In: Statistics and Probability
A random sample of 5994 physicians in Colorado showed that 3062 provided at least some charity care.
(a) Let p represent the proportion of all Colorado
physicians who provide some charity care. Find a point estimate for
p. (Round your answer to four decimal places.)
(b) Find a 99% confidence interval for p. (Round your
answers to three decimal places.)
Lower Limit = ?
Upper Limit=?
(c) Give a brief explanation of the meaning of your answer in the context of this problem.
A- We are 99% confident that the true proportion of Colorado physicians providing at least some charity care falls within this interval.
B-We are 1% confident that the true proportion of Colorado physicians providing at least some charity care falls within this interval.
C-We are 1% confident that the true proportion of Colorado physicians providing at least some charity care falls above this interval.
D-We are 99% confident that the true proportion of Colorado physicians providing at least some charity care falls outside this interval.
(d) Is the normal approximation to the binomial justified in this
problem? Explain.
A-Yes; np < 5 and nq < 5.
B-No; np > 5 and nq < 5.
C-Yes; np > 5 and nq > 5.
D-No; np < 5 and nq > 5.
a)
Number of Items of Interest, x =
3062
Sample Size, n = 5994
Sample Proportion , p̂ = x/n =
0.5108
b)
Level of Significance, α =
0.01
Number of Items of Interest, x =
3062
Sample Size, n = 5994
Sample Proportion , p̂ = x/n =
0.511
z -value = Zα/2 = 2.576 [excel
formula =NORMSINV(α/2)]
Standard Error , SE = √[p̂(1-p̂)/n] =
0.0065
margin of error , E = Z*SE = 2.576
* 0.0065 = 0.0166
99% Confidence Interval is
Interval Lower Limit = p̂ - E = 0.511
- 0.0166 = 0.4942
Interval Upper Limit = p̂ + E = 0.511
+ 0.0166 = 0.5275
99% confidence interval is (
0.4942 < p < 0.5275
)
c)
A- We are 99% confident that the true proportion of Colorado physicians providing at least some charity care falls within this interval.
d)
C-Yes; np > 5 and nq > 5
THANKS
revert back for doubt
please upvote