According to a Pew Research Center study, in May 2011, 35% of all American adults had a smart phone (one which the user can use to read emai

Question

According to a Pew Research Center study, in May 2011, 35% of all American adults had a smart phone (one which the user can use to read email and surf the Internet). A communications professor at a university believes this percentage is higher among community college students.

She selects 300 community college students at random and finds that 120 of them have a smart phone. In testing the hypotheses H0: p = 0.35 versus Ha: p > 0.35, she calculates the test statistic as Z = 1.82. Assume the significance level is ? = 0.10.

Which of the following is an appropriate conclusion for the hypothesis test?

There is enough evidence to show that more than 35% of community college students own a smart phone (P?value = 0.034).

There is enough evidence to show that more than 35% of community college students own a smart phone (P?value = 0.068).

There is not enough evidence to show that more than 35% of community college students own a smart phone (P?value = 0.966).

There is not enough evidence to show that more than 35% of community college students own a smart phone (P?value = 0.034).

Does secondhand smoke increase the risk of a low weight birth? A baby is “low birth weight” if it weighs less than 5.5 pounds at birth. According to the National Center of Health Statistics, about 7.8% of all babies born in the U.S. are categorized as low birth weight. Researchers randomly select 1200 babies whose mothers had extensive exposure to secondhand smoke during pregnancy. 10.4% of the sample are categorized as low birth weight.

Answer the following:

Which of the following are the appropriate null and alternative hypotheses for this research question.

H0: p = 0.078; Ha: p ? 0.078

H0: p = 0.078; Ha: p > 0.078

H0: p = 0.104; Ha: p ? 0.104

H0: ? = 0.104; Ha: ? > 0.104

in progress 0
Aaliyah 2 weeks 2021-10-01T16:33:30+00:00 1 Answer 0

Answers ( )

    0
    2021-10-01T16:35:00+00:00

    Answer:

    z=\frac{0.4 -0.35}{\sqrt{\frac{0.35(1-0.35)}{300}}}=1.82  

    p_v =P(z>1.82)=0.034  

    So the p value obtained was a very low value and using the significance level given \alpha=0.1 we have p_v<\alpha so we can conclude that we have enough evidence to reject the null hypothesis

    And the best conclusion would be:

    There is enough evidence to show that more than 35% of community college students own a smart phone (Pvalue = 0.034).

    And for the second case the correct system of hypothesis is:

    H0: p = 0.078; Ha: p > 0.078

    Step-by-step explanation:

    Data given and notation

    n=300 represent the random sample taken

    \hat p=\frac{120}{300}=0.4 estimated proportion of college students that have a smart phone

    p_o=0.35 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)  

    Concepts and formulas to use  

    We need to conduct a hypothesis in order to test the claim that the proportion is >0.35.:  

    Null hypothesis:p\leq 0.35  

    Alternative hypothesis:p > 0.35  

    When we conduct a proportion test we need to use the z statistic, and the is given by:  

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

    The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

    Calculate the statistic  

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

    z=\frac{0.4 -0.35}{\sqrt{\frac{0.35(1-0.35)}{300}}}=1.82  

    Statistical decision  

    It’s important to refresh the p value method or p value approach . “This method is about determining “likely” or “unlikely” by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed”. Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

    The significance level provided \alpha=0.1. The next step would be calculate the p value for this test.  

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

    p_v =P(z>1.82)=0.034  

    So the p value obtained was a very low value and using the significance level given \alpha=0.1 we have p_v<\alpha so we can conclude that we have enough evidence to reject the null hypothesis

    And the best conclusion would be:

    There is enough evidence to show that more than 35% of community college students own a smart phone (Pvalue = 0.034).

    And for the second case the correct system of hypothesis is:

    H0: p = 0.078; Ha: p > 0.078

Leave an answer

45:7+7-4:2-5:5*4+35:2 =? ( )