Part B: identify the test statistic ____ (Round your answer to two decimal places as needed)Part C: Identify the P-Value____ (Round your answer to three decimal places as needed) Assume that adults were randomly selected for a poll. They were asked if they "favor or oppose using federal tax dollars to fund medical research using stem cells obtained from human embryos." Of those polled, 483 were in favor,398 were opposed, and 116 were unsure. A politician claims that people don't really understand the stem cell issue and their responses to such questions are random responses equivalent to a coin toss. Exclude the 116 subjectswho said that they were unsure, and use a 0.01 significance level to test the claim that the proportion of subjects who respond in favor is equal to 0.5. What does the result suggest about the politician's claim?Identify the null and alternative hypotheses for this test. Choose the correct answer below.O A. Ho: p=0.5H1: p>0.5О В. Но: р20.5H1: p=0.5Ос. Но: р3D0.5H1: p<0.5O D. Ho: p=0.5H1: p+0.5

Name: Part B: identify the test statistic ____ (Round your answer to two dec
Uploaded: 2024-03-20T07:07:12.000Z
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Description: Part B: identify the test statistic ____ (Round your answer to two decimal places as needed)Part C: Identify the P-Value____ (Round your answer to three decimal places as needed) Assume that adults were randomly selected for a poll. They were asked i

Denote p as the true proportion of adults who is in favor.It is needed to test, whether the proportion of subjects who respond in favor is equal to 0.5 or not. The null and alternative hypotheses are: H0: p = 0.5, vs. H1: p ≠ 0.5. Correct option is D. The direction of the test is both sided. Hence, it is two-tailed test. The test must is performed at 5% or 0.05 level of significance. If x denotes the number of cases favourable to an event in a sample of size n, then the sample proportion is, p̂ = x/n. It is given that, 483 adults were in favor, 398 opposed and 116 were unsured. It is not needed to consider the unsure adults. Hence, the sample size is n = 483 + 398 = 881. Hence, x = 482 and p̂ = 0.548241 (= 483/881). The standardized sample test statistic is, z = (p̂ – p)/ √ [p (1 – p)/n], where p̂ is the point estimate of the population proportion, p is the population proportion and n is the sample size. The test statistic is: z = (p̂ – p)/ √ [p (1 – p)/n] [Where p is the proportion value suggested in H0] = (0.548241 – 0.5) / √ [(0.5) (1 – 0.5)/881] = 0.048241/0.016845428 ≈ 2.86. The test statistic is 2.86. Using the Excel formula =2*(1-NORM.S.DIST(2.86,TRUE)), the P-value is 0.004. Decision rule: If P-value is less than or equal to the level of significance, then reject the null hypothesis . Otherwise, fail to reject the null hypothesis. Conclusion: Here, the P-value is 0.004, which is less than the level of significance of 0.01. Hence, reject the null hypothesis. Thus, there is sufficient evidence against the claim that the proportion of subjects who respond in favor is equal to 0.5, at 1% level of significance.