Question

1. John has obtained two independent samples form two populations, where the sample statistics are shown in the table below. Assuming equal variances, he can construct a 95% confidence interval for the difference of the population means to be

	Sample 1	Sample 2
Mean	22.7	20.5
Variances (s^2)	5.4	3.6
Observations (sample size)	9	9

1. [0.08, 4.32]
2. [1.17,5.08]
3. [2.44,6.19]
4. [-0.09,3.19]

2. True or False? A new policy of “flex hours” is proposed. Random Sampling showed that 40 of 50 female workers favored the change, while 25 of 50 male workers favored the change. It seems the approval rate among female workers are greater than males. By a right-tailed test with alpha= 0.025 where H0: pie1-pie2 ≤0.1 and H1:pie1-pie2 greater than 0.1, we can reject the null hypothesis.

Answer 1

We have the following sample information

$\begin{align*} &\bar{x}_1=22.7\quad\text{sample mean of sample 1}\\ &s^2_1=5.4\quad\text{sample variance of sample 1}\\ &n_1=9\quad\text{sample size of sample 1}\\ &\bar{x}_2=20.5\quad\text{sample mean of sample 2}\\ &s^2_2=3.6\quad\text{sample variance of sample 2}\\ &n_2=9\quad\text{sample size of sample 2}\\ \end{align*}$

Since the sample sizes of 2 samples are less than 30, we will do a small sample analysis. We will use t distribution for the test statistics.

Since the population variance are equal, we will get an estimate of the variance using

$\begin{align*} s^2_p=\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}=\frac{(9-1)5.4+(9-1)3.6}{9+9-2}=4.5 \end{align*}$

The estimated standard error of the difference between sample means is

$\begin{align*} \hat{\sigma}_{\bar{x}_1-\bar{x}_2}=\sqrt{s^2_p\left(\frac{1}{n_1}+\frac{1}{n_2} \right )}=\sqrt{4.5\times \left(\frac{1}{9}+\frac{1}{9} \right )}=1 \end{align*}$

95% confidence interval corresponds to $\begin{align*} \alpha=1-95/100=0.05 \end{align*}$ .

The critical value of t is $\begin{align*} P(T>t_{\alpha/2})=\alpha/2=0.05/2=0.025 \end{align*}$ . The degrees of freedom is  $\begin{align*} n_1+n_2-2=9+9-2=16 \end{align*}$

Using the t tables we can get the the critical value of t as $\begin{align*}t_{\alpha/2}=2.120 \end{align*}$

Now the 95% confidence interval for the difference in population means is

$\begin{align*}&(\bar{x}_1-\bar{x}_2)\pm t_{\alpha/2}\hat{\sigma}_{\bar{x}_1-\bar{x}_2}=\\ \implies &(22.7-20.5)\pm 2.120\times 1\\ \implies &[0.08, 4.32] \end{align*}$

ans: a [0.08, 4.32]

True or False

Let $\begin{align*}\pi_1 \end{align*}$ be the true proportion of female workers favoring the change and $\begin{align*}\pi_2 \end{align*}$ be the true proportion of male workers favoring the change

The hypotheses are

$\begin{align*}&H_0:\pi_1-\pi_2\le 0.1\leftarrow\text{null hypothesis}\\ &H_a:\pi_1-\pi_2> 0.1\leftarrow\text{alternative hypothesis}\\ &\alpha=0.025\leftarrow\text{level of significance to test the hypotheses} \end{align*}$

The sample information

$\begin{align*}&\bar{\pi}_1=\frac{40}{50}=0.8\quad\text{sample approval rating among female workers}\\ &\bar{\pi}_2=\frac{25}{50}=0.5\quad\text{sample approval rating among male workers}\\ \end{align*}$

The estimated overall approval rating is

$\begin{align*}\hat{\pi}=\frac{40+25}{50+50}=0.65 \end{align*}$

The standard error of the difference in proportions is

$\begin{align*}\hat{\sigma}_{\bar{\pi}_1-\bar{\pi}_2}=\sqrt{\frac{\hat{\pi}(1-\hat{\pi})}{n_1}+\frac{\hat{\pi}(1-\hat{\pi})}{n_2}}=\sqrt{\frac{0.65(1-0.65)}{50}+\frac{0.65(1-0.65)}{50}}=0.0954 \end{align*}$

the hypothesized value of the difference in 0.1 (from the null hypothesis). That is

$\begin{align*}({\pi}_1-{\pi}_2)_{H_0}=0.1 \end{align*}$

The test statistics is

$\begin{align*}z=\frac{(\bar{\pi}_1-\bar{\pi}_2)-({\pi}_1-{\pi}_2)_{H_0}}{\hat{\sigma}_{\bar{\pi}_1-\bar{\pi}_2}}=\frac{(0.8-0.5)-0.1}{0.0954}=2.10 \end{align*}$

This is a right tailed test (the alternative hypothesis has ">")

The critical value corresponding to alpha=0.025 is $\begin{align*}P(Z>z_{\alpha})=0.025\implies P(Z<z_{\alpha})=1-0.025=0.975 \end{align*}$

Using the standard normal tables we get for z=1.96, P(Z<1.96) = 0.5+0.4750=0.9750. Hence the critical value is 1.96

We will reject the null hypothesis if the test statistics is greater than the critical value corresponding to alpha=0.025.

Here the test statistics is 2.10 and it is greater than the critical value of 1.96.

Hence we will reject the null hypothesis. We can conclude that there is sufficient evidence to support the  claim that  the approval rate among female workers are greater than males by more than 0.1.

ans:True