Question

# A company would like its questionnaire to be completed in 35 minutes on average. 20 people...

A company would like its questionnaire to be completed in 35 minutes on average. 20 people are randomly sampled and it is found that their mean time to complete the questionnaire is 29 minutes with a variance of 64 minutes. Is there sufficient evidence to conclude that the completion time differs from its intended duration? Using the 99% degree of confidence, the rejection rule is:

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Lets calculate the confidence interval :

Xbar +/- t*stdev/sqrt(n)

The t for 99% CI and df = n-1=20-1 = 19 is 2.861 ( we looked up the t table. or you may use the T.DIST(.995, 19, TRUE) formula to get the critical t. It is 2.861.

Xbar +/- t*stdev/sqrt(n)

= 29 +/- 2.861*8/sqrt(20)
= 23.8821 to 34.1179

99% Confidence interval is : (23.8821, 34.1179)

Does this include 35? No, therefore there is sufficient evidence to conclude that the completion time differs from its intended duration.

The rejection rule is :

If t > 2.861 or t < -2.861 we reject the null hypothesis and accept the claim that completion time differs from its intended duration.

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