Question

all answers please

3. The teachers union is concerned about the amount of time teachers spend each week doing schoolwork at home. A simple random sample of 56 teachers had a mean of 8.0 hours per week working at home after school. Assume that the population standard deviation is 1.5 hours per week. a) Construct a 95% confidence interval estimate for the mean number of hours per week a teacher spends working at home. b) Does the confidence interval support that teachers work at home after school more than 10 hours a week? Why or why not? 4. A survey of 85 randomly selected homeowners finds that they spend a mean of $67 per month on internet. Assume that the population standard deviation is $14 per month. a) Construct and interpret a 99% confidence interval estimate for the mean amount of money spent per month on internet by all homeowners. b) Does the confidence interval support that internet costs more than $60 per month? Why or why not? 5. The actual time it takes to cook a ten-pound turkey is a normally distributed. Suppose that a random sample of 20 ten pound turkeys is taken with a sample mean of 2.9 hours and a sample standard deviation of 0.24 hours. Calculate a 95% confidence interval for the average cooking time of a ten pound turkey.

Answer 1

3)

a)

Level of Significance ,    α =    0.05          
'   '   '          
z value=   z α/2=   1.9600   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   1.5000   / √   56   =   0.2004
margin of error, E=Z*SE =   1.9600   *   0.2004   =   0.3929
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    8.00   -   0.392867   =   7.6071
Interval Upper Limit = x̅ + E =    8.00   -   0.392867   =   8.3929
95%   confidence interval is (   7.61   < µ <   8.39   )

b)

since, confidence interval is less than 10, there is no enough evidence that teachers work more than 10 hours a week

4)

a)

Level of Significance ,    α =    0.01          
'   '   '          
z value=   z α/2=   2.5758   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   14.0000   / √   85   =   1.5185
margin of error, E=Z*SE =   2.5758   *   1.5185   =   3.9114
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    67.00   -   3.911431   =   63.0886
Interval Upper Limit = x̅ + E =    67.00   -   3.911431   =   70.9114
99%   confidence interval is (   63.09   < µ <   70.91   )

b)

yes, it support the claim that internet cost is more than 60, because both ends of confidence interval are more than 60