Question #253485

performance of 40 teachers in the Mathematics department showed a mean of 95 with a standard deviation of 7.3.


1
Expert's answer
2021-11-30T11:01:45-0500

The performance of 50 teachers from the faculty of engineering showed a mean of 98 with a standard deviation of 9.2, while, a sample of 40 teachers from the college of science showed an average performance of 95 with a standard deviation of 7.3. Is there a difference in performance between the two samples? Use α = 0.05

The following null and alternative hypotheses need to be tested:

H0:μ1=μ2H1:μ1μ2H_0: \mu_1 = \mu_2 \\ H_1: \mu_1 ≠ \mu_2

This corresponds to a two-tailed test.

Level of significance =0.05

Sincesample sizes are greater than 30 so z-test for two means, with unknown population standard deviations will be used.

Under null hypothesis the test statistic is obtained as:

Z=x1ˉx2ˉs12/n1+s22/n2=9895(9.2)2/50+(7.3)2/40=1.725Z = \frac{\bar{x_1} - \bar{x_2}}{\sqrt{s^2_1/n_1 + s^2_2/n_2}} \\ = \frac{98-95}{\sqrt{(9.2)^2/50 + (7.3)^2/40}} \\ = 1.725

Based on the information provided, the significance level is 0.05, and the critical value for a two-tailed test is Zc=1.96Z_c =1.96

The rejection region for this two-tailed test is

R=Z:Z>1.96R = {Z:|Z|>1.96}

Since it is observed that

Z=1.725<Zc=1.96|Z| = 1.725 < Z_c = 1.96

It is then concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean μ1\mu_1 is different than μ2\mu_2 at 0.05 significance level.


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