Answer to Question #170152 in Statistics and Probability for PIP

Question

1. For a random sample of 75 British entrepreneurs, the mean number of job changes was 1.92 and the sample standard deviation was 1.41. For an independent random sample of 115 British corporate managers, the mean number of job changes was 0.71 and the sample standard deviation was 0.68. Test the null hypothesis that the population means are equal against the alternative that the mean number of job changes is higher for British entrepreneurs than for British corporate managers. Are your findings significant, if so, at what level?

Answer 1

$μ_1$ = population mean of British entrepreneurs

$μ_2$ = population mean of British corporate managers

$H_0 : μ_1 – μ_2 = 0 \\ H_1 : μ_1 – μ_2 > 0$

Sample size for British entrepreneurs $n_1 = 75$

Sample size for British corporate managers $n_2 = 115$

Sample mean for British entrepreneurs $\bar{x} = 1.92$

Sample mean for British corporate managers $\bar{y} = 0.71$

Sample standard deviation for British entrepreneurs $s_1 = 1.41$

Sample standard deviation for British corporate managers $s_2 = 0.68$

We have taken α=0.05

We can use the following decision rule

reject $H_0$ if

$\frac{\bar{x}-\bar{y}}{\sqrt{ \frac{s^2_p}{n_x} + \frac{s^2_p}{n_y} }}>t_{n_x+n_y,-2,α} \\ \frac{1.92-0.71}{\sqrt{ \frac{(1.41)^2}{75} + \frac{(0.68)^2}{115} }}=6.92 \\ 6.92>t_{190,-2,α}$

Reject $H_0$