Question

all answers please!!

1. A company that produces brendis concerned about the distribution of the amount of sodium in its bread. The company takes a simple random sample of 100 slices of bread and compute the sample mean to be 103 milligrams of sodium per slice. Assume that the population standard deviation is 10 milligrams a) Construct a 95% confidence interval estimate for the mean sedium level. b) Construct a 99% confidence intervalestime for the mean sodium level. 2. Fill in the Marks with on of the following: imases de me sa e saw where a) As the sample size (n) increases, the margin of error (E) b) As the confidence level ( increases, the margin of error c) As the standard deviation ( increases the margin of error The Irachers union is concemed a t the amount of time teachers spend cach weck doing schoolwork at home. A simple random sample of 56 teachers had a mean of 8 hours per week working at home after school. Assume that the population standard deviation is 1.5 hours per week Construct a 95% confidence interval estimate for the mean number of hours per week a teacher spends working at home Does the confidence interval support that cachers work a home after school more than 10 hours a week? Why or why not? A survey of 85 randomly sclected homeowners finds that they spend a man of 567 per month on internet. Assume that the population standard deviation is 514 per month Construct and interpreta 99% confidence interval estimate for the mean amount of money spent per month on internet by all h omes. Does the confidence interval support that intemet 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

1)

a)

Level of Significance ,    α =    0.05          
'   '   '          
z value=   z α/2=   1.9600   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   10.0000   / √   100   =   1.0000
margin of error, E=Z*SE =   1.9600   *   1.0000   =   1.9600
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    103.00   -   1.959964   =   101.0400
Interval Upper Limit = x̅ + E =    103.00   -   1.959964   =   104.9600
95%   confidence interval is (   101.04   < µ <   104.96   )

b)

Level of Significance ,    α =    0.01          
'   '   '          
z value=   z α/2=   2.5758   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   10.0000   / √   100   =   1.0000
margin of error, E=Z*SE =   2.5758   *   1.0000   =   2.5758
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    103.00   -   2.575829   =   100.4242
Interval Upper Limit = x̅ + E =    103.00   -   2.575829   =   105.5758
99%   confidence interval is (   100.42   < µ <   105.58   )

2)
a) decreases
b) increases
c) increases