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:-
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)
Ho : µ = 100
Ha : µ < 100 (Left tail test)
Level of Significance , α = 0.05
sample std dev , s = 5.0000
Sample Size , n = 5000
Sample Mean, x̅ = 99.8000
degree of freedom= DF=n-1= 4999
Standard Error , SE = s/√n = 5.0000 / √ 5000 = 0.0707
t-test statistic= (x̅ - µ )/SE = ( 99.800 - 100 ) / 0.0707 =
-2.83
p-Value = 0.002 [Excel formula =t.dist(t-stat,df) ]
Decision: p-value<α, Reject null hypothesis
b)
x̅ = 99.8%
c)
yes,part a was statistically significant. we reject the null hypothesis
the claim of 100% was not true. there was not as much vitamin as they claim
e)
Standard Deviation , σ = 5
sampling error , E = 0.7
Confidence Level , CL= 90%
alpha = 1-CL = 10%
Z value = Zα/2 = 1.645 [excel formula =normsinv(α/2)]
Sample Size,n = (Z σ / E )² = ( 1.645 5 / 0.7 ) ² = 138.038
So,Sample Size needed= 139.
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