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coldgirl [10]
3 years ago
9

Kids with cell phones: A marketing manager for a cell phone company claims that less than 58% of children aged 8-12 have cell ph

ones. 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
Mathematics
1 answer:
Vikentia [17]3 years ago
5 0

Answer:

z=\frac{0.54 -0.58}{\sqrt{\frac{0.58(1-0.58)}{820}}}=-2.32  

p_v =P(z  

Since the p value is lower than the significance level \alpha=0.1 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

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

p_o=0.58 is the value that we want to test

\alpha=0.1 represent the significance level

Confidence=90% or 0.90

z would represent the statistic (variable of interest)

p_v 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:p\geq 0.58  

Alternative hypothesis:p < 0.58  

The statistic is given by:

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

And replacing we got:

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

z=\frac{0.54 -0.58}{\sqrt{\frac{0.58(1-0.58)}{820}}}=-2.32  

Decision

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

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

p_v =P(z  

Since the p value is lower than the significance level \alpha=0.1 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

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