Employees in a large company are entitled to 15-minute water breaks. A random sample of the duration
of water breaks for 10 employees was taken with the times shown as: 12, 16, 14, 18, 21, 17, 19, 15, 18,
and 16. Assuming that the times are normally distributed at the 5% significance level
What test do you need to use?
State the null and alternative hypotheses.
What is the p-value?
What is the conclusion (2-part)?
Find the 95% Confidence Interval
To answer these questions, we need to perform a hypothesis test for the mean duration of water breaks.
1. Test to use:
Since we are comparing the mean duration of water breaks to a specific value (15 minutes), we will use a one-sample t-test.
2. Null and alternative hypotheses:
The null hypothesis (H0) is that the mean duration of water breaks is equal to 15 minutes.
The alternative hypothesis (Ha) is that the mean duration of water breaks is not equal to 15 minutes.
H0: μ = 15 (mean duration of water breaks = 15 minutes)
Ha: μ ≠ 15 (mean duration of water breaks ≠ 15 minutes)
3. P-value:
To calculate the p-value, we need to find the t-statistic and compare it to the critical value in the t-distribution table or use statistical software.
4. Conclusion:
To make a conclusion, we compare the p-value with the significance level (5% in this case). If the p-value is less than 0.05, we reject the null hypothesis. Otherwise, if the p-value is greater than or equal to 0.05, we fail to reject the null hypothesis.
The two-part conclusion will be:
a) If the p-value < 0.05, we reject the null hypothesis and conclude that there is evidence to suggest that the mean duration of water breaks is different from 15 minutes.
b) If the p-value ≥ 0.05, we fail to reject the null hypothesis and conclude that there is not enough evidence to suggest that the mean duration of water breaks is different from 15 minutes.
5. 95% Confidence Interval:
To find the 95% Confidence Interval, we need to calculate the mean and standard deviation of the sample and then use the t-distribution with appropriate degrees of freedom to find the margin of error.
The formula for the 95% Confidence Interval is:
Confidence Interval = mean ± (t-value * standard error)
where the t-value is obtained from the t-distribution table or using statistical software, and the standard error is the standard deviation divided by the square root of the sample size.