A personnel manager has found that historically the scores on aptitude test given to applicants for entry-level positions follow a normal distribution with a standard deviation of 32.4 points. A random sample of nine test scores from the current group of applicants had a mean score of 187.9 points. a) Find an 80% confidence interval for the population mean score of the current group of applicants. 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.

Using the z-distribution, it is found that

a) The 80% confidence interval for the population mean score of the current group of applicants is (174.1, 201.7).

b) The confidence level is of approximately 96%.

Item a:

We are given the standard deviation for the population, which is why the z-distribution is used to solve this question.

The information given is:

The confidence interval is:

The critical value, using a z-distribution calculator, for a two-tailed 80% confidence interval is z = 1.28.

The 80% confidence interval for the population mean score of the current group of applicants is (174.1, 201.7).

Item b:

The margin of error is given by:

Hence:

Using a calculator, it is found that the confidence level is of approximately 96%.

A similar problem is given at link

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%

hi there!

the answer to this question is d: 35

step-by-step explanation:

5 to the power of 2 is 25

2 multiplied by 5 is 10

then you add the 2 numbers and you get 35