Using the z-distribution, we have that:
a)
The hypothesis test is:
The p-value is of 0.0668.
The critical value is
b) Since the <u>test statistic is less than the critical value</u> for the left-tailed test, there is enough evidence to conclude that the null hypothesis will be rejected.
Item a:
At the null hypothesis, it is <u>tested if the mean is of 7.4 minutes</u>, that is:
At the alternative hypothesis, it is <u>tested if the mean is lower than 7.4 minutes</u>, that is:
We have the <u>standard deviation for the population</u>, hence, the z-distribution is used.
The test statistic is given by:
The parameters are:
is the sample mean.
is the value tested at the null hypothesis.
is the standard deviation of the sample.
- n is the sample size.
For this problem, the values of the <em>parameters </em>are: .
Hence, the value of the <em>test statistic</em> is:
Using a z-distribution calculator, the p-value is of 0.0668.
Also using a calculator, considering a <u>left-tailed test</u>, as we are testing if the mean is less than a value, the critical value with a <u>significance level of 0.1</u> is of .
Item b:
Since the <u>test statistic is less than the critical value</u> for the left-tailed test, there is enough evidence to conclude that the null hypothesis will be rejected.
You can learn more about the use of the z-distribution to test an hypothesis at brainly.com/question/16313918