Question

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

Answer 1

The variance of the population is 180², hence, the standard deviation is 180.

Sample Size = 400

Sample mean = 550 (Saw this in the previous question but maybe the student has not provided in this part, so assuming it to be the same.)

Significance Level of the confidence interval = 95% (Assumed as no information is provided)

Confidence interval = [Sample MeanZ score*(Standard Deviation/√sample size)]

Lower Confidence interval = [550-1.96*(180/√400)] (Z score at 95% confidence interval=1.96)

Lower Confidence interval = (550-17.64)

Lower Confidence interval = 532.36

Now,

Upper Confidence interval = 550+1.96*(180/√400)]

Upper Confidence interval = 567.64

Hence, the confidence interval at 95% significance level is [532.36 , 567.64]

With the provided sample information, we can conclude that 500 does not lie in our confidence interval.

There, it is false.