Question

Use the standard normal distribution or the t-distribution to construct a 99% confidence interval for the population mean. Justify your decision. If neither distribution can be used, explain why. Interpret the results. In a recent season, the population standard deviation of the yards per carry for all running backs was 122. The yards per carry of 25 randomly selected running backs are shown below. Assume the yards per carry are normally distributed 22 2.9 3.9 4.6 5.5 5.5 62 75 14 21 64 4.3 99 55 5.7 6.6 37 2.8 53 725 .2 6.3 6 6. 5 .1 6.9 64 Which distribution should be used to construct the confidence interval? OA. Use a t-distribution because n < 30 and is unknown OB. Use a normal distribution because o is known and the data are normally distributed OC. Use a normal distribution because n< 30 the data are normally distributed and is unknown OD. Use a t-distribution because n< 30 and is known O E. Cannot use the standard normal distribution or the t-distribution because o is unknown, n< 30, and the data are not normally distributed Select the correct choice below and, if necessary, fill in any answer boxes to complete your choice. O A. The 99% confidence interval is ). (Round to two decimal places as needed) B. Neither distribution can be used to construct the confidence interval O Interpret the results. Choose the correct answer below. O A. If a large sample of players are taken approximately 99% of them will have yards per carry between the bounds of the confidence interval Click to select your answer S ONO DUE TO constructor conce EPORTISO C23541

Answer 1

use a normal distribution because σ is known and the data are normally distributed.

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population std dev ,    σ =    1.2200
Sample Size ,   n =    25
Sample Mean,    x̅ = ΣX/n =    5.0880
Level of Significance ,    α =    0.01          
'   '   '          
z value=   z α/2=   2.5758   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   1.2200   / √   25   =   0.244000
margin of error, E=Z*SE =   2.5758   *   0.24400   =   0.628502
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    5.09   -   0.628502   =   4.4595
Interval Upper Limit = x̅ + E =    5.09   -   0.628502   =   5.7165
99%   confidence interval is (   4.46   < µ <   5.72   )
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interpret :

answer: option a)