Question

A screening procedure was designed to measure attitudes towards minorities as managers. High scores indicate negative attitudes and low scores indicate positive attitudes. Independent random samples were taken of 151 male financial analysts and 108 female analysts. For the former group, the sample mean and standard deviation scores were 85.8 and 19.13, whereas the corresponding statistics for the latter group were 71.5 and 12.2. Test the null hypothesis that the two population means are equal against the alternative that the true mean score is higher for male than for female financial analysts.

Answer 1

The summary of t Male Female 85.871.5 mean std. 19.1312.2dev 151 108 n The pooled standard deviation is: n+2-2 (151-1)(19.13)+(108-1)(1222) 151+108-2 16.6001 ##Since the ratio ofstandard deviations is less than 2, so we use two sample (pooled t-test)# Null hypothesis, Ho -2 Alternative hypothesis, H1:4>μ. [one tailed test]

Significance level: a 0.05 Test statistic: 85.8-71.5 16.6001, 151 108 6.84 The degrees of freedom is (n, +n, -2) 151 +108-2 257 1 of significance at Critical value: Using the t-table, the critical value at 5% leve df257 for a right tailed test is 1.651. Compare the test statistic value with the critical value: Since the calculated value of the test statistic is greater 51 so we reject H than critical value.ie6.84 >1.6 Conclusion:Hence,there is sufficient evidenceto support the claimthat the true mean score is higher for male than for female financial analysts.