Answer:
There is not convincing evidence at the a = 0.05 significance level that the true mean service time during the early morning shift is less than 4 minutes
Step-by-step explanation:
The null hypothesis is:
[tex]H_{0} = 4[/tex]
The alternate hypothesis is:
[tex]H_{1} < 4[/tex]
The test statistic is:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.
From a random sample of 41 orders, the mean time was 3.75 minutes with a standard deviation of 1.2 minutes.
This means that [tex]n = 41, \mu = 3.75, \sigma = 1.2[/tex]
The test-statistic is:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{3.75 - 4}{\frac{1.2}{\sqrt{41}}}[/tex]
[tex]z = -1.33[/tex]
[tex]z = -1.33[/tex] has a pvalue of 0.0918, looking at the z-table.
0.0918 > 0.05, which means that the null hypothesis is accepted, and that there is not convincing evidence at the a = 0.05 significance level that the true mean service time during the early morning shift is less than 4 minutes
