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: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 [tex]\mu_{1}[/tex] be the mean number of ring tones for girls, and [tex]\mu_{2}[/tex] the mean number of ring tones for boys. We want to test [tex]H_{0}: \mu_{1}-\mu_{2} = 0[/tex] vs [tex]H_{1}: \mu_{1}-\mu_{2} > 0[/tex] (upper-tail alternative). The test statistic is T = [tex]\frac{\bar{X}_{1}-\bar{X}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}}[/tex] where [tex]S_{p} = \sqrt{\frac{(n_{1}-1)S_{1}^{2}+(n_{2}-1)S_{2}^{2}}{n_{1}+n_{2}-2}}[/tex]. For this case, [tex]n_{1}=n_{2}=20[/tex], [tex]\bar{x}_{1}=3.2[/tex], [tex]s_{1}=1.5[/tex], [tex]\bar{x}_{2}=2.2[/tex], [tex]s_{2}=0.8[/tex]. [tex]s_{p} = \sqrt{\frac{(19)(1.5)^{2}+(19)(0.8)^{2}}{38}} = 1.2021[/tex] and the observed value ist = [tex]\frac{3.2-2.2}{1.2021\sqrt{1/20+1/20}} = \frac{1}{0.3801} = 2.6309[/tex]. 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.