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.

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asked 2021-08-22
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?

Expert Answers (1)

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

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