Question

An admissions officer for a law school has determined that historically applicants have undergraduate grade point averages that are normally distributed with standard deviation 0.3. From a random sample of 25 applications from the current year, the sample mean grade point average is 3.2. (a)Find a 99% confidence interval for the population mean. (b)Find a 95% confidence interval for the population mean.(c)Find a 90% confidence interval for the population mean.(d)Based on these sample results, a statistician computes for the population mean a confidence interval extending from 3.15 to 3.25. Find the confidence level associated with this interval.

Answer 1

Part A:

interval=(3.0452,3.3548,)

Part B:

Interval=(3.0824,3.3176)

Part C:

Interval=(3.100763.29924,)

Part D:

Z=0.833

CI=91%

Explanation:

CI Z

90% 1.645

95% 1.96

99% 2.58

The formula we are going to use is:

Interval=X±
`(Z*S)/(√(n) )`

Where

X is the mean value

S is the standard deviation

n is the sample size

Z is the distribution

Part A:

Interval=3.2±
`(2.58*0.3)/(√(25) )`

Interval=3.2±0.1548

interval=(3.0452,3.3548)

Part B:

Interval=3.2±
`(1.96*0.3)/(√(25) )`

Interval=3.2±0.1116

Interval=(3.0824,3.3176)

Part C:

Interval=3.2±
`(1.654*0.3)/(√(25) )`

Interval=3.2±0.09924

Interval=(3.10076,3.29924)

Part D:

3.2-3.15=3.25-3.2=0.05

0.05=
`(Z*S)/(√(n) )`

0.05=
`(Z*0.3)/(√(25) )`

Z=0.833

CI=91%