Question

1.) Assume that a sample is used to estimate a population proportion p. Find the margin of error m that corresponds to the given statistics and confidence level. Round the margin of error to four decimal places.

95% confidence; n = 380, x = 50

Group of answer choices

0.0340

0.0408

0.0306

0.0357

2.) Formulate the indicated conclusion in nontechnical terms. Be sure to address the original claim.

A researcher claims that 62% of voters favor gun control. Assuming that a hypothesis test to reject this claim has been conducted and that the conclusion is failure to reject the null hypothesis, state the conclusion in nontechnical terms.

Group of answer choices

There is not sufficient evidence to support the claim that 62% of voters favor gun control.

There is not sufficient evidence to warrant rejection of the claim that 62% of voters favor gun control.

There is sufficient evidence to support the claim that more than 62% of voters favor gun control.

There is sufficient evidence to warrant rejection of the claim that 62% of voters favor gun control.

3.) Use the given information to find the P-value. Also, use a 0.05 significance level and state the conclusion about the null hypothesis (reject the null hypothesis or fail to reject the null hypothesis).

The test statistic in a two-tailed test is z = 1.95.

Group of answer choices

0.0256; reject the null hypothesis

0.9744; fail to reject the null hypothesis

0.0512; reject the null hypothesis

0.0512; fail to reject the null hypothesis

4.) Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p.

n = 56, x = 30; 95% confidence

Group of answer choices

0.404 < p < 0.668

0.426 < p < 0.646

0.405 < p < 0.667

0.425 < p < 0.647

5.) Use the confidence level and sample data to find a confidence interval for estimating the population μ. Round your answer to the same number of decimal places as the sample mean.

A laboratory tested 82 chicken eggs and found that the mean amount of cholesterol was 228 milligrams with σ=19.0 milligrams. Construct a 95% confidence interval for the true mean cholesterol content, μ, of all such eggs.

Group of answer choices

225 mg < μ < 233 mg

223 mg < μ < 231 mg

223 mg < μ < 232 mg

224 mg < μ < 232 mg

Answer 1

1. Given : n=380 , x=50

The estimate of the sample proportion is , $\hat{p}=\frac{x}{n}=\frac{50}{380}=0.1316$

Now , the critical value is , $Z_{\alpha/2}=Z_{0.05/2}=1.96$ ; From Z-table

Therefore , the margin of error is ,

$ME=Z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=1.96*\sqrt{\frac{0.1316(1-0.1316)}{380}}=0.0340$