Hey, Guys, can someone help me with this
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Answer:
a) The 98% confidence interval would be given (0.182;0.218).
b) We are 98% confident that the true proportion of of England people who are deficient in Vitamin D is between 0.182 and 0.218.
c) If repeated samples were taken and the 98% confidence interval computed for each sample, 98% of the intervals would contain the population proportion.
d) Yes since the confidence interval not contains the value 0.25, we can refute the claim at 2% of significance.
Step-by-step explanation:
Data given and notation
n=2700 represent the random sample taken
X represent the people in England who are deficient in Vitamin D
estimated proportion of England people who are deficient in Vitamin D
represent the significance level
Confidence =0.98 or 98%
z would represent the statistic (variable of interest)
p= population proportion of England people who are deficient in Vitamin D
Part a
The confidence interval would be given by this formula
For the 98% confidence interval the value of and
, with that value we can find the quantile required for the interval in the normal standard distribution.
And replacing into the confidence interval formula we got:
And the 98% confidence interval would be given (0.182;0.218).
Part b
We are 98% confident that the true proportion of of England people who are deficient in Vitamin D is between 0.182 and 0.218.
Part c
If repeated samples were taken and the 98% confidence interval computed for each sample, 98% of the intervals would contain the population proportion.
Part d
Yes since the confidence interval not contains the value 0.25, we can refute the claim at 2% of significance.