Question

A company claims that a certain vitamin contains 100% of your recommended daily allowance of vitamin C. You suspect that they are not putting as much vitamin C as they claim in their vitamins. The standard deviation of percent of daily allowance of vitamin C is 5%. You take an SRS of 5000vitamins and find the average percent of daily allowance to be 99.8%. a) Use the appropriate method to determine if your fears are true. b) Estimate the mean percent of your daily allowance of vitamin C in these vitamins c)Was part (a) statistically significant? why or why not? d)Was part (a) practically significant? why or why not? e) You want to create a confidence interval with a confidence level of 90% and a margin of error no more than 0.7%. What is he smallest sample size you can use and get these requirements?

Answer 1

SOLUTION A: NULL HYPOTHESIS H0:

u = 100%

ALTERNATIVE HYPOTHESIS Ha:$\mu< 100\%$

The Population standard deviation is already known that is $\sigma=5\%$

We have n=5000

Level of significance=0.1

$\overline{x}=99.8\%$

Under null hypothesis H0 test statistic is

$Z= \overline{x}-\mu/\sigma/\sqrt(n)$

$Z= 99.8-100/5/\sqrt(5000)$

Z critical= -1.28

Since Z Calculated is SMALLER than the Z critical therefore we can REJECT NULL HYPOTHESIS H0.

DECISION: REJECT NULL HYPOTHESIS H0.

CONCLUSION: We have sufficient evidence to conclude that they are not putting as much vitamin C as they claim in their vitamins.

b)  The mean percent of your daily allowance of vitamin C in these vitamins c is 99.8%

C) Yes it is statistically significant. Since the P-Value is .002342.The result is significant because p < 0.1

d) YES practically significant. Because few percent of vitamins get lost during handling of product.

e) n=(Zc*S.D/E)^2

= (1.65*5/0.7)^2

= 138.9

n= 139