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?

Step-by-step explanation:

(a)

The population under study are the adults of United States.

(b)

A parameter the population characteristic that is under study.

In this case the researcher is interested in the proportion of US adults who say they could not cover a $400 unexpected expense without borrowing money or going into debt.

So the parameter is the population proportion of US adults who say this.

(c)

A point estimate is a numerical value that is the best guesstimate of the parameter. It is computed using the sample values.

For example, sample mean is the point estimate of population mean.

The point estimate of the population proportion of US adults who say cover a $400 unexpected expense without borrowing money or going into debt, is the sample proportion, .

(d)

The uncertainty of the point estimate can be measured by the standard error.

The standard error tells us how closer the sample statistic is to the parameter value.

(e)

The standard error is:

(f)

The sample proportion of US adults who say they could not cover a $400 unexpected expense without borrowing money or going into debt, is approximately 42.1%.

As the sample size is quite large this value can be used to estimate the population proportion.

If the proportion is believed to be 50% then she will be surprised because the estimated percentage is quite less than 50%.

(g)

Compute the standard error using p = 0.40 as follows:

The standard error does not changes much.

i dont know

step-by-step explanation: