Respuesta :
Answer:
(a) H₀: μ = 87 vs. Hₐ: μ ≠ 87.
(b) [tex]z=-1.52>-z_{\alpha/2}=-1.96[/tex]. The average grade on the statistics final examination is 87.
(c) The p-value = 0.1286 > α = 0.05. The average grade on the statistics final examination is 87.
(d) The 95% confidence interval for the average grade on the statistics final examination is (80.04, 87.88).
Explanation:
A statistical hypothesis test is to be performed to determine whether the average grade on the statistics final examination is 87.
(a)
The hypothesis can be defined as follows:
H₀: The average grade on the statistics final examination is 87, i.e. μ = 87.
Hₐ: The average grade on the statistics final examination is not 87, i.e. μ ≠ 87.
(b)
The information provided is:
[tex]n=36\\\bar x=83.96\\\sigma^{2}=144[/tex]
As the population variance is provided, we will use a z-test for single mean.
Compute the test statistic value as follows:
[tex]z=\frac{\bar x-\mu}{\sqrt{\sigma^{2}/n}}=\frac{83.96-87}{\sqrt{144/36}}=-1.52[/tex]
The test statistic value is -1.52.
Decision rule:
Reject H₀ if:
[tex]z<-z_{\alpha/2}\ \text{or}\ z<z_{\alpha/2}\\\\\Rightarrow z<-z_{0.05/2}\ \text{or}\ z<z_{0.05/2}\\\\\Rightarrow z<-1.96\ \text{or}\ z<1.96[/tex]
The calculated value of [tex]z=-1.52>-z_{\alpha/2}=-1.96[/tex].
The null hypothesis will not be rejected.
Conclusion:
The average grade on the statistics final examination is 87.
(c)
Decision Rule:
If the p-value of the test is less than the significance level then the null hypothesis will be rejected.
Compute the p-value for the two-tailed test as follows:
[tex]p-value=2\cdot P(Z<-1.52)=2\times 0.0643=0.1286[/tex]
*Use a z-table for the probability.
The p-value of the test is 0.1286.
p-value = 0.1286 > α = 0.05
The null hypothesis will not be rejected.
Thus, it can be concluded that average grade on the statistics final examination is 87.
(d)
Compute the 95% confidence interval for the average grade on the statistics final examination as follows:
[tex]CI=\bar x\pm z_{\alpha/2}\cdot \sqrt{\frac{\sigma^{2}}{n}}[/tex]
[tex]=83.96\pm 1.96\cdot \sqrt{\frac{144}{36}}\\\\=83.96\pm 3.92\\\\=(80.04, 87.88)[/tex]
The 95% confidence interval for the average grade on the statistics final examination is (80.04, 87.88).
As the 95% confidence interval consists of the null value, i.e. 87, the null hypothesis will not be rejected.
Hence, concluding that the average grade on the statistics final examination is 87.
