A poll asked the following question: “If the military draft were reinstated, would you favor or oppose drafting women as well as men?” 45 pe

Question

A poll asked the following question: “If the military draft were reinstated, would you favor or oppose drafting women as well as men?” 45 percent of the 1000 people responding said that they would favor drafting women if the draft were reinstated. Using a 0.05 significance level, carry out a test to determine if there is convincing evidence that fewer than half of adult Americans would favor the drafting of women. (For z give the answer to two decimal places. For P give the answer to four decimal places.)

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Sarah 2 weeks 2021-10-08T01:12:20+00:00 1 Answer 0

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    2021-10-08T01:13:24+00:00

    Answer:

    Null hypothesis: H0 = 0.50

    Alternative hypothesis: Ha < 0.50

    z = -3.16

    P value = P(Z<-3.16) =  0.0008

    Decision we reject the null hypothesis and accept the alternative hypothesis. That is, there is convincing evidence that fewer than half of adult Americans would favor the drafting of women.

    Rule

    If;

    P-value > significance level  — accept Null hypothesis

    P-value < significance level  — reject Null hypothesis

    Z score > Z(at 95% confidence interval) —- reject Null hypothesis

    Z score < Z(at 95% confidence interval) —— accept Null hypothesis

    Step-by-step explanation:

    Given;

    n=1000 represent the random sample taken

    Null hypothesis: H0 = 0.50

    Alternative hypothesis: Ha < 0.50

    Test statistic z score can be calculated with the formula below;

    z = (p^−po)/√{po(1−po)/n}

    Where,

    z= Test statistics

    n = Sample size = 1000

    po = Null hypothesized value = 0.50

    p^ = Observed proportion = 0.45

    Substituting the values we have

    z = (0.45-0.50)/√{0.50(1-0.50)/1000}

    z = -3.16

    z = -3.16

    To determine the p value (test statistic) at 0.05 significance level, using a one tailed hypothesis.

    P value = P(Z<-3.16) =  0.0008

    Since z at 0.05 significance level is between -1.96 and +1.96 and the z score for the test (z = -3.16) which doesn’t falls with the region bounded by Z at 0.05 significance level. And also the one-tailed hypothesis P-value is 0.0008 which is lower than 0.05. Then we can conclude that we have enough evidence to FAIL or reject the null hypothesis, and we can say that at 5% significance level the null hypothesis is invalid, therefore we accept the alternative hypothesis.

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