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.

Lipossig

Lipossig

Answered question

2021-08-22

At Western University the historical mean of scholarship examination scores for freshman applications 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 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 xx= 935?
c. Use the confidence interval to conduct a hypothesis test. Using αα=.05, what is your conclusion?
d. What is the p-value?

Answer & Explanation

au4gsf

au4gsf

Skilled2021-08-23Added 95 answers

a. Determine the hypotheses H0:μ=900
Ha:μ900


b. For confidence level 1α=0.95, determine za2=z0.025 using table II (look up 0.025 in the table, the z-score is then the found z-score with opposite sign): za2=1.96
The margin of error is then: E=za2σn=1.9618020024.9467
The confidence interval: 910.0533=93524.9467=xE<μ<x+E=935+24.9467=959.9467
 

c. The sampling distribution of the sample mean has mean μ and standard deviation σn The z-value is the sample mean decreased by the population mean, divided by the standard deviation: z=xμσn=9359001802002.75
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. P=P(Z<2.75 or Z>2.75)=2P(Z<2.75)=2(0.0030)=0.0060
If the P-value is smaller than the significance level alpha, then the null hypothesis is rejected. P<0.05RejectH0
 

a. H0:μ=900,Ha:μ900
b. 910.0533 to 959.9467

c. Reject the null hypotheses H0

d. P=0.0060

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