Question

A random sample of fifty si ​200-meter swims has a mean time of 3.06 minutes and the population standard deviation is 0.08 minutes. Construct a 95​% confidence interval for the population mean time. Interpret the results.In a random sample of 50 ​refrigerators, the mean repair cost was $136.00 and the population standard deviation is​$19.1019.10. A 90​% confidence interval for the population mean repair cost is (131.56,140.44). Change the sample size to n=100. Construct a 90% confidence interval for the population mean repair cost. Which confidence interval is​ wider? Explain.

Construct a 90% confidence interval for the population mean repair cost.

The 95​% confidence interval isA random sample of thirty-seven ​200-meter swims has a mean time of 3.591 minutes. The population standard deviation is 0.080 minutes. A 90​% confidence interval for the population mean time is (3.569,3.613). Construct a 90​% confidence interval for the population mean time using a population standard deviation of 0.03 minutes. Which confidence interval is​ wider? Explain.

The 90​% confidence interval is

The 95​% confidence interval is

You are given the sample mean and the population standard deviation. Use this information to construct the​ 90% and​ 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If​ convenient, use technology to construct the confidence intervals.

A random sample of 35 home theater systems has a mean price of ​$128.00. Assume the population standard deviation is ​$15.90. Find the 90% and 95% of confidence interval.

Answer 1

Level of Significance ,    α =    0.05          

z value=   z α/2=   1.9600   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   0.0800   / √   50   =   0.011314
margin of error, E=Z*SE =   1.9600   *   0.01131   =   0.022174
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    3.06   -   0.022174   =   3.037826
Interval Upper Limit = x̅ + E =    3.06   -   0.022174   =   3.082174
95%   confidence interval is (   3.04   < µ <   3.08   )

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REFRIGERATOR 100 SAMPLE SIZE 90% CONFIDENCE

Level of Significance ,    α =    0.1          
'   '   '          
z value=   z α/2=   1.6449   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   19.1000   / √   100   =   1.910000
margin of error, E=Z*SE =   1.6449   *   1.91000   =   3.141670
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    136.00   -   3.141670   =   132.858330
Interval Upper Limit = x̅ + E =    136.00   -   3.141670   =   139.141670
90%   confidence interval is (   132.86   < µ <   139.14   )

Sample size 50 is more wider

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Level of Significance ,    α =    0.1          
'   '   '          
z value=   z α/2=   1.6449   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   0.0300   / √   37   =   0.004932
margin of error, E=Z*SE =   1.6449   *   0.00493   =   0.008112
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    3.59   -   0.008112   =   3.582888
Interval Upper Limit = x̅ + E =    3.59   -   0.008112   =   3.599112
90%   confidence interval is (   3.582 < µ <   3.599 )

Standard deviation 0.08 is more wider

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