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12 August, 11:18

A local cable company claims that the proportion of people who have Internet access is less than 63%. To test this claim, a random sample of 800 people is taken and its determined that 478 people have Internet access. The following is the setup for this hypothesis test: H0:p=0.63 Ha:p<0.63 Find the p-value for this hypothesis test for a proportion and round your answer to 3 decimal places.

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  1. 12 August, 13:32
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    Step-by-step explanation:

    For the null hypothesis,

    H0 : p = 0.63

    For the alternative hypothesis,

    Ha : p < 0.63

    This is a left tailed test

    Considering the population proportion, probability of success, p = 0.63

    q = probability of failure = 1 - p

    q = 1 - 0.63 = 0.37

    Considering the sample,

    Sample proportion, P = x/n

    Where

    x = number of success = 478

    n = number of samples = 800

    P = 478/800 = 0.6

    We would determine the test statistic which is the z score

    z = (P - p) / √pq/n

    z = (0.6 - 0.63) / √ (0.63 * 0.37) / 800 = - 1.76

    From the normal distribution table, the area below the test z score in the left tail 0.039

    Thus

    p = 0.039
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