Answer to Question #217699 in Statistics and Probability for Reine

Question

A cosmetics company claims that teenage ladies consume an average of P 150 for cosmetic products. In order to verify this claim, a researcher conducted a survey to 40 teenage ladies and found that their mean monthly cosmetic products is P 146 with a standard deviation of P12. Using a=0.05, test whether the companies claim a realistic or not?

Answer 1

$n=40 \\ \bar{x}=146 \\ s=12 \\ H_0: \mu=150 \\ H_1: \mu ≠150$

When dealing with a large sample of size n >30 from a population which need not be normal but has finite variance, we can use the central limit theorem to justify using the test for normal populations. Even when $\sigma^2$ is unknown we can approximate its value with $s^2$ in the computation of the test statistic.

Test-statistic:

$Z= \frac{\bar{x}- \mu}{s / \sqrt{n}} \\ Z= \frac{146-150}{12 / \sqrt{40}}= -2.1 \\ α=0.05$

Two-tailed test

$Z_{crit}=1.96$

The decision rule is: Reject H₀ if Z ≤ -1.96 or if Z ≥ 1.96.

$Z= -2.1 < Z_{crit} = -1.96$

Reject H₀.

There is enough evidence to conclude that the companies claim is NOT realistic at the 0.05 significance level.