Question

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" or those polled, 483 were in favor, 398 were opposed and 123 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 loss. Exclude the 123 subjects who 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?​

Answer 1

Testing the hypothesis, it is found that since the p-value of the test is 0.0042 < 0.01, it can be concluded that the proportion of subjects who respond in favor is different of 0.5.

At the null hypothesis, it is tested if the proportion is of 0.5, that is:

$H_0: p = 0.5$

At the alternative hypothesis, it is tested if the proportion is different of 0.5, that is:

$H_1: p \neq 0.5$

The test statistic is given by:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1 - p)}{n}}}$

In which:

• $\overline{p}$ is the sample proportion.
• p is the value tested at the null hypothesis.
• n is the sample size.

In this problem, the parameters are given by:

$p = 0.5, n = 483 + 398 = 881, \overline{p} = \frac{483}{881} = 0.5482$

The value of the test statistic is:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1 - p)}{n}}}$

$z = \frac{0.5482 - 0.5}{\sqrt{\frac{0.5(0.5)}{881}}}$

$z = 2.86$

Since we have a two-tailed test(test if the proportion is different of a value), the p-value of the test is P(|z| > 2.86), which is 2 multiplied by the p-value of z = -2.86.

Looking at the z-table, z = -2.86 has a p-value of 0.0021.

2(0.0021) = 0.0042

Since the p-value of the test is 0.0042 < 0.01, it can be concluded that the proportion of subjects who respond in favor is different of 0.5.

A similar problem is given at https://brainly.com/question/24330815