Question

Marketing companies have collected data implying that teenage girls use more ring tones on their cellular phones than

teenage boys do. In one particular study of 40 randomly chosen teenage girls and boys (20 of each) with cellular phones,

the mean number of ring tones for the girls was 3.2 with a standard deviation of 1.5. The mean for the boys was 1.7 with

a standard deviation of 0.8. Conduct a hypothesis test to determine if the means are approximately the same or if the girls’

mean is higher than the boys’ mean.

Short answer

If we take $\alpha =0.05,$we can see that $\alpha >\rho -value.$this means we reject the null hypothesis. Thus, at the 5% significance level, there is sufficient evidence to conclude that the mean number of ringtones for girls is higher than that for boys.

Step-by-step solution

Given Information

The data is given by:

we need to test these hypotheses:

$H_{0:{\mu }_{G={\mu }_{\beta }},H_{1:{\mu }_G>{\mu }_{\beta }}}$

$\overline{x_{1-\overline{X_2}}}$is the difference between the mean number of ringtones for girls and boys.

We are using Student t-test for testing our hypothesis.

 Explanation 

$SE=\sqrt{\frac{{s_1}^2}{n_1}}+\frac{{s_2}^2}{n_2}=\sqrt{0.1445}=0.38$.

The test statistic is given by:

${\mu }_{1\frac{{\mu }_{1-{\mu }_2}}{SE}=3·9474}$

Finding the number degrees of freedom :

df=$df=SE=\frac{S{E}^4}{{\displaystyle \frac{s{1}^4}{n{1}^2(n_1-1)}}+{\displaystyle \frac{{s_2}^2}{{n_2}^2(n_2-1)}}}=\frac{S{E}^4}{0.00072}=28.96$

Using the above formula we get$p-value$is$0.00023$

Graphical representation 

following are the graph