Question

Let's say you want to test (at 0.05 significance level) whether female students, on average, scores more than 600 on the verbal SAT exam. 20 female students were selected randomly and their scores (such as in the table): 650 730 510 670 480 800 690 530 590 620 710 670 640 780 650 490 800 600 510 700

a) Suppose σ = 100. Calculate the test statistic value is used.

b) Suppose σ is unknown. Calculate the test statistic value is used.

Answer 1

The mean 600 510, 800, 800, 780, 730, 710, 700, 690, 670, 670, 650, 650, 640, 620, 590, 530, 510, 490, 480 is calculated as

Standard deviation is calculated as

S=100.10The hypotheses are:

;

Reject ho if t_obs>t_0.05

Test statistic :

the test statistic is calculated as

P- value:

P value associated with t score will be calculated either by Z table or by excel tool as

0.0336

Conclusion:

Since t calculated is > t critical value and P-value is less than of the significance level 0.05 hence we reject the null hypothesis and conclude that we have enough evidence to support the claim.

a) If the σ = 100.

b) if test statistic is unknown the we use sample test statistic