Answer to Question #150165 in Statistics and Probability for Srihari

Question

4. A buyer of electric bulbs purchases 400 bulbs; 200 bulbs of each brand. Upon testing these bulbs be found that brand A has an average of 1225 hrs with a S.D of 42 hrs.

Whereas brand B had a mean life of 1265 hrs with a S.D of 60hrs. Can the buyer be certain that brands B is superior than brand A in equality

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu_1\leq\mu_2$

$H_1:\mu_1>\mu_2$

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

We have two different brands. Then we assume that the population variances are unequal, so then the number of degrees of freedom is computed as follows:

$$df=\dfrac{(s_1^2/n_1+s_2^2/n_2)^2}{\dfrac{(s_1^2/n_1)^2}{n_1+1}+\dfrac{(s_2^2/n_2)^2}{n_2+1}}=359.842$$

It is found that the critical value for this right-tailed test is $t_c=1.6491,$ for $\alpha=0.05$

and $df=359.842.$  

The rejection region for this right-tailed test is $R=\{t:t>1.6491\}.$

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

$$t=\dfrac{\bar{X_1}-\bar{X_2}}{\sqrt{\dfrac{s_1 ^2}{n_1}+\dfrac{s_2^2}{n_2}}}=\dfrac{1265-1225}{\sqrt{\dfrac{60 ^2}{200}+\dfrac{42^2}{200}}}=7.7238$$

Since it is observed that $t=7.7238>1.6491=t_c,$ it is then concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that population mean $\mu_1$ is greater than $\mu_2,$ at the 0.05 significance level.

Using the P-value approach: The p-value for the right-tailed test with $df=359.842$ is $p\approx0,$ and since $p=0<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that population mean $\mu_1$ is greater than $\mu_2,$ at the 0.05 significance level.

Therefore, there is enough evidence to claim that brands B is superior than brand A in equality at the 0.05 significance level.