Question

Two types of chemical solutions treams, A and B, were tested for their alkalinity levels. Analysis of the 6 samples taken from solution A showed that the mean alkalinity level is 7.52 with a standard deviation of 0.024 while the 5 samples taken at Solution B showed an alkalinity level of 7.49 with a standard deviation of 0.032. Using a (a) 0.05 and (b) 0.01 significance level, determine whether the streams have different alkalinity levels. Use F-distribution.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\sigma_1^2=\sigma_2^2$

$H_1:\sigma_1^2\not=\sigma_2^2$

This corresponds to a two-tailed test, for which a F-test for two population variances needs to be used.

Based on the information provided, the significance level is $\alpha=0.05,$ and the the rejection region for this two-tailed test is $R=\{F:F<0.135\ or\ F>9.364\}$.

The F-statistic is computed as follows:

$$F={s_1^2 \over s_2^2}={0.000576 \over 0.001024}=0.5625$$

Since from the sample information we get that  $F_L=0.135<F=0.5625<9.364=F_U,$it is then concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population variance $\sigma_1^2$ is different than the population variance $\sigma_2^2,$ at the $\alpha=0.05$ significance level.

Based on the information provided, the significance level is $\alpha=0.01,$ and the the rejection region for this two-tailed test is $R=\{F:F<0.064\ or\ F>22.456\}.$

The F-statistic is computed as follows:

$$F={s_1^2 \over s_2^2}={0.000576 \over 0.001024}=0.5625$$

Since from the sample information we get that $F_L=0.064<F=0.5625<22.456=F_U,$ it is then concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population variance $\sigma_1^2$ s different than the population variance $\sigma_2^2,$ at the $\alpha=0.01$ significance level.

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

Based on the information provided, the significance level is $\alpha=0.05,$ and the degrees of freedom are $df=6+5-2=9.$ In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal.

Hence, it is found that the critical value for this two-tailed test is $t_c=2.262,$ for $\alpha=0.05$ and $df=9.$

The rejection region for this two-tailed test is $R=\{t:|t|>2.262\}$

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

$$t={\bar{X_1}-\bar{X_2}\over \sqrt{\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}(\dfrac{1}{n_1}+\dfrac{1}{n_2})}}=$$

$$={7.52-7.49\over \sqrt{\dfrac{(6-1)(0.024)^2+(5-1)(0.032)^2}{6+5-2}(\dfrac{1}{6}+\dfrac{1}{5})}}=$$

$$=1.78$$

Since it is observed that $|t|=1.78<2.262=t_c,$ 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 0.05 significance level.

Using the P-value approach: The p-value is $p=0.1089,$ and since $p=0.1089>0.05,$ 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 0.05 significance level.

b) Based on the information provided, the significance level is $\alpha=0.01,$ and the degrees of freedom are  $df=6+5-2=9.$ In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal.

Hence, it is found that the critical value for this two-tailed test is $t_c=3.25,$ for $\alpha=0.01$ and $df=9.$

The rejection region for this two-tailed test is $R=\{t:|t|>3.25\}$

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

$$t={\bar{X_1}-\bar{X_2}\over \sqrt{\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}(\dfrac{1}{n_1}+\dfrac{1}{n_2})}}=$$

$$={7.52-7.49\over \sqrt{\dfrac{(6-1)(0.024)^2+(5-1)(0.032)^2}{6+5-2}(\dfrac{1}{6}+\dfrac{1}{5})}}=$$

$$=1.78$$

Since it is observed that $|t|=1.78<3.35=t_c,$ 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 0.05 significance level.

Using the P-value approach: The p-value is $p=0.1089,$ and since $p=0.1089>0.01,$ 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 0.05 significance level.