A report from a Gallup poll 1 in 2011 started by saying, "Forty-five percent of American adults reported getting their health insurance from an employer..." Later in the article we find information on the sampling method, "a random sample of adults, aged and over, living in the US," and a sentence about the accuracy of the results, "the maximum margin of sampling error is percentage point." Use the margin of error 2 to give an interval estimate for the parameter of interest.

The interval estimate for the population proportion of American adults who got their health insurance from an employer is (0.43, 0.47).

Step-by-step explanation:

The confidence interval is the interval estimate of the population parameter.

The confidence interval has a certain probability that the true value of the parameter is contained in the interval.

The general form of the confidence interval is:

Here,

SS = sample statistic.

MOE = margin of error

The sample statistic is an unbiased estimator of the population parameter. If the sample size is large enough then the sample statistic can be used to estimate the population parameter value.

In this case the parameter of interest is the population proportion of American adults who got their health insurance from an employer.

The information provided is:

SS = p = 0.45.

MOE = 0.02.

Compute the confidence interval for the population proportion p as follows:

Thus, the interval estimate for the population proportion of American adults who got their health insurance from an employer is (0.43, 0.47).

f(x) has value 150 at x= 0

and 0 at x= 3

so d(f(x))/dt = slope = -150/3 = -50

idk

step-by-step explanation: