Question

The mayor is interested in finding a 90% confidence interval for the mean number of pounds of trash per person per week that is generated in city. The study included 248 residents whose mean number of pounds of trash generated per person per week was 34.6 pounds and the standard deviation was 8.1 pounds.

a. To compute the confidence interval use a ? t z  distribution.

b. With 90% confidence the population mean number of pounds per person per week is between  and   pounds.

c. If many groups of 248 randomly selected members are studied, then a different confidence interval would be produced from each group. About  percent of these confidence intervals will contain the true population mean number of pounds of trash generated per person per week and about  percent will not contain the true population mean number of pounds of trash generated per person per week.

Answer 1

Solution :

a) We using t-distribution,

b) degrees of freedom = n - 1 = 248 - 1 = 247

t/2,df = t_0.05,247 = 1.651

Margin of error = E = t_/2,df * (s /n)

= 1.651 * (8.1 / 248)

Margin of error = E = 0.849

The 90% confidence interval estimate of the population mean is,

  ± E  

= 34.6 ± 0.849

= ( 33.751, 35.449 )

With 90% confidence the population mean number of pounds per person per week is between 33.751 and 35.449 pounds.

c) If many groups of 248 randomly selected members are studied, then a different confidence interval would be produced from each group. About 90 percent of these confidence intervals will contain the true population mean number of pounds of trash generated per person per week and about 10 percent will not contain the true population mean number of pounds of trash generated per person per week.