Question

1)     Recent business graduates currently employed in full-time positions were surveyed. Family backgrounds were self-classified as relatively high or low socioeconomic status. For a random sample of 16 high- socioeconomic status recent business graduates, mean total compensation was $34500 and the sample standard deviation was $8520. For an independent random sample of 9 low-socioeconomic status recent business graduates, mean total compensation was $31,499 and the sample standard deviation was $7,521.

A)   Find a 90% confidence interval for the difference between the two populations means. 

B)    Test at 1% level of significance the means of the two populations are not equal.

Answer 1

A) Assume that the population variances are unequal, so then the number of degrees of freedom is computed as follows:

$$df=\dfrac{(s_1^2/n_1+s_2^2/n_2)^2}{\dfrac{(s_1^2/n_1)^2}{n_1-1}+\dfrac{(s_2^2/n_2)^2}{n_2-1}}=18.56027$$

The critical value for $\alpha=0.1$ and $df=18.56027$ degrees of freedom is $t_{1-\alpha/2, n-1}=1.7312.$

Since we assume that the population variances are unequal, the standard error is computed as follows:

$$se=\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}}$$

$$=\sqrt{\dfrac{(8520)^2}{16}+\dfrac{(7521)^2}{9}}=3289.673$$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x_1}-\bar{x_2}-t_c\times se, \bar{x_1}-\bar{x_2}+t_c\times se)$$

$$=(34500-31499-1.7312\times 3289.673,$$

$$34500-31499+1.7312\times 3289.673)$$

$$=(-2693.424, 8695.424)$$

Therefore, based on the data provided, the 90% confidence interval for the difference between the population means $\mu_1-\mu_2$ is $-2693.424<\mu_1-\mu_2<8695.424,$ which indicates that we are 90%

confident that the true difference between population means is contained by the interval $(-2693.424, 8695.424)$

B) The provided sample means are $\bar{x}_1=34500$ and $\bar{x}_2=31499.$

The provided sample standard deviations are $s_1=8520, s_2=7521.$

The sample sizes are $n_1=16$ and $n_2=9.$

The following null and alternative hypotheses need to be tested:

$H_0: \mu_1=\mu_2$

$H_1: \mu_1\not=\mu_2$

This corresponds to two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

The number of degrees of freedom are $df=18.56027,$ and the significance level is $\alpha=0.01.$

Based on the information provided, the critical value for a two-tailed test is $t_c=2.868.$

The rejection region for this two-tailed test is $R=\{t:|t|>2.868\}$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}_1-\bar{x}_2}{\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}}}=\dfrac{34500-31499}{\sqrt{\dfrac{8520^2}{16}+\dfrac{7521^2}{9}}}=0.912$$

Since it is observed that $t=0.912<2.868=t_c,$ it is then concluded that the null hypothesis is not rejected. Therefore, there is not enough evidence to claim that the population mean $\mu_1$ is different than $\mu_2,$ at the 0.01 significance level.

Using the P-value approach: The p-value $df=18.56027, t=0.912,$ $two-tailed,$ is $p=0.3733,$ and since $p=0.3733\geq0.01,$ it is concluded that the null hypothesis is not rejected. Therefore, there is not enough evidence to claim that the population mean $\mu_1$ is different than $\mu_2$ at the 0.01 significance level.