Question

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

Answer 1

(910.053 ; 959.947)

Pvalue = 0.00596

Explanation:

Given :

Population mean, μ = 900

Sample size, n = 200

Population standard deviation, σ = 180

The hypothesis :

H0 : μ = 900

H0 : μ ≠ 900

The 95% confidence interval:

Xbar ± Margin of error

Margin of Error = Zcritical * σ/√n

Since the σ is known, we use the z- distribution

Zcritical at 95% confidence = 1.96

Hence,

Margin of Error = 1.96 * 180/√200

Margin of Error = 24.947

95% confidence interval is :

935 ± 24.947

Lower boundary = 935 - 24.947 = 910.053

Upper boundary = 935 + 24.947 = 959.947

(910.053 ; 959.947)

Hypothesis test :

Test statistic

(935- 900) ÷ (180/√(200))

Test statistic = 2.750

Pvalue from Test statistic ;

Pvalue = 0.00596

Pvalue < α ; Reject H0 and conclude that score has changed

Hence, we can conclude that the score has changed