Question

In: Statistics and Probability

Consider common applications and misconceptions about a confidence interval of a proportion. Explain how you arrived...

Consider common applications and misconceptions about a confidence interval of a proportion.

Explain how you arrived at your answers and conclusions, showing computations as appropriate.

Below is the scenario.

From a survey that Mika conducted for this veterinary medicine class, he found that 11 households out of a total of 30 households claimed to own at least one dog or cat. Mika constructed a 95% confidence interval in order to predict the population proportion, π, leading to 0.1867 ≤π ≤ 0.5466. (He used the Excel function CONFIDENCE.NORM.)

Mika concluded that:

1) It was Ok to use the sample proportion p = 11/30 = 0.3667 to construct this confidence interval;

2) the proportion of households in this whole state that would claim to own a dog or cat would be in the range of 36.67% +/- 5% = 31.67% - 41.67%; and

3) He was glad that he had not chosen a larger sample because a sample greater than n = 30 would have caused the confidence interval to become wider and less precise.

Solutions

Expert Solution

Solution1:

95% confidence interval for population proportion is

p^-z*sqrt(p^*(1-p^)/n,p^+z*sqrt(p^*(1-p^)/n

p^=x/n=11/30=0.3667

therefore

0.3667-1.96*sqrt(0.3667*(1-0.3667)/30,0.3667+1.96*sqrt(0.3667*(1-0.3667)/30

0.1943,0.5391

It was Ok to use the sample proportion p = 11/30 = 0.3667 to construct this confidence interval but

the 95% confidence interval constructed is wrong.

Solution2:

let u do hypothesis test for popualtion proportion

left tailed test

H0:p=0.50

Ha:p<0.50

alpha=0.05

z=p^-p/sqrt(p*(1-p)/n

=0.3667-0.5/sqrt(0.5*(1-0.5)/30)

=-1.460

p value=NORM.S.DIST(-1.46;TRUE)

p=0.072145

p>0.05

Fail to reject Null hypothesis

Accept null hypthesis

There is no sufficient statisitcal evidence at 5% level of significance to support the claim

he proportion of households in this whole state that would claim to own a dog or cat would be in the range of 36.67% +/- 5%

Solution3:

n>30 large samples

follows normal distribution according to the central limit theorem

the greater the sample size ,better is the approximation

As sample size increases,confidence interval becomes wider

and more precise.


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