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.

- Set up the null hypothesis and the alternative hypothesis.
- Calculate the test statistic and find the p-value.
- 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

At Stony Brook University, the historical mean of scholarship
examination scores for freshman is 500. A historical population
variance is assumed to be known as 1802. Each year, the
assistant dean uses a sample of applications to determine whether
the mean examination score for the new freshman applications has
changed.
Now the sample mean is given by , and the number of samples are
400. Provided the sample mean, is 500 in your confidence
interval?
True
False

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