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Question 424:



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We'd conduct a 1-sample t-test or z-test on the summary data against the criteria of 40 hours. We are making the assumption  that the time to learn data are roughly normally distributed.

a. Null hypothesis is the mean of the 100 users is not different than 40 hours. Alternative Hypothesis is the mean is less than 40 hours.

b. We will conclude there is a difference if the mean difference has less than a 5% chance its difference from 40 hours is due to chance.

c. The sample is sufficiently large that you could use the z-test, but I'll take the slightly more conservative approach and use the t-test. The test statistics t is -1.25.

  1. To get this value we first subtract the observed mean from the test mean 39-40 = -1 and this becomes the numerator for our test statistic.
  2. Find the standard error of the sample mean (SEM) which is the standard deviation divided by the square root of the sample size 8/SQRT(100) =.8 This is the denominator for the test statistic.
  3. Dividing the fraction we get -1/.8 = -1.25, which is our test t-statistic.

d. The p-value associated with the t-statistic of -1.25 for 99 degrees of freedom (1-sided) is .107, which is greater than the decision threshold of .05. To find that p-value we can use the excel function =TDIST(1.25,99,1) and we get the p-value less than .1071. Where the parameters in the excel function are the test statistic, the degrees of freedom (100-1) and 1 tail probability. We can also use the calculator available here http://www.usablestats.com/calcs/1samplet&summary=1 and enter the data and we get the same result p = .1071.

Result: There is not enough evidence to reject the NULL hypothesis that the mean time to learn is less than 40 hours.  The observed difference is what we'd expect due to chance fluctuations given our decision threshold of .05.

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