Question

The National Assessment of Educational Progress (NAEP) gave a test of basic arithmetic and the ability to apply it in everyday life to a sample of 840 men 21 to 25 years of age. Scores range from 0 to 500; for example, someone with a score of 325 can determine the price of a meal from a menu. The mean score for these 840 young men was x¯¯¯x¯ = 272. We want to estimate the mean score μμ in the population of all young men. Consider the NAEP sample as an SRS from a Normal population with standard deviation σσ = 60.

(a) If we take many samples, the sample mean x¯¯¯x¯ varies from sample to sample according to a Normal distribution with mean equal to the unknown mean score μμ in the population. What is the standard deviation of this sampling distribution?
(b) According to the 95 part of the 68-95-99.7 rule, 95% of all values of x¯¯¯x¯ fall within _______ on either side of the unknown mean μμ. What is the missing number?
(c) What is the 95% confidence interval for the population mean score μμ based on this one sample? Note: Use the 68-95-99.7 rule to find the interval.

2.

The National Institute of Standards and Technology (NIST) supplies "standard materials" whose physical properties are supposed to be known. For example, you can buy from NIST a liquid whose electrical conductivity is supposed to be 5. (The units for conductivity are microsiemens per centimeter. Distilled water has conductivity 0.5.) Of course, no measurement is exactly correct. NIST knows the variability of its measurements very well, so it is quite realistic to assume that the population of all measurements of the same liquid has the Normal distribution with mean μμ equal to the true conductivity and standard deviation σσ = 0.2. Here are 6 measurements on the same standard liquid, which is supposed to have conductivity 5:

5.32   4.88   5.10   4.73   5.15   4.75

NIST wants to give the buyer of this liquid a 96% confidence interval for its true conductivity. What is this interval?

3.

Here are the IQ test scores of 31 seventh-grade girls in a Midwest school district:

114	100	104	89	102	91	114	114	103	105
108	130	120	132	111	128	118	119	86	72
111	103	74	112	107	103	98	96	112	112	93

These 31 girls are an SRS of all seventh-grade girls in the school district. Suppose that the standard deviation of IQ scores in this population is known to be σσ = 15. We expect the distribution of IQ scores to be close to Normal. Estimate the mean IQ score for all seventh-grade girls in the school district, using a 98% confidence interval.

3.

How heavy a load (pounds) is needed to pull apart pieces of Douglas fir 4 inches long and 1.5 inches square? Here are data from students doing a laboratory exercise:

33,190     31,860     32,590     26,520     33,280

32,320     33,020     32,030     30,460     32,700

23,040     30,930     32,720     33,650     32,340

24,050     30,170     31,300     28,730     31,920

Suppose that the strength of pieces of wood like these follows a Normal distribution with standard deviation 3000 pounds. Give a 99% confidence interval for the mean load required to pull the wood apart.

to  lb

Answer 1

a) = = 60/ = 2.07

b) According to emperical rule 95% of the data falls within two standard deviaions from the mean.

2 * = 2 * 2.07 = 4.14

95% of all values of fall within 4.14 on either side of the unknown mean .

c) At 95% confidence interval the critical value is z_0.025 = 1.96

The 95% confidence interval for is

+/- z_0.025 *

= 272 +/- 1.96 * 2.07

= 272 +/- 4.0572

= 267.9428, 276.0572

2) = 4.9883

At 96% confidence interval the critical value is z_0.02 = 2.05

The 96% confidence interval for is

+/- z_0.02 *

= 4.9883 +/- 2.05 * 0.2/

= 4.9833 +/- 0.1674

= 4.8159, 5.1507

3) = 105.8387

At 98% confidence interval the critical value is z_0.01 = 2.33

The 98% confidence interval for is

+/- z_0.01 *

= 105.8387 +/- 2.33 * 15/

= 105.8387 +/- 6.2772

= 99.5615, 112.1159

4) = 30841

At 99% confidence interval the critical value is z_0.005 = 2.58

The 99% confidence interval for is

+/- z_0.005 *

= 30841 +/- 2.58 * 3000/

= 30841 +/- 1730.72

= 29110.28, 32571.72