Answer:
A) An 80% confidence interval for the population mean score of the current group of applicants is (172.8124,202.9876)
B)The confidence level of this interval is 96%
Step-by-step explanation:
Mean =
Standard deviation = s= 32.4
n = 9
A) Find an 80% confidence interval for the population mean score of the current group of applicants.
Degree of freedom = n-1=9-1=8
Significance level =
So,
Formula of confidence interval :
CI :
So, CI:
CI:(172.8124,202.9876)
So, an 80% confidence interval for the population mean score of the current group of applicants is (172.8124,202.9876)
B)Based on these sample results, a statistician found for the population mean a confidence interval extending from 165.8 to 210.0 points. Find the confidence level of this interval.
Standard error of sample mean = se =
Margin of error = 210 - 165.8 = 44.2
Critical test statistic =
Using excel
Confidence interval =
Confidence level= 1 - 0.04 = 0.96
Confidence interval is 96%.
Hence the confidence level of this interval is 96%