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

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.

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{((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{((19)(1.5)^(2)+(19)(0.8)^(2))/(38)} = 1.2021` and the observed value is

t =
`(3.2-2.2)/(1.2021√(1/20+1/20)) = (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.