The sample taken was n=66 companies
The sample mean obtained is 7.5 years
The sample standard deviation obtained was 5.3 years
You have to work with a significance level of α= 0.05
And test the statistic hypotheses:
[tex]\begin{gathered} H_0\colon\mu\ge8 \\ H_1\colon\mu<8 \end{gathered}[/tex]
a) The test statistic to use, considering that we can assume a normal distribution and the sample size is greater than n=30, is the standard normal
[tex]Z=\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt[]{n}}}[/tex]
You can calculate the statistic under the null hypothesis as follows:
[tex]\begin{gathered} Z_{H0}=\frac{7.5-8}{\frac{5.3}{\sqrt[]{66}}} \\ Z_{H0}=-0.7664\approx-0.766 \end{gathered}[/tex]
b) The p-value is the probability corresponding to the calculated statistic if possible under the null hypothesis.
This test is one-tailed to the left, so the p-value is the accumulated probability to the statistic:
[tex]p-value=P(Z\leq-0.766)=0.2218[/tex]
c) The p-value is greater than α, so the decision is to not reject the null hypothesis.
