Suppose average pizza delivery times are normally distributed with an unknown population mean and a population standard deviation of 6 minutes. A random sample of 28 pizza delivery restaurants is taken and has a sample mean delivery time of 36 minutes. What parameters should be used to calculate a confidence interval at the 95% confidence level for this population mean?

Answer :

Answer:

95% confidence interval = [ 33.78 , 38.22 ]

Step-by-step explanation:

We are given that average pizza delivery times are normally distributed with an unknown population mean and a population standard deviation of 6 minutes.

A random sample of 28 pizza delivery restaurants is taken and has a sample mean delivery time of 36 minutes.

The Pivotal quantity for 95% confidence interval is given by;

             [tex]\frac{Xbar - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)

where, X bar = sample mean = 36

             [tex]\sigma[/tex] = population standard deviation = 6

             n = sample size = 28

So, 95% confidence interval for population mean, [tex]\mu[/tex] is given by;

P(-1.96 < N(0,1) < 1.96) = 0.95

P(-1.96 < [tex]\frac{Xbar - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < 1.96) = 0.95

P(-1.96 * [tex]{\frac{\sigma}{\sqrt{n} }[/tex] < [tex]{Xbar - \mu}[/tex] < 1.96 * [tex]{\frac{\sigma}{\sqrt{n} }[/tex] ) = 0.95

P(X bar - 1.96 * [tex]{\frac{\sigma}{\sqrt{n} }[/tex] < [tex]\mu[/tex] < X bar + 1.96 * [tex]{\frac{\sigma}{\sqrt{n} }[/tex] ) = 0.95

95% confidence interval for [tex]\mu[/tex] = [ X bar - 1.96 * [tex]{\frac{\sigma}{\sqrt{n} }[/tex] , X bar + 1.96 * [tex]{\frac{\sigma}{\sqrt{n} }[/tex] ]

                                               = [ 36 - 1.96 * [tex]{\frac{6}{\sqrt{28} }[/tex] , 36 + 1.96 * [tex]{\frac{6}{\sqrt{28} }[/tex] ]

                                               = [ 33.78 , 38.22 ]

Therefore, 95% confidence interval for population mean is [33.78 , 38.22] .

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