Families USA, a monthly magazine that discusses issues related to health and health costs, surveyed 20 of its subscribers. It found that the annual health insurance premiums for a family with coverage through an employer averaged $ 10,979. The standard deviation of the sample was $ 1,000.
a. Based on this sample information, develop a 90 percent confidence interval for the population mean yearly premium.
b. How large a sample is needed to find the population mean within $ 250 at 99 percent confidence?
During recent seasons, Major League Baseball has been criticized for the length of the games. A report indicated that the average game lasts 3 hours and 30 minutes. A sample of 17 games revealed the following times to completion. (Note that the minutes have been changed to fractions of hours, so that a game that lasted 2 hours and 24 minutes is reported at 2.40 hours.)
2.98, 2.40, 2.70, 2.25, 3.23, 3.17, 2.93, 3.18, 2.80
2.38, 3.75, 3.20, 3.27, 2.52, 2.58, 4.45, 2.45
Can we conclude that the mean time for a game is less than 3.50 hours? Use the .05 significance level.
The amount of income spent on housing is an important component of the cost of living. The total costs of housing for homeowners might include mortgage payments, property taxes, and utility costs (water, heat, electricity). An economist selected a sample of 20 homeowners in New England and then calculated these total housing costs as a percent of monthly income, five years ago and now. The information is reported below. Is it reasonable to conclude the percent is less now than five years ago?
Homeowner :Five Years Ago, Now Homeowner, Five Years Ago, Now
1:17%, 10%,11, 35%, 32%
2: 20, 39, 12, 16, 32
3: 29, 37, 13, 23, 21
4: 43, 27, 14, 33, 12
5: 36, 12, 15, 44, 40
6: 43, 41, 16, 44, 42
7: 45, 24, 17, 28, 22
8: 19, 26, 18, 29, 19
9: 49, 28, 19, 39, 35
10: 49, 26, 20, 22, 12
a. ANSWER: 90% Resulting Confidence Interval for ‘true mean’: = [10599, 11359]
Why???
SMALL SAMPLE, CONFIDENCE INTERVAL, NORMAL POPULATION DISTRIBUTION
x-bar: Sample mean = 10979
s: Sample standard deviation = 1000
n: Number of samples = 20
df: degrees of freedom = 19
Confidence Level = 90
“Look-up” Table (‘t-critical value’) = 1.7
Look-up Table of (‘t critical values’) for confidence and prediction intervals. Central two-sided area = 90% with df = 19. Another Look-up method is to utilize Microsoft Excel function: TINV(probability,degrees_freedom) Returns the inverse of the Student’s t-distribution 90% Resulting Confidence Interval for ‘true mean’: x-bar +/- (‘t critical value’) * s/SQRT(n) = 10979 +/- 1.7 * 1000/SQRT(20) = [10599, 11359]
b. ANSWER: Sample Size = 107 for 99% level of confidence
Why???
SMALL SAMPLE, LEVEL OF CONFIDENCE, NORMAL POPULATION DISTRIBUTION
Margin of Error (half of confidence interval) = 250
The margin of error is defined as the “radius” (or half the width) of a confidence interval for a particular statistic.
Level of Confidence = 99
σ: population standard deviation = 1000
(‘z critical value’) from Look-up Table for 99% = 2.58
The Look-up in the Table for the Standard Normal Distribution utilizes the Table’s cummulative ‘area’ feature. The Table shows positve and negative values of (‘z critical’) but since the Standard Normal Distribution is symmetric, only the magnitude of (‘z critical’) is important.
For a Level of Confidence = 99% the corresponding LEFT ‘area’ = 0.5. And due to Table’s symmetric nature, the corresponding RIGHT ‘area’ = 0.5 The (‘z critical’) value Look-up is 2.58
significant digits = 2
Margin of Error = (‘z critical value’) * σ/SQRT(n)
n = Sample Size
Algebraic solution for n:
n = [('z critical value') * σ/Margin of Error]²
= [ (2.58 * 1000)/250 ]²
Sample Size = 107 for 99% level of confidence