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.
Accepted Solution
A:
Answer:Part A:interval=(3.0452,3.3548,)Part B:Interval=(3.0824,3.3176)Part C:Interval=(3.100763.29924,)Part D:Z=0.833CI=91%Step-by-step explanation:CI Z90% 1.64595% 1.9699% 2.58The formula we are going to use is:Interval=X±[tex]\frac{Z*S}{\sqrt{n} }[/tex]Where X is the mean valueS is the standard deviationn is the sample sizeZ is the distribution Part A:Interval=3.2±[tex]\frac{2.58*0.3}{\sqrt{25} }[/tex]Interval=3.2±0.1548interval=(3.0452,3.3548)Part B:Interval=3.2±[tex]\frac{1.96*0.3}{\sqrt{25} }[/tex]Interval=3.2±0.1116Interval=(3.0824,3.3176)Part C:Interval=3.2±[tex]\frac{1.654*0.3}{\sqrt{25} }[/tex]Interval=3.2±0.09924Interval=(3.10076,3.29924)Part D:3.2-3.15=3.25-3.2=0.050.05=[tex]\frac{Z*S}{\sqrt{n} }[/tex]0.05=[tex]\frac{Z*0.3}{\sqrt{25} }[/tex]Z=0.833CI=91%