The Pew Research Center reported that 73% of Americans who own a cell phone also use text messaging. In a recent local survey, 155 out of 20

Question

The Pew Research Center reported that 73% of Americans who own a cell phone also use text messaging. In a recent local survey, 155 out of 200 cell phone owners used text messaging.
Since a Z test is appropriate, test whether the population proportion of Americans who use text messaging is different from 73%. Use level of significance α = 0.10.
Hint: Do you need to conduct a t-test or a z-test? Next, find the p-value, using p-value, and level of significance, you can see if the decision (Reject or Do Not reject H0.) You can also find the critical value(s) to finalize your decision.

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Luna 1 week 2021-10-03T20:22:00+00:00 1 Answer 0

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    2021-10-03T20:23:30+00:00

    Answer:

    z=\frac{0.775 -0.73}{\sqrt{\frac{0.73(1-0.73)}{200}}}=1.433  

    Now we can find the p value. Since we have a bilateral test the p value would be:  

    p_v =2*P(z>1.433)=0.152  

    Since the p value is higher than the significance level of 0.1 we have enough evidence to FAIL to reject the null hypothesis and the best conclusion for this case would be:

    Do Not reject H0

    Step-by-step explanation:

    Information provided

    n=200 represent the sample size slected

    X=155 represent the cell phone owners used text messaging

    \hat p=\frac{155}{200}=0.775 estimated proportion of cell phone owners used text messaging

    p_o=0.73 is the value to verify

    \alpha=0.1 represent the significance level

    We need to conduct a z test for a proportion

    z would represent the statistic

    p_v represent the p value

    System of hypothesis

    We want to verify if the true proportion of cell phone owners used text messaging is different from 0.73 so then the system of hypothesis are:

    Null hypothesis:p=0.73  

    Alternative hypothesis:p \neq 0.73  

    The statistic to check this hypothesis is given by:

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

    Replacing the data given we got:

    z=\frac{0.775 -0.73}{\sqrt{\frac{0.73(1-0.73)}{200}}}=1.433  

    Now we can find the p value. Since we have a bilateral test the p value would be:  

    p_v =2*P(z>1.433)=0.152  

    Since the p value is higher than the significance level of 0.1 we have enough evidence to FAIL to reject the null hypothesis and the best conclusion for this case would be:

    Do Not reject H0

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