Answer :
Answer:
The value of the test statistic and its associated p-value at the 5% significance level are -1.54 and 0.9382, respectively.
Step-by-step explanation:
A university interested in tracking its honors program believes that the proportion of graduates with a GPA of 3.00 or below is less than 0.16.
This means that the null hypothesis is:
[tex]H_{0}: p \geq 0.16[/tex]
Testing this hypothesis, means that the alternate hypothesis is:
[tex]H_{a}: p > 0.16[/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.
0.16 is tested at the null hypothesis:
This means that [tex]\mu = 0.16, \sigma = \sqrt{0.16*0.84}[/tex]
In a sample of 200 graduates, 24 students have a GPA of 3.00 or below.
This means that [tex]n = 200, X = \frac{24}{200} = 0.12[/tex]
Value of the test statistic:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{0.12 - 0.16}{\frac{\sqrt{0.16*0.84}}{\sqrt{200}}}[/tex]
[tex]z = -1.54[/tex]
Pvalue:
The pvalue is the probability of finding a sample mean above 0.12, which is 1 subtracted by the pvalue of z = -1.54.
Looking at the z-table, z = -1.54 has a pvalue of 0.0618
1 - 0.0618 = 0.9382
The value of the test statistic and its associated p-value at the 5% significance level are -1.54 and 0.9382, respectively.