Kids with cell phones: A marketing manager for a cell phone company claims that less than 58% of children aged 8-12 have cell phones. In a survey of 820 children aged 8-12 by a national consumers group, 443 of them had cell phones. Can you conclude that the manager's claim is true? Use the =α0.10 level of significance and the P-value method with the TI-84 Plus calculator. Part: 0 / 40 of 4 Parts Complete

Since the p value is lower than the significance level [tex] \alpha=0.1[/tex] we have enough evidence to reject the null hypothesis and the claim for the manager makes sense.

For the Ti84 preocedure we need to do this:

STAT> TESTS> 1-Z prop Test

And then we need to input the following values:

po= 0.58

x = 443 , n= 820

prop <po

And then calculate and we will get the same results

Step-by-step explanation:

Data given and notation

n=820 represent the random sample taken

X=443 represent the people that had cell phones

[tex]\hat p=\frac{443}{820}=0.540[/tex] estimated proportion of people that had cell phones

[tex]p_o=0.58[/tex] is the value that we want to test

[tex]\alpha=0.1[/tex] represent the significance level

Confidence=90% or 0.90

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value (variable of interest)

System of hypothesis

We need to conduct a hypothesis in order to test the claim that the true proportion is less than 0.58.:

Null hypothesis:[tex]p\geq 0.58[/tex]

Alternative hypothesis:[tex]p < 0.58[/tex]

The statistic is given by:

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)

And replacing we got:

Since we have all the info requires we can replace in formula (1) like this:

[tex]z=\frac{0.54 -0.58}{\sqrt{\frac{0.58(1-0.58)}{820}}}=-2.32[/tex]

Decision

The significance level provided [tex]\alpha=0.1[/tex]. Now we can calculate the p value

Since is a left tailed test the p value would be:

[tex]p_v =P(z<-2.32)=0.0102[/tex]