Question

Marketing companies have collected data implying that teenage girls use more ring tones on their cell phones than teenage boys do. In a random sample of 20 teenage girls, the mean number of ring tones was 3.2 with a standard deviation of 1.5. In a random sample of 20 teenage boys, the mean was 2.2 ring tones, with a standard deviation of 0.8. Conduct a hypothesis test to determine if the girls’ mean is higher than the boys’ mean.

Answer 1

Answer:

The p-value here is 0.0061, which is very small and we have evidence that the girls' mean is higher than the boys' mean.

Step-by-step explanation:

We suppose that the two samples are independent and normally distributed with equal variances. Let $\mu_{1}$ be the mean number of ring tones for girls, and $\mu_{2}$ the mean number of ring tones for boys.

We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} > 0$ (upper-tail alternative).

The test statistic is

T = $\frac{\bar{X}_{1}-\bar{X}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}}$

where

$S_{p} = \sqrt{\frac{(n_{1}-1)S_{1}^{2}+(n_{2}-1)S_{2}^{2}}{n_{1}+n_{2}-2}}$.

For this case, $n_{1}=n_{2}=20$, $\bar{x}_{1}=3.2$, $s_{1}=1.5$, $\bar{x}_{2}=2.2$, $s_{2}=0.8$.

$s_{p} = \sqrt{\frac{(19)(1.5)^{2}+(19)(0.8)^{2}}{38}} = 1.2021$ and the observed value is

t = $\frac{3.2-2.2}{1.2021\sqrt{1/20+1/20}} = \frac{1}{0.3801} = 2.6309$.

We can compute the p-value as P(T > 2.6309) where T has a t distribution with 20 + 20 - 2 = 38 degrees of freedom, so, the p-value is 0.0061. Because the p-value is very small, we can reject the null hypothesis for instance, at the significance level of 0.05.