Question

Suppose production specifications for a new energy drink require exactly 40 mg of vitamins per serving. Fred, a quality control engineer, suspects that the production process does not ensure each serving contains 40 mg of vitamins, on average To test his suspicions, Fred collects a random sample of 8 servings and determines their average vitamin content to be 34.9985 mg with a standard deviation of 6.9871 mg. Fred uses a 95% confidence interval to test the null hypothesis Ho: μ-μο versus the alternative Hi : μヂμο, where the hypothesized population mean μο is 40 mg. What are the upper and lower limits of Fred's interval? Enter your results to two decimal places. lower limit mg upper limit: mg the null hypothesis Based on the confidence interval computed above, Fred should -at a significance level of 0.05 because the

Answer 1

The given null and alternative hypothesis is as follows:

$H_{o}:\mu=\mu_{o}$

$H_{1}:\mu\neq \mu_{o}$

The formula to calculate the 95% confidence interval for population average vitamin content$(\mu):$

/İ) t(a/2.dfen-l)

Given:

34.9985mq, s= 6.9871774, n-8

t(00S5-1=7) 2.364624 2.3646 n-I

Lower limit:

34.9985-2.3646 (-n)- T-t(a/2.df=n-1)

                                             

29.15719813 29.16 mg

Upper limit:

71) twadfan--G)-34.9985 + 2.3646(

                                              

- 40.83980187 40.84 mg

The 95% confidence interval for the population average vitamin content$(\mu)$ is (29.16 mg, 40.84 mg)

Based on the confidence interval we are FTR(fail to reject) H_o . Since the confidence interval does contains the hypothesized value of $\mu_{o}=40\:mg$ for population mean $(\mu)$. We conclude that at $\alpha=0.05$ level of significance we does not have evidence to support alternative hypothesis, i.e., $H_{1}:\mu\neq \mu_{o}$