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Today, the professor claims the mean student performance on his midterm has significantly improved from last year. Last year, the mean midterm score for all of his students was 70. The professor uses a random sample of 25 student’s midterm scores (column 2). Use a hypothesis test (t-test) to test the professors claim using a 5% significance level.

1) What is the value of the test statistic?

2) What is the critical value from the t-table?

3) What is the p-value?

4) Can the professor conclude this year’s mean midterm score has improved? Explain!

Answer 1

Answer: The null and alternative hypotheses for the given problem is H0: mu = 70 vs Ha: mu > 70. Where mu is the unknown population mean.

The test statistic for the above is T = (xbar - mu0)/(s/sqrt(n)), where xbar = sample mean, mu0 is the hypothesized value of the unknown population mean, s = sample standard deviation, n = sample size.

1) The value of the test statistic is T(observed) =  2.846.

We reject H0 if T(observed) > t(alpha,n-1) where t(alpha,n-1) is the upper alpha point of a t distribution with (n-1) degrees of freedom. Alpha is the level of significance. t(alpha,n-1) is termed as the critical value from the t-table.

2) The critical value from the t-table = 1.7108

3) The p-value is the probability of observing a sample statistic as extreme as the test statistic. Thus, here the P-value is the probability that a t statistic having 24 degrees of freedom is more extreme than 2.846. Thus, the p-value of the test statistic is =  P( t > 2.486) = 1- P( t < 2.486) =  0.004459

4) Since the p-value is less than the level of significance and T(observed) > t(alpha,n-1), then we can say that we reject H0 and conclude on the basis of the given sample at 5% level of significance that the professor might say that this year's mean midterm score has improved.