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Peace Awoke

A random of 200 school managers were administered a developed leadership skill test. The sample mean and standard deviation were 78 and 4.2 respectively. In the standardization of test, the mean was 73 and the standard deviation was 8. Test for significant difference using alpha = 0.05

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n_1=n_2=200\\\bar x_1=78\\\bar x_2=73\\s_1=4.2\\s_2=8

Let us first perform a test for their variability using F test.

We test,


The test statistic is,

F_c={s_2^2\over s_1^2}={64\over17.64}=3.6281(4dp)

The table value is,

F_{\alpha,n_2-1,n_1-1}=F_{0.05,199,199}=1.26334 ​and we reject the null hypothesis ifF_c\gt F_{0.05,199,199}

Since F_c=3.6281\gt F_{0.05,199,199}=1.26334, we reject the null hypothesis that the population variances are equal. Hence they are not equal.

Next, we perform the test for difference in means.,

The hypothesis tested are,


The test statistic is,

t_c={(\bar x_1-\bar x_2)\over \sqrt{({s_1^2\over n_1}+{s^2_2\over n_2})}}={78-73\over\sqrt{{17.64\over200}+{64\over200}}}={5\over0.63890531}=7.8259

t_c is compared with the table value at \alpha=0.05 with v degrees of freedom. The number of degrees of freedom v is given as,

v={({s_1^2\over n_1}+{s_2^2\over n_2})^2\over {({s_1^2\over n_1})^2\over n_1-1}+{({s_2^2\over n_2})^2\over n_2-1}}={0.4082^2\over 0.0005145729+3.909166e-5}={0.16663\over0.000554}=300.95\approx. 301

The table value is,


The null hypothesis is rejected if |t_c|\gt t_{0.025,301}.

Since |t_c|=7.8259\gt t_{0.025,301}=1.649932, we reject the null hypothesis and conclude that there is sufficient evidence to show that the means are significantly different at 5% level of significance.

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Question ID: mtid-4-stid-47-sqid-4839-qpid-558