Statistics and Probability

Question

A math teacher wants to know whether the true average time of his students who do a Math test is 98 minutes. A random sample of 18 students have an average of 98.2 minutes with a standard deviation of 0.4 minutes. Is there enough evidence to reject the null hypothesis at the 0.05 level of significance?

What is the test value of the problem? (round off your answer to three decimal places)

Answer 1

This Is a 2 sided test with null hypothesis of mu = 98 and alternate hypothesis of mu /= 98.

tstat = (98.2 - 98)/(3/sqrt(18)) = 2.121 with 18-1=17 degrees of freedom. This translates to a p value of 0.0488 so you can reject the null hypothesis at 0.05 level of significance.

Answer 2

Since we only know the sample standard deviation, not the populations standard deviation, and the sample size is relatively small, our test statistic will be t, not z.

t = (xbar - x₀)/(s/√n) = (98.2 - 98)/(.4/√18) ≈ 2.121

We can compare this t-statistic with a t-table (such as this one https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf), looking for the row where the degrees of freedom = n - 1 = 17.

If we're doing a two-tailed test, we look for where at the top it says "two tails, 0.05," then we look for the row where the degrees of freedom is 17, and we find a critical value of 2.11. It's extremely close, but our test statistic is bigger than the critical value, i.e. it is in the rejection region. Thus there is evidence to suggest that the true average time his students take on a test is significantly different than 98 minutes.