Question

You are looking to estimate the proportion of pet owners that have pet insurance. You know from experience that at least 10% have pet insurance. A statistician takes a random sample of 1000 pet owners and creates a 95% confidence interval for the proportion of pet owners that have pet insurance. The confidence interval is [0.23, 0.31]. For each part below, if you can answer the question, provide the answer and an explanation. If you cannot answer the question, provide details on why you cannot. a. What is the probability that the true population proportion of pet owners that have pet insurance is above 0.1? b. What is the probability that the true population proportion of pet owners that have pet insurance is between [0.23, 0.31]? c. David Bergman has decided to create a procedure for building interval estimates. He takes a random sample of 100 pet owners. If the proportion of pet owners in the sample that have pet insurance is less than 0.1, he reports an interval estimate of [0.0, 0.05]. If the proportion of pet owners in the sample that have pet insurance is greater than or equal to 0.1, he reports an interval estimate of [0.1, 1.0]. Ignoring the statistician’s test, but taking the rest of the problem statement into account, what is the maximum probability that David Bergman’s test reports an interval of [0.0, 0.05]? d. Suppose the true proportion of per owners that have pet insurance is 0.3. What is the confidence level of David Bergman’s test?

Answer 1

a.

We know from experience that at least 10% have pet insurance. So, the probability that the true population proportion of pet owners that have pet insurance is above 0.1 is close to 1.

b.

Point estimate of p = (0.23 + 0.31)/2 = 0.27

Standard error of proportion , SE =

= 0.014

Probability that the true population proportion of pet owners that have pet insurance is between [0.23, 0.31]

= P(0.23 < p < 0.31)

= P(p < 0.31) - P(p < 0.23)

= P[Z < (0.31 - 0.27)/0.014] - P[Z < (0.23 - 0.27)/0.014]

= P[Z < 2.86] - P[Z < -2.86]

= 0.9979 - 0.0021

= 0.9958

c.

Standard error of proportion , SE =

= 0.0444

Probability that David Bergman’s test reports an interval of [0.0, 0.05] = P(p < 0.1)

= P(Z < (0.1 - 0.27)/0.0444]

= P[Z < -3.83]

= 0.000064

d.

Standard error of proportion , SE =

= 0.0458

Confidence level of David Bergman’s test = P(p 0.1)

= P(Z (0.1 - 0.3)/0.0458]

= P[Z -4.37]

= 0.9999938

99.99938 %