Question

A TV provider has 300,000 customers and is considering unlimited movie streaming services. A random sample of 256 customers is asked what they would be willing to pay per month for the unlimited movie streaming services. the same average of the responses is $15 and the standard deviation is $16.

1. Consider all 300,000 customers. What is the average price that they would be willing to pay for unlimited movie streaming? What is the associated standard error (in $)?

2. What is the 90% confidence interval for the average of what all the customers (300,000) would pay?

3. Will the histogram of the prices given by customers in the sample be a good fit for the normal curve? What about the histogram of the prices if the TV provider company asked all 300,000 customers? What about the probability histogram of the sample average price?

Answer 1

1)

Average price = 15

Standard Error , SE = s/√n =   16.0000   / √   300000   =   0.029212

2)

Level of Significance ,    α =    0.1          
degree of freedom=   DF=n-1=   299999          
't value='   tα/2=   1.6449   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   16.0000   / √   300000   =   0.029212
margin of error , E=t*SE =   1.6449   *   0.02921   =   0.048049
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    15.00   -   0.048049   =   14.951951
Interval Upper Limit = x̅ + E =    15.00   -   0.048049   =   15.048049
90%   confidence interval is (   14.95   < µ <   15.05   )

3)

Will the histogram of the prices given by customers in the sample be a good fit for the normal curve?

No because standard deviation is too large and sample is small

What about the histogram of the prices if the TV provider company asked all 300,000 customers?

No because standard deviation is too small and sample is very large.

Please revert back in case of any doubt.

Please upvote. Thanks in advance.