Question

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 that 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 pˆ, does the resulting value change much?

Answer 1

Answer:

a) The population under consideration in the data set is "adults in the United States".

b) The parameter being estimated is the proportion of adults in the U.S. that could not cover a $400 unexpected expense without borrowing money or going into debt.

c) The point estimate for the parameter of the population is p-hat.

d) The standard deviation of the sampling distribution

e)

$\hat p=322/765=0.421\\\\s=\sqrt{\frac{\hat p(1-\hat p)}{n} } =\sqrt{\frac{0.421\cdot 0.579}{765} } =\sqrt{0.00032} =0.018$

f) Yes, 50% is a value not probable if the estimations are right.

g) No, it does not change.

Step-by-step explanation:

f) We can calculate the plausability of the value 50% with our estimations.

The P-value of a proportion of 50% is very low, so the value should surprise.

$z=(x-\hat p)/s=(0.5-0.421)/0.018=0.079/0.018=4.38\\\\P(z>4.38)=0$

g) We re-calculate with p=0.4

$z=(x-\hat p)/s=(0.5-0.4)/0.018=0.100/0.018=5.55\\\\P(z>5.55)=0$