Answer to Question #264170 in Statistics and Probability for Nathan

Question

The daily revenue of Modern Style Hair Salon is approximately normally distributed. Over a random sample of 20 days, the standard deviation of daily revenue is R1,329. At the 5% significance level, we can conclude that the data provide sufficient evidence that the average daily revenue is greater than R6,831 if the total revenue generated over the 20 days is greater than (Rands only, no cents)

Answer 1

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

$H_{0}: \mu \leq 6831\\ H_{a}: \mu>6831$

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

(2) Rejection Region

Based on the information provided, the significance level is $\alpha=0.05$ , and the critical value for a right-tailed test is $t_{c}=1.729.$

The rejection region for this right-tailed test is $R=\{t: t>1.729\}$

(3) Test Statistics

The t-statistic is computed as follows:

$t=\frac{\bar{X}-\mu_{0}}{s / \sqrt{n}}$

Since, we can conclude that the data provide sufficient evidence that the average daily revenue is greater than R6831. Then t > 1.729

$t= (\bar X- 6831) /(1329/\sqrt{20}) >1.729$

Now, $\bar X > 7344.8$

Then $\bar X >7345$ in Rands

Hence, total revenue generated over the 20 days is greater than = 20×7345=146900