Question

1.) 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

2.) 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 )

Wth z = 1.95 ( two-tailed test )

we get

P value =0.0512

Decision :

P value > L.O.S

i.e., 0.0512 > 0.05

fail to reject the null hypothesis

Answer : 0.0512; fail to reject the null hypothesis

( 2 )

The provided sample mean is X = 228 and the population standard deviation is o = 19. The size of the sample is n = 82 and the required confidence level is 95%. Based on the provided information, the critical z-value for a = 0.05 is te = 1.96. The 95% confidence for the population mean is computed using the following expression CI= (x - *** X + vn 228 + Therefore, based on the information provided the 95 % confidence for the population mean siis CI = 228 1.96 x 19 1.96 x 19 V82 82 (228 - 4.112,228 + 4.112) = (223.888, 232.112)

Answer :  224 mg < μ < 232 mg