Question

1) Human Resource Consulting (HRC) surveyed a random sample of 80 Twin Cities construction companies to find information on the costs of their health care plans. One of the items being tracked is the annual deductible that employees must pay. The Minnesota Department of Labor reports that historically the mean deductible amount per employee is $505 with a standard deviation of $105. (Round your z value to 2 decimal places and final answers to 4 decimal places. Leave no cells—blankbe certain to enter "0" if required.)

A) Compute the standard error of the sample mean for HRC.

B) What is the chance HRC finds a sample mean between $477 and $527?

C) Calculate the likelihood that the sample mean is between $492 and $512.

D) What is the probability the sample mean is greater than $550?

2) Families USA, a monthly magazine that discusses issues related to health and health costs, surveyed 21 of its subscribers. It found that the annual health insurance premiums for a family with coverage through an employer averaged $10,350. The standard deviation of the sample was $990

1. Based on this sample information, develop a 90% confidence interval for the population mean yearly premium. (Round your answers to the nearest whole dollar amount.)

2. How large a sample is needed to find the population mean within $260 at 95% confidence? (Round up your answer to the next whole number.)

Answer 1

(1):

The given information is:

The sample size (n) is 80.

The population mean is $505.

The standard deviation is $105.

Let X be the number of samples of Twin Cities construction companies selected to determine information related to the cost of their health care plans. So, the random variable X follows normal distribution such that,

A): The standard error of the sample mean (S.E.) for HRC can be calculated as:

Therefore, the standard error of the sample mean (S.E.) for HRC is 11.7394.

B): The chance HRC finds a sample mean between $477 and $527 can be calculated as:

Using standard normal table, the obtained P(Z<1.87) is 0.9693 and P(Z<2.39) is 0.9916. So,

Therefore, the chance HRC finds a sample mean between $477 and $527 is 0.9609.

C): The likelihood that the sample mean is between $492 and $512 can be calculated as:

Using standard normal table, the obtained P(Z<0.60) is 0.7257 and P(Z<1.11) is 0.8665. So,

Therefore, the likelihood that the sample mean is between $492 and $512 is 0.5922.

D): The probability the sample mean is greater than $550 can be calculated as:

Using standard normal table, the obtained P(Z<3.83) is 0.9999. So,

Therefore, the probability the sample mean is greater than $550 is 0.0001.