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 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

estimated proportion of people that had cell phones

is the value that we want to test

represent the significance level

Confidence=90% or 0.90

z would represent the statistic (variable of interest)

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:  

Alternative hypothesis:  

The statistic is given by:

(1)  

And replacing we got:

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

Decision

The significance level provided . Now we can calculate the p value

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

Since the p value is lower than the significance level 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

an advantage of the standard deviation is that it increases as the dispersion of the data increases. e. the interquartile range is preferred when the data are not skewed or no have outliers. an advantage of the standard deviation is that it uses all the observations in its computation.

a and c

step-by-step explanation: