. In a survey of 1005 adult Americans, 46% indicated that they were somewhat interested or very interested [(5)] in having web access in the

Question

. In a survey of 1005 adult Americans, 46% indicated that they were somewhat interested or very interested [(5)] in having web access in their cars. Suppose that the marketing manager of a car manufacturer claims that the 46% is based only on a sample and that 46% is close to half, so there is no reason to believe that the proportion of all adult Americans who want car web access is less than 0.50. Is the marketing manager correct in his claim

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Emery 4 days 2021-11-24T04:38:22+00:00 1 Answer 0 views 0

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    2021-11-24T04:39:43+00:00

    Answer:

    We conclude that the the proportion of all adult Americans who want car web access is less than 0.50.

    This means that the marketing manager was not correct in his claim.

    Step-by-step explanation:

    We are given that in a survey of 1005 adult Americans, 46% indicated that they were somewhat interested or very interested [(5)] in having web access in their cars.

    Let p = proportion of all adult Americans who want car web access.

    So, Null Hypothesis, H_0 : p \geq 0.50     {means that the proportion of all adult Americans who want car web access is more than or equal to 0.50}

    Alternate Hypothesis, H_A : p < 0.50    {means that the proportion of all adult Americans who want car web access is less than 0.50}

    The test statistics that would be used here One-sample z proportion statistics;

                             T.S. =  \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

    where, \hat p  = sample proportion of Americans who are interested or very interested in having web access in their cars = 46%

                n = sample of adult Americans = 1005

    So, test statistics  =  \frac{0.46-0.50}{\sqrt{\frac{0.46(1-0.46)}{1005} } }  

                                    =  -2.544

    The value of z test statistics is -2.544.

    Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the z table gives critical value of -1.645 for left-tailed test. Since our test statistics is less than the critical value of z, so we have sufficient evidence to reject our null hypothesis as it will in the rejection region due to which we reject our null hypothesis.

    Therefore, we conclude that the the proportion of all adult Americans who want car web access is less than 0.50.

    This means that the marketing manager was not correct in his claim.

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