Question # At Western University the historical mean of scholarship examination scores for freshmanapplications is 900. Ahistorical population standard deviation σσ= 180 is assumed known.Each year, the assistant dean uses a sample of applications to determine whether the meanexamination score for the new freshman applications has changed.a. State the hypotheses.b.

Probability
ANSWERED At Western University the historical mean of scholarship examination scores for freshman applications is 900. Ahistorical population standard deviation $$\displaystyleσσ={180}$$ is assumed known.
Each year, the assistant dean uses a sample of applications to determine whether the mean examination score for the new freshman applications has changed.
a. State the hypotheses.
b. What is the 95% confidence interval estimate of the population mean examination score if a sample of 200 applications provided a sample mean $$\displaystyle{x}‾{x}$$= 935?
c. Use the confidence interval to conduct a hypothesis test. Using $$\displaystyleαα={.05}$$, what is your conclusion?
d. What is the p-value? 2021-08-23

$$\displaystyle\mu={900}$$
$$\displaystyle\overline{{x}}={935}$$
$$\displaystyle\sigma={180}$$
$$n=200$$

$$\displaystyle\alpha={0.05}$$
a. Determine the hypotheses $$\displaystyle{H}_{{0}}:\mu={900}$$
$$\displaystyle{H}_{{a}}:\mu\ne{900}$$
b. For confidence level $$\displaystyle{1}-\alpha={0.95}$$, determine $$\displaystyle{z}_{{\frac{{a}}{{2}}}}={z}_{{0.025}}$$ using table II (look up 0.025 in the table, the z-score is then the found z-score with opposite sign): $$\displaystyle{z}_{{\frac{{a}}{{2}}}}={1.96}$$
The margin of error is then: $$\displaystyle{E}={z}_{{\frac{{a}}{{2}}}}\cdot{\frac{{\sigma}}{{\sqrt{{n}}}}}={1.96}\cdot{\frac{{{180}}}{{\sqrt{{200}}}}}\approx{24.9467}$$
The confidence interval then becomes: $$\displaystyle{910.0533}={935}-{24.9467}=\overline{{x}}-{E}{<}\mu{<}\overline{{x}}+{E}={935}+{24.9467}={959.9467}$$
c. The sampling distribution of the sample mean has mean $$\displaystyle\mu$$ and standard deviation $$\displaystyle{\frac{{\sigma}}{{\sqrt{{n}}}}}$$ The z-value is the sample mean decreased by the population mean, divided by the standard deviation: $$\displaystyle{z}={\frac{{\overline{{x}}-\mu}}{{\frac{\sigma}{\sqrt{{n}}}}}}={\frac{{{935}-{900}}}{{\frac{{180}}{\sqrt{{200}}}}}}\approx{2.75}$$
The P-value is the probability of obtaining a value more extreme or equal to the standardized test statistic z. Determine the probability using table 1. $$\displaystyle{P}={P}{\left({Z}{<}-{2.75}{\quad\text{or}\quad}{Z}{>}{2.75}\right)}={2}{P}{\left({Z}{<}-{2.75}\right)}={2}{\left({0.0030}\right)}={0.0060}$$
If the P-value is smaller than the significance level alpha, then the null hypothesis is rejected. $$\displaystyle{P}{<}{0.05}\Rightarrow{R}{e}{j}{e}{c}{t}{H}_{{0}}$$
Result:

a. $$\displaystyle{H}_{{0}}:\mu={900},{H}_{{a}}:\mu\ne{900}$$
b. 910.0533 to 959.9467

c. Reject the null hypotheses $$\displaystyle{H}_{{0}}$$

d. P=0.0060