Question

(20 pts)2. According to records, the average length of all long-distance calls placed in 1999 was 12.44 minutes. The company's management wanted to check if the mean length of the current long-distance calls is different from 12.44 minutes. A sample of 150 produced a mean length of 13.71minutes with a standard deviation of 2.65 minutes. Using the 5% significance level, can you conclude that the mean length of all current long-distance calls is different from 12.44 minutes? a. State the null and alternative hypotheses. b. Construct a statistic to test the hypotheses and select the distribution to use. c. Find the critical value, and determine the rejection and non-rejection regions for this statistic. d. Make the decision and interpret.

Answer 1

0 fol: briven Sample Size (n) = 150 Sample mean ( ² ) = 1371 minutes sample Sod (s)= 2.65 minutes. Sequired to test the hypothesis that peph 12.44 minutes. meahh is @ Null hypothesis (Ho): 4= 12.44. . Alternative Hypothesie CHI): la 12.44 (Two tailed det? b) since the sample Size (n) is large, the Sample mean ) is approximately 'mor melly distributed with moan (i) and SE(x) = The test Ho is given by el uN (0,1). S under Z= sur IZ = 1.27 x 12.247 2.65 = 5.869 13.11-12.44 2.65 1150 5.869 Oral 0.025 Acceptance regpon. 0.025 Z cal criticat -1.96 1.96 c) Critical for value two at 5% level of significance is 1.96 Zcritical we call Since Izall> 1.96, so reject Ho and Conclude length of all current long distance Salls is different from 12. 44 minutes that the mean