Question

10 A survey of 25 grocery stores revealed that the average price of a gallon of milk was $2.98, with a standard error of $0.10. What is the 98% confidence interval to estimate the true cost of a gallon of milk?

Multiple Choice

• $2.85 to $3.11

• $2.73 to $3.23

• $2.95 to $3.01

• $2.94 to $3.02

11. A sample of 100 is selected from a known population of 350 elements. The population standard deviation is 15. Using the finite correction factor, what is the standard error of the sample means?

Multiple Choice

• 0.8452

• 12.6773

• Cannot be determined

• 1.2695

12. A survey of 50 retail stores revealed that the average price of a microwave was $375 with a sample standard deviation of $20. Assuming the population is normally distributed, what is the 99% confidence interval to estimate the true cost of the microwave?

Multiple Choice

• $315.00 to $415.00

• $335.82 to $414.28

• $323.40 to $426.60

• $367.42 to $382.58

13. A local company wants to evaluate their quality of service by surveying their customers. Their budget limits the number of surveys to 100. What is their maximum error of the estimated mean quality for a 95% level of confidence and an estimated standard deviation of 5?

Multiple Choice

• 0.98

• 1.96

• 5%

• 0.9604

14. A group of statistics students decided to conduct a survey at their university to find the average (mean) amount of time students spent studying per week. They sampled 240 students and found a mean of 22.3 hours per week. Assuming a population standard deviation of six hours, what is the 99% level of confidence?

rev: 11_15_2017_QC_CS-109448

Multiple Choice

• [21.30, 23.30]

• [16.3, 28.3]

• [21.80, 22.80]

• [20.22, 22.0]

Answer 1

10)

Level of Significance ,    α =    0.02          
degree of freedom=   DF=n-1=   24          
't value='   tα/2=   2.4922   [Excel formula =t.inv(α/2,df) ]     

Standard error , SE = 0.1
                  
margin of error , E=t*SE =   2.4922   *   0.10000   =   0.249
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    2.98   -   0.249=   2.73
Interval Upper Limit = x̅ + E =    2.98 + 0.249 =   3.229
98%   confidence interval is (   2.73   < µ <   3.23   )

11)

n = 100

N= 350

σ=15

standard error is given by correction factor =

= 15/ 10 * √ (350-100) / (350-1)

= 1.5 * √(0.7163)

= 1.5*0.8463 =1.2695

option(d)

12)

Level of Significance ,    α =    0.01          
degree of freedom=   DF=n-1=   49          
't value='   tα/2=   2.6800   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   20.0000   / √   50   =   2.8284
margin of error , E=t*SE =   2.6800   *   2.82843   =   7.5800
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    375.00   -   7.580049   =   367.4200
Interval Upper Limit = x̅ + E =    375.00 + 7.580049   =   382.5800
99%   confidence interval is (   367.42   < µ <   382.58   )

13)

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   99          
't value='   tα/2=   1.9842   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   5.0 / √   100   =   0.5
maximum error of estimated mean  =   1.9842   *   0.5 =   0.992

14)

Level of Significance ,    α =    0.01          
degree of freedom=   DF=n-1=   239          
't value='   tα/2=   2.5966   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   6.0000   / √   240   =   0.387
margin of error , E=t*SE =   2.5966   *   0.38730   =   1.006
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    22.30   -   1.006   =   21.294
Interval Upper Limit = x̅ + E =    22.30   +   1.006   =   23.306
99%   confidence interval is (   21.3   < µ <   23.3   )