Question

In: Statistics and Probability

The TIV Telephone Company provides long-distance telephone service in an area. According to the company’s records,...

The TIV Telephone Company provides long-distance telephone service in an area. According to the company’s records, the average length of all long-distance phone calls placed through this company in 2015 was 12.44 minutes. The company’s management wants to check if the mean length of the current long- distance calls is different from 12.44 minutes. A sample of 150 such calls placed through the company produced a mean length of 13.71 minutes. The standard deviation of all such calls is 2.65 minutes. Using a 2% level of significance, can you conclude that the mean length of all current long-distance calls is different from 12.44 minutes? Hypothesis to be tested (in words): Hypothesis to be tested (using symbols): H0 : H1 : Is this problem a (circle one): left-tailed test, right tailed test, or a two-tailed test? a= ,x= ,sors= ,n= .

Test statistic =

The p-value =

Reject or fail to reject H0 (circle one) Rejection region (picture): Conclusion (in words):

Solutions

Expert Solution

H0 : The mean length is 12.44 minutes

H1: The mean length is not equal to 12.44 minutes

As we are interested only in finding whether the mean is different from 12.44, it is two tailed test. The mean could be smaller or greater.

therefore area= level of significance/2 (both tails)= 0.01

X (sample mean from 150 calls)= 13.71

Next, we have to determine whether the population standard deviation is available. The sample mean was calculated from 150 calls. However, the question says that standard deviation of all calls is 2.65. It seems a bit ambiguous, but looks like the given value is population standard deviation. The sample is also considerably large, considering law of large numbers. Consequently, we can use normal distribution for the hypothesis test.

Let A be average which is 12.44.

S be the population standard deviation=2.65

Z(test statistic)= (X - 12.44) / (2.65/150)  

= ( X - 12.44) /0.2163

P Value= ( 13.71 -12.44)/0.2163= 5.871

Our critical values from normal distribution tables are -2.33 and 2.33 (approximately)

As our Z value is greater than critical value, 5.871> 2.33, we can safely conclude that mean is different from 12.44.

orchestra answered 1 year ago
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