GMAT Scores: Total scores on the GMAT are normally distributed and historically

Question

GMAT Scores:

Total scores on the GMAT are normally distributed and historically have a population standard deviation of 113.The Graduate Management Admission Council (GMAC), who administers the test, claims that the mean total score is 529. Suppose a random sample of 8 students took the test, and their scores are given below: (hint: assume s = 113)

699, 560, 414, 570, 521, 663, 727, 416

PLEASE TYPE ANSWER FOR LEGIBILITY. ANSWER ALL QUESTIONS OR LEAVE IT FOR SOMEONE WHO WILL.

a)Find a point estimate for the population mean

b)Construct a 95% confidence interval for the true mean score for the population.

c)Does this interval contain the value reported by GMAC?

d)How many students should be surveyed to estimate the mean score within 25 points with 90% confidence?

e)How many students should be surveyed to estimate the mean score within 25 points with 95% confidence?

f)How many students should be surveyed to estimate the mean score within 25 points with 99% confidence?

Explanation / Answer

a)Find a point estimate for the population mean

Answer : Here the point estimate of the population mean is equals to sample mean x = 571.25

b)Construct a 95% confidence interval for the true mean score for the population.

Answer : 95% confidence interval = x +- Z95% ( /sqrt(n) = 571.25 +- 1.96 * (113/sqrt(8))

= (492.945, 649.555)

c)Does this interval contain the value reported by GMAC?

Answer : Yes the interval contain the value reported by GMAC.

d)How many students should be surveyed to estimate the mean score within 25 points with 90% confidence?

Answer : HEre margin of error = 25

Margin of error = Critical test statistic * standard error of sample mean

25 = Z90% *(/sqrt(n)

25 = 1.645 * (113/n)

n = 55.285 or 56

e)How many students should be surveyed to estimate the mean score within 25 points with 95% confidence?

HEre margin of error = 25

Margin of error = Critical test statistic * standard error of sample mean

25 = Z95% *(/sqrt(n)

25 = 1.96* (113/n)

n = 78.48 or 79

f)How many students should be surveyed to estimate the mean score within 25 points with 99% confidence?

HEre margin of error = 25

Margin of error = Critical test statistic * standard error of sample mean

25 = Z99% *(/sqrt(n)

25 = 2.576 * (113/n)

n = 135.57 or 136

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