Question

Unexpected expense. In a random sample 765 adults in the United States, 322 say they could not cover a $400 unexpected expense without borrowing money or going into debt.

(a) What population is under consideration in the data set?

(b) What parameter is being estimated?

(c) What is the point estimate for the parameter?

(d) What is the name of the statistic can we use to measure the uncertainty of the point estimate?

(e) Compute the value from part (d) for this context.

(f) A cable news pundit thinks the value is actually 50%. Should she be surprised by the data?

(g) Suppose the true population value was found to be 40%. If we use this proportion to recompute the value in part (e) using p = 0:4 instead of ^p, does the resulting value change much?

Answer 1

(a) What population is under consideration in the data set?
Population under consideration = 765 adults in the United states

(b) What parameter is being estimated?
Parameter estimated is the proportion of people who cannot cover an unexpected expense of $400 without borrowing money or going into debt.

(c) What is the point estimate for the parameter?
point_estimate = 322/765 = 0.420915

(d) What is the name of the statistic can we use to measure the uncertainty of the point estimate?
Point estimate measure the variablity in a point estimate.

(e) Compute the value from part (d) for this context.
$se= \sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.43(1-0.42)}{765}}= 0.017845$

(f) A cable news pundit thinks the value is actually 50%. Should she be surprised by the data?
Yes, since there is difference of about 0.08 or 8%

(g) Suppose the true population value was found to be 40%. If we use this proportion to recompute the value in part (e) using p = 0:4 instead of ^p, does the resulting value change much?

$se= \sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.40(1-0.40)}{765}}= 0.01771$

It changes very slightly.