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.

Where as 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.

Since the brand A and brand B are the different populations, 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.