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Fantom [35]
3 years ago
8

During a recent election, a clerk was responsible for the placement of candidates names on election ballots for a particular vot

ing precinct. Party A candidates were selected for the first position in 30 of 40 ballots. Because the clerk was supposed to use a method of random selection, Party B claimed that instead of using randomness, the clerk used a method favoring Party A. Use a 0.05 significance level to test the claim that the method favored Party A. [Note: Favored means a proportion greater than 50%.]
Mathematics
1 answer:
hjlf3 years ago
3 0

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, the clerk used a method favoring Party A.

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 = 40 represent the number of samples 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 = 40

po = Null hypothesized value = 0.50

p^ = Observed proportion = 30/40 = 0.75

Substituting the values we have

z = (0.75-0.50)/√(0.50(1-0.50)/40)

z = 3.16227766

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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