Question

A personnel manager has found that historically the scores on aptitude test given to applicants for entry level positions follow a normal distribution. A random sample of nineteen test scores from the current group of applicants had a sample mean score of 187.9 points and sample standard deviation as 32.4 points.

a. Find the margin of error.

b. Construct a confidence interval estimate for population mean with 80 percent level of confidence.

c. Construct a confidence interval estimate for population mean with 85 percent level of confidence.

Answer 1

SOLUTION:

From given data,

A personnel manager has found that historically the scores on aptitude test given to applicants for entry level positions follow a normal distribution. A random sample of nineteen test scores from the current group of applicants had a sample mean score of 187.9 points and sample standard deviation as 32.4 points.

mean = =187.9

standard deviation =s = 32.4

Sample size = n = 19

(a). Find the margin of error.

standard error = s / sqrt( n ) = 32.4 / sqrt(19) = 7.433069

margin of error (ME) = Critical value x Standard error

For 80% level of confidence

To find the critical value, we take the following steps.

• Compute alpha (α):

  α = 1 - (confidence level / 100) = 1 - 0.80 = 0.2

• Find the critical probability :

  α/2 = 0.2/2 = 0.1

• Find the degrees of freedom (df):

  df = n - 1 = 19 -1 = 18

Critical value : t_α/2,df = t_0.1,18 = 1.33039

margin of error (ME) = 1.33039 x 7.433069 = 9.888

For 85% level of confidence

To find the critical value, we take the following steps.

• Compute alpha (α):

  α = 1 - (confidence level / 100) = 1 - 0.85 = 0.15

• Find the critical probability :

  α/2 = 0.15/2 = 0.075

• Find the degrees of freedom (df):

  df = n - 1 = 19 -1 = 18

Critical value : t_α/2,df = t_0.075,18 = 1.50370

margin of error (ME) = 1.50370 x 7.433069 = 11.177

(b). Construct a confidence interval estimate for population mean with 80 percent level of confidence.

80 percent confidence interval

ME

- ME <   <   + ME

187.9 - 9.888<   <  187.9  + 9.888

178.012 <   <  197.788

c. Construct a confidence interval estimate for population mean with 85 percent level of confidence.

85 percent confidence interval

ME

- ME <   <   + ME

187.9 - 11.177 <   <  187.9  + 11.177

176.723 <   <  199.077