Since an instant replay system for tennis was introduced at a major​ tournament, men challenged 1406 referee​ calls, with the result that 41

Question

Since an instant replay system for tennis was introduced at a major​ tournament, men challenged 1406 referee​ calls, with the result that 416 of the calls were overturned. Women challenged 778 referee​ calls, and 217 of the calls were overturned. Use a 0.05 significance level to test the claim that men and women have equal success in challenging calls. Complete parts​ (a) through​ (c) below. A. Test the claim using a hypothesis test. Consider the first sample to be the sample of male tennis players who challenged referee calls and the second sample to be the sample of female tennis players who challenged referee calls. What are the null and alternative hypotheses for the hypothesis​ test?

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Aaliyah 2 weeks 2021-11-18T05:14:45+00:00 1 Answer 0 views 0

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    2021-11-18T05:15:59+00:00

    Answer:

    Step-by-step explanation:

    This is a test of 2 population proportions. Let 1 and 2 be the subscript for men and women players. The population proportions of men and women challenges for calls would be p1 and p2

    P1 – P2 = difference in the proportion of men and women challenges for calls.

    The null hypothesis is

    H0 : p1 = p2

    p1 – p2 = 0

    The alternative hypothesis is

    Ha : p1 ≠ p2

    p1 – p2 ≠ 0

    it is a two tailed test

    Sample proportion = x/n

    Where

    x represents number of success(number of complaints)

    n represents number of samples

    For old dough

    x1 = 416

    n1 = 1406

    P1 = 416/1406 = 0.3

    For new dough,

    x2 = 217

    n2 = 778

    P2 = 217/778 = 0.28

    The pooled proportion, pc is

    pc = (x1 + x2)/(n1 + n2)

    pc = (416 + 217)/(1406 + 778) = 0.29

    1 – pc = 1 – 0.29 = 0.71

    z = (P1 – P2)/√pc(1 – pc)(1/n1 + 1/n2)

    z = (0.3 – 0.28)/√(0.29)(0.71)(1/1406 + 1/778) = 0.02/√0.00553857907

    z = 0.99

    Since it is a two tailed test, the curve is symmetrical. We will look at the area in both tails. Since it is showing in one tail only, we would double the area

    From the normal distribution table, the area below the test z score to the left of 0.99 is 1 – 0.84 = 0.16

    We would double this area to include the area in the left tail of z = – 0.99 Thus

    p = 0.16 × 2 = 0.32

    Since 0.1 < 0.5, we would accept the null hypothesis.

    p value = 0.337

    Since 0.05 < 0.32, we would accept the null hypothesis

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