Question

You are given the sample mean and the population standard deviation. Use this information to construct the​ 90% and​ 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If​ convenient, use technology to construct the confidence intervals. A random sample of 50 home theater systems has a mean price of ​$113.00. Assume the population standard deviation is ​$15.20.

Answer 1

Answer:

90% confidence interval, (109.47,116.53)

For 95% confidence interval (108.79,117.21)

Step-by-step explanation:

n =50

mean = 113

standard deviation = 15.20

a.) For 90% confidence interval, z=1.645

90% confidence interval is

$(113 - \frac{1.645*15.20}{sqrt(50)},113 + \frac{1.645*15.20}{sqrt(50)}}$

=(113 - 3.53, 113 + 3.53)

=(109.47,116.53)

b.)For 95% confidence interval, z = 1.96

95% confidence interval is

$(113 - \frac{1.96*15.20}{\sqrt(50)},113 + \frac{1.96*15.20}{\sqrt(50)}$

=(113 - 4.21, 113 + 4.21)

=(108.79,117.21)