Lipossig

2021-08-22

At Western University the historical mean of scholarship examination scores for freshman applications is 900. Ahistorical population standard deviation $\sigma \sigma =180$ 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 $x‾x$= 935?
c. Use the confidence interval to conduct a hypothesis test. Using $\alpha \alpha =.05$, what is your conclusion?
d. What is the p-value?

au4gsf

a. Determine the hypotheses ${H}_{0}:\mu =900$
${H}_{a}:\mu \ne 900$

b. For confidence level $1-\alpha =0.95$, determine ${z}_{\frac{a}{2}}={z}_{0.025}$ using table II (look up 0.025 in the table, the z-score is then the found z-score with opposite sign): ${z}_{\frac{a}{2}}=1.96$
The margin of error is then: $E={z}_{\frac{a}{2}}\cdot \frac{\sigma }{\sqrt{n}}=1.96\cdot \frac{180}{\sqrt{200}}\approx 24.9467$
The confidence interval: $910.0533=935-24.9467=\stackrel{―}{x}-E<\mu <\stackrel{―}{x}+E=935+24.9467=959.9467$

c. The sampling distribution of the sample mean has mean $\mu$ and standard deviation $\frac{\sigma }{\sqrt{n}}$ The z-value is the sample mean decreased by the population mean, divided by the standard deviation: $z=\frac{\stackrel{―}{x}-\mu }{\frac{\sigma }{\sqrt{n}}}=\frac{935-900}{\frac{180}{\sqrt{200}}}\approx 2.75$
The P-value is the probability of obtaining a value more extreme or equal to the standardized test statistic

d. Determine the probability using table 1.
If the P-value is smaller than the significance level alpha, then the null hypothesis is rejected. $P<0.05⇒Reject{H}_{0}$

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

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

d. P=0.0060

Do you have a similar question?