Answer to Question #237561 in Statistics and Probability for suvi

Question

The operations manager of a motor vehicle accessories firm with branches in Limpopo and Durban
wants to establish whether their Limpopo branch is performing better than their Durban branch
in terms of the average size of orders received. Both branches have been operating for only one
year. A random sample of 31 orders from the Limpopo branch had an average value of R355.20
with a standard deviation equal to R151.50. Orders received by the Durban branch of the firm
were also randomly sampled. The average of the 35 orders sampled was R265.60 with a sample
standard deviation of R162.20. Can the operations manager conclude, at the 3% significance level,
that the Limpopo branch is performing better than the Durban branch?
Show the null and alternative hypotheses for the test and draw the appropriate conclusion.
Compute the p-value.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0: \mu_1=\mu_2$

$H_1: \mu_1>\mu_2$

Or

$H_0: \mu_1\leq\mu_2$

$H_1: \mu_1>\mu_2$

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.

The degrees of freedom are computed as follows, assuming that the population variances are equal:

$$df_{total}=df_1+df_2=31-1+35-1=64$$

The critical value for this right-tailed test is $t_c= 1.915,$ for $\alpha=0.03$ and $df=64.$

The rejection region for this right-tailed test is $R=\{t:t>1.915\}$

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

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

$$=\dfrac{355.20-265.60}{\sqrt{\dfrac{(31-1)151.50^2+(35-1)162.20^2}{31+35-2}(\dfrac{1}{31}+\dfrac{1}{35})}}$$ $$=2.31$$

Since it is observed that $t=2.31>1.915=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value for right-tailed, $\alpha=0.03, df=64,t=2.31$ is $p=0.012062,$ and since $p=P(T>2.31)=0.012062<0.03=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu_1$ is greater than $\mu_2,$ at the $\alpha=0.03$ significance level.

The degrees of freedom are computed as follows, assuming that the population variances are unequal:

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

$$=63.806678180021$$

The critical value for this right-tailed test is $t_c= 1.915,$ for $\alpha=0.03$ and $df=63.806678180021.$

The rejection region for this right-tailed test is $R=\{t:t>1.915\}.$

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

$$t=\dfrac{\bar{X_1}-\bar{X_2}}{\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}}}=\dfrac{355.20-265.60}{\sqrt{\dfrac{151.50^2}{31}+\dfrac{162.20^2}{35}}}=2.32$$

Since it is observed that $t=2.32>1.915=t_c,$ it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value for right-tailed, $\alpha=0.03,$ $df=63.806678180021,t=2.32$ is $p=P(T>2.32)=0.011771,$ and since $p=0.011771<0.03=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu_1$ is greater than $\mu_2,$ at the $\alpha=0.03$ significance level.