Question

STA 2221 Examples on CI Stesting PART 1 PARTI MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answer the question. Provide an appropriate response. 1) Find the critical value ze that corresponds to a 99% confidence level. A) *1.645 B) 233 +1.88 D) 41.96 2) 2)Determine the sampling error if the grade point averages for 10 randomly selected students from a class of 125 students has a mean of x=28. Assume the grade point average of the 125 students has a mean of 3.5 A) 245 B)-07 0907 D) 3.15 3) Find the margin of error for the given values of co, and n. C-0.98, 0.78, n = 150 A) 0.11 B) 0.12 0.08 D) 0.15 4) 4) A random sample of 150 students has a grade point average with a mean of 2.86. Assume the population standard deviation is 0.78. Construct the confidence interval for the population mean, w, ifc-0.98 A) (2.51, 3.53) B) (2.31, 3.88) C) (2.43, 3.79) D) (2.71, 3.01) 5) 5) A random sample of 56 fluorescent light bulbs has a mean life of 645 hours. Assume the population standard deviation is 31 hours. Construct a 95% confidence interval for the population mean. A) (112.0, 118.9) B) (712.0, 768.0) (539.6, 551.2) D) (636.9, 653.1) 6) 6) A group of 40 bowlers showed that their average score was 192. Assume the population standard deviation is 8. Find the 95% confidence interval of the mean score of all bowlers. A) (186.5, 1975) B) (189.5, 194.5) (188.5, 195.6) D) (1873, 196.1) SHORT ANSWER. Write the word or phrase that best completes each statement or answer the question. 7) In a random sample of 60 computers, the mean repair cost was $150. Assume the 7)_ population standard deviation is $36. a) Construct the 99% confidence interval for the population mean repair cost. b) If the level of confidence was lowered to 95%, what will be the effect on the confidence interval? 8) – 8) A certain confidence in interval is 7.75<<945. Find the sample mean x and the error of estimate E.

Answer 1

1)z value=   z α/2= ±1.88 [Excel formula =NORMSINV(α/2) ]
3) Level of Significance ,    α =    0.02          
'   '   '          
z value=   z α/2=   2.3263   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   0.7800   / √   150   =   0.0637
margin of error, E=Z*SE =   2.3263   *   0.0637   =   0.15

4)

confidence interval is                   
Interval Lower Limit = x̅ - E =    2.86   -   0.15 =   2.7118
Interval Upper Limit = x̅ + E =    2.86   -   0.15 =   3.0082
98%   confidence interval is (   2.71   < µ <   3.01   )

5)

Level of Significance ,    α =    0.05          
'   '   '          
z value=   z α/2=   1.9600   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   31.0000   / √   56   =   4.1425
margin of error, E=Z*SE =   1.9600   *   4.1425   =   8.1192
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    645.00   -   8.119247   =   636.8808
Interval Upper Limit = x̅ + E =    645.00   -   8.119247   =   653.1192
95%   confidence interval is (   636.9 < µ <   653.1 )