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meriva
3 years ago
5

Identify thenull hypothesis, alternative hypothesis, test statistic, P-value or critical value(s), then state the conclusion abo

ut the null hypothesis, as well as the final conclusion that addresses the original claim. Since the Hawk-Eye instant replay system for tennis was introducedat the U.S. Open in 2006, men challenged 2441 referee calls, with the result that 1027 of the calls were overturned. Women challenged 1273 referee calls, and 509 of the calls were overturned.We want to use a 0.05 significance level to test the claim that men and women haveequal success in challenging calls. Test the claim by constructing an appropriate confidence interval.

Mathematics
1 answer:
miv72 [106K]3 years ago
3 0

Answer:

Step-by-step explanation:

This is a test of 2 population proportions. Let m and w be the subscript for men and women players. The population proportions would be pm and pw

Pm - Pw = difference in the proportion of male and female players.

The null hypothesis is

H0 : pm = pw

pm - pw = 0

The alternative hypothesis is

Ha : pm ≠ pw

pm - pw ≠ 0

it is a two-tailed test

Sample proportion = x/n

Where

x represents number of success

n represents number of samples

For men,

xm = 1027

nm = 2441

Pm = 1027/2441 = 0.42

For women,

xw = 509

nw = 1273

Pw = 509/1273 = 0.4

The pooled proportion, pc is

pc = (xm + xw)/(nm + nw)

pc = (1027 + 509)/(2441 + 1273) = 0.41

1 - pc = 1 - 0.41 = 0.59

z = (Pm - Pw)/√pc(1 - pc)(1/nm + 1/nw)

z = (0.42 - 0.4)/√(0.41)(0.59)(1/2441 + 1/1273) = 0.02/√0.0002891223

z = 1.18

Since it is a 2 tailed test, we would find the p value by doubling the area to the right of the z score to include the area to left.

Area to the right from the normal distribution table is

1 - 0.881 = 0.119

P value = 0.119 × 2 = 0.238

Since 0.05 < 0.238, we would accept the null hypothesis

Therefore, there is no sufficient evidence to conclude that that men and women have unequal success in challenging calls.

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CLASS     FREQUENCIES     RELATIVE FREQUENCIES

A                        60                                 0.5

B                        12                                  0.1

C                        48                                 0.4

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Given that;

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Now to determine our relative frequency, we divide each frequency by the total sum of the given frequencies;

Relative Frequency A = Frequency A / total = 60 / 120 = 0.5

Relative Frequency B = Frequency B / total = 12 / 120 = 0.1

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CLASS     FREQUENCIES     RELATIVE FREQUENCIES

A                        60                                 0.5

B                        12                                  0.1

C                        48                                 0.4

TOTAL              120                                  1

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