Respuesta :
Using the z-distribution, it is found that the p-value for this test is of 0.0375.
At the null hypothesis, it is tested if the expenses are the same as the national average, that is, of $5,423. Hence:
[tex]H_0: \mu = 5423[/tex]
At the alternative hypothesis, it is tested if it is higher, hence:
[tex]H_1: \mu > 5423[/tex]
We have the standard deviation for the population, thus, the z-distribution is used. The test statistic is given by:
[tex]z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
The parameters are:
- [tex]\overline{x}[/tex] is the sample mean.
- [tex]\mu[/tex] is the value tested at the null hypothesis.
- [tex]\sigma[/tex] is the standard deviation of the sample.
- n is the sample size.
For this problem, the values of the parameters are: [tex]\overline{x} = 5516, \mu = 5423, \sigma = 979, n = 352[/tex]
The value of the test statistic is:
[tex]z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{5516 - 5423}{\frac{979}{\sqrt{352}}}[/tex]
[tex]z = 1.78[/tex]
The p-value is the probability of finding a sample mean above $5,516, which is 1 subtracted by the p-value of z = 1.78.
- Looking at the z-table, z = 1.78 has a p-value of 0.9625.
1 - 0.9625 = 0.0375
Hence, the p-value for this test is of 0.0375.
A similar problem is given at https://brainly.com/question/25549474
