Question

Problem 1

The time for an employee to complete a certain manufacturing task was was approximately Normally distributed with a mean of 138 seconds and a standard deviation of 39 seconds. Suppose you plan to take an SRS of 250 employees and look at the average manufacturing time.

a)       Determine the mean and standard deviation of the distribution of sample means.

b)      Calculate the probability of coming up with a sample that shows a manufacturing time of 114 seconds or less.

Problem 2

You want to research the cost of a certain trendy ergonomic chair. In your online search of the cost of that chair on 20 different sites, you note that the mean cost of the chair is $562. Assume that the SD of the chair among all sites is $74. Find the 90%, 95%, and 99% confidence intervals for the mean cost of this chair.

Look at the 95% confidence interval and say whether the following statement is true or false. “This interval describes the price of 95% for all chairs of this model that are being sold.” If you think this is incorrect, be sure to explain why.   

Problem 3

A credit card company wants to estimate the average length of client churn (turning over their card for a different one). In a sample of 5000 customers obtained at random from the company’s database, the mean churn is 418. The the standard deviation of the calls from that sample is 93. Provide the mean and SD of the distribution of this sample.

Problem 4

A web-based travel search engine is designed to complete searches for cruise ships involving price retrieval and scheduling information. You need to analyze how long it takes to complete a typical search. The search time for 2500 randomly selected searches is recorded and detreemined as sample mean = 0.94 seconds and sample standard deviation = 0.36 seconds. Choose the most typical value for C and provide the confidence interval.

Answer 1

1)

mean =138

and standard deviation of the distribution of sample means=σ/√n=39/√250=2.47

b)

µ =    138                                      
σ =    39                                      
n=   250                                      
                                          
X =   114                                      
                                          
Z =   (X - µ )/(σ/√n) = (   114   -   138.00   ) / (   39.000   / √   250   ) =   -9.730  
                                          
P(X ≤   114   ) = P(Z ≤   -9.730   ) =   0.0000                       (answer)

2)

Level of Significance ,    α =    0.1          
degree of freedom=   DF=n-1=   19          
't value='   tα/2=   1.729   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   74.0000   / √   20   =   16.5469
margin of error , E=t*SE =   1.7291   *   16.5469   =   28.6118
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    562.00   -   28.611793   =   533.3882
Interval Upper Limit = x̅ + E =    562.00   -   28.611793   =   590.6118
90.0%   confidence interval is (   533.3882   < µ <   590.6118   )
----------------

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   19          
't value='   tα/2=   2.093   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   74.0000   / √   20   =   16.5469
margin of error , E=t*SE =   2.0930   *   16.5469   =   34.6331
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    562.00   -   34.633066   =   527.3669
Interval Upper Limit = x̅ + E =    562.00   -   34.633066   =   596.6331
95.0%   confidence interval is (   527.3669   < µ <   596.6331   )
======================

Level of Significance ,    α =    0.01          
degree of freedom=   DF=n-1=   19          
't value='   tα/2=   2.861   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   74.0000   / √   20   =   16.5469
margin of error , E=t*SE =   2.8609   *   16.5469   =   47.3396
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    562.00   -   47.339608   =   514.6604
Interval Upper Limit = x̅ + E =    562.00   -   47.339608   =   609.3396
99.0%   confidence interval is (   514.6604   < µ <   609.3396   )

3)

mean=418

std dev = σ/√n = 93/√5000 = 1.3152

4)

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   2499          
't value='   tα/2=   1.961   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   0.3600   / √   2500   =   0.0072
margin of error , E=t*SE =   1.9609   *   0.0072   =   0.0141
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    0.94   -   0.014119   =   0.9259
Interval Upper Limit = x̅ + E =    0.94   -   0.014119   =   0.9541
95%   confidence interval is (   0.9259   < µ <   0.9541   )