Answer to Question #147045 in Statistics and Probability for Yashasvi

Question

the average number of articles produced by two machines per day are 200 and 250 with standard deviation of 20 and 25 respectively on the basis of 25 days production. Can you regard both the machines equally efficient at 1% level of significance?Also construct 99% confidence limit

Answer 1

The provided sample means are shown below: $\bar{x}_1=200, \bar{x}_2=250.$

Also, the provided sample standard deviations are: $s_1=20, s_2=25,$

and the sample sizes are $n_1=25$  and $n_2=25.$

Based on the information provided, we assume that the population variances are unequal, so then the number of degrees of freedom is computed as follows:

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

$$=\dfrac{(20^2/25+25^2/25)^2}{\dfrac{(20^2/25)^2}{25-1}+\dfrac{(25^2/25)^2}{25-1}}\approx46$$

The critical value for $\alpha=0.01$ and $df=46$ degrees of freedom is $t_c=t_{1-\alpha/2, df}=2.687014$

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

$$se=\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}}=\sqrt{\dfrac{20^2}{25}+\dfrac{25^2}{25}}\approx6.403$$

 Compute the confidence interval:

$$CI=(\bar{x}_1-\bar{x}_2-t_c\times se, \bar{x}_1-\bar{x}_2+t_c\times se)$$

$$=(200-250-2.687\times 6.403, 200-250+2.687\times 6.403)$$

$$\approx(-67.205, -32.795)$$

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.

The rejection region for this two-tailed test is $R=\{t:|t|>2.687014\}$

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

$$se=\dfrac{\bar{x}_1-\bar{x}_2}{\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}}}$$

$$=\dfrac{200-250}{\sqrt{\dfrac{20^2}{25}+\dfrac{25^2}{25}}}\approx-7.808688$$

Since it is observed that $|t|=7.808688>2.687014,$ it is then concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that population mean $\mu_1$ is different than $\mu_2,$ at the 0.01 significance level.

Using the P-value approach: The p-value is $p=0,$ and since $p=0<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that population mean $\mu_1$ is different than $\mu_2,$ at the 0.01 significance level.