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:

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
![Z=\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt[]{n}}}](https://img.qammunity.org/2023/formulas/mathematics/high-school/h5idfem2zuffllcnvn9svhd4q5gv3lzy9r.png)
You can calculate the statistic under the null hypothesis as follows:
![\begin{gathered} Z_(H0)=\frac{7.5-8}{\frac{5.3}{\sqrt[]{66}}} \\ Z_(H0)=-0.7664\approx-0.766 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/nis3xn0416i1j0i5rgu5f81zr5cecxytrd.png)
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:

c) The p-value is greater than α, so the decision is to not reject the null hypothesis.