Question

5. At Western University the average scholarship examination scores for freshman applications has been 900. Each year, the assistant dean uses a sample of applications to determine whether the average examination score for the new freshman applications has increased (from the previous year). A random sample of 200 applications was drawn and the average score was found to be 935. Historically, it is known that the score has a Normal distribution and a standard deviation of 180.

a) Identify the parameter of interest.

2. Set up the null hypothesis and the alternative hypothesis.
3. Calculate the test statistic and find the p-value.
4. Using 5% level of significance level to make an appropriate conclusion so that the Dean could understand.

Answer 1

(a) Parameter of interest :Average scholarship examination scores for freshman applications

(b) H0: Null Hypothesis: = 900

HA: Alternative Hypothesis: > 900

(c)
SE = /

= 180/ = 12.7279

Test statistic is given by:

Z =(935 - 900)/12.7279 = 2.75

So,

Test statistic is:

Z = 2.75

Table of Area Under Standard Normal Curve gives area = 0.4970

So,

P - Value= 0.5 - 0.4970 = 0.0030

So,

P - value = 0.0030

(d)

ince p - value = 0.0030 is less than = 0.05, the difference is significant. Reject null hypothesis.

Conclusion:
The data support the claim that the average examination score for the new freshman applications has increased