Question

# Historically, evening long-distance calls from a particular city have averaged 15.2 minutes per call.  In a random...

Historically, evening long-distance calls from a particular city have averaged 15.2 minutes per call.  In a random sample of 35 calls, the sample mean time was 14.3 minutes.  Assume the standard deviation is known to be 5 minutes.  Using a 0.05 level of significance, is there sufficient evidence to conclude that the average evening long-distance call has decreased? Use the six-steps (clearly labeled )

Solution:

1)

The null and alternative hypothesis are

H0 : = 15.2 vs Ha : < 15.2

2)

Test statistic z = = [14.3- 15.2]/[5/35] = -1.06

Test statistic z = -1.06

3)

Given , 0.05 level of significance,

Here , left tailed test

So , critical value is

Critical value is -1.65

Rejection rule : Reject H0 if z < -1.65

4)

p value = P(Z < -1.06) = 0.1446

5)

Decision : Fail to reject H0

Because p value is greater than 0.05 level of significance,

6)

Conclusion: There is not sufficient evidence to conclude that the average evening long-distance call has decreased

#### Earn Coins

Coins can be redeemed for fabulous gifts.