **Calculate the test statistic**:
\[
t = \frac{72 - 60}{20 / \sqrt{100}} = \frac{12}{2} = 6
\]
5. **Determine the critical value**:
We need the critical value for a one-tailed t-test with \( \alpha = 0.05 \) and \( n - 1 = 99 \) degrees of freedom.
Using a t-distribution table or calculator for \( \alpha = 0.05 \) in the left tail and \( 99 \) degrees of freedom (df), the critical value \( t_{\text{critical}} \) is approximately \(-1.660\).
6. **State the rejection region**:
The rejection region for this one-tailed test is:
\[
t < -1.660
\]
7. **Conclusion**:
If the calculated t-statistic is less than \(-1.660\), we reject the null hypothesis \( H_0 \).
Given the calculated \( t = 6 \) is greater than \(-1.660\), we do not reject the null hypothesis in this case. Thus, the rejection region for \( \alpha = 0.05 \) is \( t < -1.660 \).