Question

2. 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 5000 vitamins and find the average percent of daily allowance to be 99.8%.

(a) Use the appropriate methods 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 the smallest sample size you can use and get these requirements?

Answer 1

Answer:

a)

Given,

Null hypothesis Ho : u = 100

Alternative hypothesis Ha : u < 100

alpha = 5% = 0.05

Standard deviation s = 5

Sample size n = 5000

Sample mean = xbar = 99.8

Degree of freedom = n - 1

= 5000 - 1

= 4999

Standard error SE = s/sqrt(n)

substitute values

= 5/sqrt(5000)

= 0.0707

consider,

test statistic z = (xbar - u)/SE

substitute values

= (99.8 - 100)/0.0707

= -2.82885

z = - 2.83

Corresponding p value = 0.002

Here we observe that, p value is less than alpha, so we reject Ho.

b)

Estimation of the mean percent = 99.8%

c)

Yes, we can clearly say that the part a was statistically significant due to p < 0.1, so we reject Ho.

And the claim isn't true.

d)

Yes, it is practically significant.

It is due to that the some percent of vitamins are lost by product handling.

e)

E = 0.7

At 90% CI, z value = 1.65

Sample size n = (z*s/E)^2

substitute values

= (1.65*5/0.7)^2

= 138.9031

Sample size n = 139