Answer to Question #345277 in Statistics and Probability for Kheti

Question

Frequent checks on the spending patterns of tourists returning from countries in Asia were made and the average amount spent by all tourists was found to be R1010 per day. To determine whether there has been a change in the average amount spent, a sample of 70 travelers was selected and the mean was determined as R1090 per day with a standard deviation of R300. A researcher wants to test for evidence of a change in the expenditure amount changed, using 0.05 level of significance. What is the appropriate distribution to use? Motivate your answer

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=1010$

$H_1:\mu\not=1010$

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is $\alpha = 0.05,$ $df=n-1=69$ and the critical value for a two-tailed test is $t_c =1.994945.$

The rejection region for this two-tailed test is $R = \{t:|t|>1.994945\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{1090-1010}{300/\sqrt{70}}=2.2311$$

Since it is observed that $|t|=2.2311>1.994945=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed, $df=69$ degrees of freedom, $t=2.2311$ is $p=0.028927,$ and since $p=0.028927<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$

is different than 1010, at the $\alpha = 0.05$ significance level.