Question

How to do this hypothesis testing question?

A business student claims that, on average, an MBA student is required to prepare more than five cases per week. To examine the claim, a statistics professor asks a random sample of 10 MBA students to report the number of cases they prepare weekly. The results are exhibited here. Can the professor conclude at the 5% significance level that the claim is true, assuming that the number of cases is normally distributed with a standard deviation of 1.5?
2 7 4 8 9 5 11 3 7 4

Answer 1

Yes, the student's claim appears to be true based upon this one sample. See explanation.

Explanation:

For this problem, it is helpful to determine the null hypothesis and the alternate hypothesis. Since the student claims that an MBA student prepares an average of more than 5 cases per week, we can specify our null hypothesis `H_0` as such:

`H_0`: The population mean `mu` is 5 or less cases per week (`mu <= 5`)

The alternate hypothesis `H_a` can be specified as the claim the student is making:

`H_a`: The population mean `mu` is > 5 cases per week (`mu > 5`)

Since we are told that the number of cases is normally distributed with a standard deviation `sigma = 1.5`, we can opt to use a One Sample Z-Test for our hypothesis testing. (I'd opt for this even with a sample size of 10, which normally would lend towards a t-test, only because we are given the population standard deviation, which is often not known...making this problem more of a "classroom" and not "real world" problem.)

To proceed, we need to know the mean of the cases from our sample of 10:

`bar(x) = (2+7+ 4+ 8+ 9+ 5+ 11+ 3+ 7+ 4)/10 = 60/10=6`

We can now calculate the z-test value according to the z-test formula:

`z=(bar(x)-mu_{bar(x)})/(sigma/sqrt(n))`

In this formula:

#{: (underline("Symbol"),underline("Meaning"),underline("Value")), (bar(x), "Sample mean", 6), (mu_{bar(x)}, "Assumed population mean in test", 5), (sigma, "Population std. dev.", 1.5), (n, "Number in sample", 10):}#

`z = (6 - 5)/(1.5/sqrt(10)) = 1 * sqrt(10)/1.5 ~~ 2.11`

Our z-test score is about 2.11. Now, since our alternate hypothesis is that we are expecting the population mean to be more than 5 cases per week, we are looking for a right-tail probability (since we are looking at an alternate hypothesis that specifies values greater than 5).

According to a z-score table, the left-tail (aka cumulative from negative infinity) probability associated with a z-score of 2.11 is about 0.9826. Thus, the right-tail probability `p` we are interested in is `p = 1 - 0.9826 = 0.0174`.

At a 5% (0.05) significance level, our `p` value of 0.0174 is definitely less than the significance level. This means we have sufficient statistical weight for us to reject the null hypothesis in favor of the alternate hypothesis. The student's claim does appear to potentially be true given this one test.