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2 July, 15:52

In a study of 11,000 car crashes, it was found that 5720 of them occurred within 5 miles of home (based on data from Progressive Insurance). Use a 0.01 significance level to test the claim that more than 50% of car crashes occur within 5 miles of home. Are the results questionable because they are based on a survey sponsored by an insurance company?

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  1. 2 July, 15:57
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    Step-by-step explanation:

    We would set up the hypothesis test.

    a) For the null hypothesis,

    p = 0.5

    For the alternative hypothesis,

    p > 0.5

    Considering the population proportion, probability of success, p = 0.5

    q = probability of failure = 1 - p

    q = 1 - 0.5 = 0.5

    Considering the sample,

    Sample proportion, P = x/n

    Where

    x = number of success = 5720

    n = number of samples = 11000

    P = 5720/11000 = 0.52

    We would determine the test statistic which is the z score

    z = (P - p) / √pq/n

    z = (0.52 - 0.5) / √ (0.5 * 0.5) / 11000 = 4.195

    Using a z test score calculator, the probability value is 0.000014

    Since alpha, 0.01 > than the p value, 0.000014, we would reject the null hypothesis.

    it means that there is sufficient evidence for us to conclude that more than 50% of car crashes occur within 5 miles of home.

    The results are not questionable
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