Question

Dr. Su is interested in whether males and females differ in their verbal ability. She selects 5 males and 5 females to participate in her study. She administers everyone a test of verbal ability (scores ranging between 0-20, higher scores represent higher ability). Here are the scores Males: 10, 10, 12, 12, 16 Females: 9, 12, 12, 15, 17 Set up non-directional hypotheses for the independent t-test. Provide the decision rule, illustrating with a drawing. Conduct an independent t-test, report your decision. What is your conclusion?

Answer 1

First we calculated mean and standard deviation for males and females verbal ability scores. MS Excel was used for calculation. Screenshot below:

Male Female 10 16 17 12 13 2.4495 3.0822 21 Mean 22 Std. Dev. 23 Function used 24 Mean 25 Std. Dev. =AVERAGE(B16:B20) =AVERAGE(C16:C20) -STDEV.S(B16:B20) ESTDEV.S(C16:C20)

Testing for Equality of Variances A F-test is used to test for the equality of variances. The following F-ratio is obtained: si 2.44952 833.08992 = 0.632 The critical values are F = 0.104 and Fu = 9.605, and since F = 0.632, then the null hypothesis of equal variances is not rejected.

Thus, based on test for equality of variances, independent t-test can be done assuming equal population variances. Following are the steps followed for hypothesis test:

The provided sample means are shown below: X1 = 12 X2 = 13 Also, the provided sample standard deviations are: $1 = 2.4495 $2 = 3.0822 and the sample sizes are ny = 5 and n2 = 5. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: L1 = 42 Ha: 41 2 This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the degrees of freedom are df = 8. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal: Hence, it is found that the critical value for this two-tailed test is te = 2.306, for a = 0.05 and df = 8. The rejection region for this two-tailed test is R = {t:t >> 2.306}.

(3) Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows: X1 - X (n -1) +(12-1) /1 1 ) 12-13 (5-12.44952+(5-1)3.08222/1 3=32 ( = = -0.568 1 ) + (4) Decision about the null hypothesis Since it is observed that t = 0.568 <te = 2.306, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p = 0.5856, and since p = 0.5856 > 0.05, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean Mi is different than 12, at the 0.05 significance level.

Thus, at 0.05 significance level, there is not enough evidence to conclude that males and females differ in their verbal ability.