Question

You are given the sample mean and the sample standard deviation. Use this information to construct the? 90% and? 95% confidence intervals for the population mean. Which interval is? wider? If? convenient, use technology to construct the confidence intervals. A random sample of 39 gas grills has a mean price of ?$643.60 and a standard deviation of ?$56.60 The? 90% confidence interval is __ , ___

Answer 1

Given: Mean,u= $ 643.60

Standard deviation,s = $ 56.60

Sample size,n = 39

Thus, standard error, e = s/?n = 56.60/?39 = 9.06

Now, for 90% confidence interval , Z = (+-) 1.645

For 95% confidence interval ,Z = (+-)1.96

Thus, 90% CI for the population mean = u (+-) 1.645*e

= { 643.60 - 1.645*9.06 , 643.60 + 1.645*9.06 } = { 628.70, 658.50}

95% CI for the population mean = u (+-) 1.96*e

= { 643.60 - 1.96*9.06, 643.60 + 1.96*9.06} = { 625.84, 661.36}

Thus, the 95% Confidence interval is wider.