Question

All house sales transactions (in R1000) concluded in two cities during a particular week are
checked. The selling price of each house is noted. The following information is obtained:
Town A Town B
à ƒ °  ‘ ›1 = 51 à ƒ °  ‘ ›2 = 34
à ƒ °  ‘ ¥Ãƒ ƒ Œ …1 = 13585 à ƒ °  ‘ ¥Ãƒ ƒ Œ …2 = 13600
à ƒ °  œ Ž1
2 = 1024 à ƒ °  œ Ž2
2 = 1024
Is it true that, on average the houses in town B are cheaper than the one in town A at 1% level of
significance? Calculate the p-value for this test.

Answer 1

Since the chosen sample sizes $n_1$ and $n_2$ are greater than 30 and the population variances are known, we consider using the Normal distribution to perform our test.

The hypothesis we test is,

$H_0:\mu_1=\mu_2$

$Against$

$H_1:\mu_1\gt\mu_2$

The test statistic is given by the formula,

$Z^*=(\bar{x}_1-\bar{x}_2)/\sqrt{(\sigma_1^2/n_1+\sigma_2^2/n_2)}$

$Z^*=(13585-13600)/\sqrt{1024/51+1024/34}=-15/7.084919=-2.1172(4\space d\space p).$

$Z^*$ is compared with the table value at $\alpha=0.01$ level of significance and this value is given by, $Z^*=2.33$ and reject the null hypothesis if $Z^*\gt Z_\alpha$

Since $Z^*=-2.1172\lt Z_\alpha=2.33$ , we fail to reject the null hypothesis and we conclude that there is no sufficient evidence to show that on average the houses in town B are cheaper than the one in town A at 1% level of significance.

To compute the p-value, we use the formula below.

$p-value=1-\phi(Z^*)$. We use this formula because the test we are performing is an upper-one tailed test.

First, $\phi(Z^*)=\phi(-2.1172)$ is obtained in normal distribution table.

Thus, $\phi(-2.1172)=0.0170$ and the p-value is 1-0.0170=0.983.