# Variances and Chi-Square Tests

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72. A random sample of 20 students obtained a mean of x = 72 and a variance of s2 = 16 on a college placement test in mathematics. Assuming the scores to be normally distributed, construct a 98% confidence interval for sigma^2.

78. Construct a 90% confidence interval for (sigma^2)1/(sigma^2)2 in Exercise 43. Were we justified in assuming that (sigma^2)1 does not equal (sigma^2)2 when we constructed our confidence interval for mu1 - mu2?

[See attached file for decription and answer to Exercise 43.]

83. Three cards are drawn from an ordinary deck of playing cards, with replacement, and the number Y of spades is recorded. After repeating the experiment 64 times, the following outcomes were recorded:

y 0 1 2 3

f 21 31 12 0

Test the hypothesis of 0.01 level of significance that the recorded data may be fitted by the binomial distribution b(y; 3, 2/4), y = 0, 1, 2, 3.

85. A coin is thrown until a head occurs and the number X of tosses recorded. After repeating the experiment 256 times, we obtained the following results:

x 1 2 3 4 5 6 7 8

f 136 60 34 12 9 1 3 1

Test the hypothesis at the 0.05 level of significance that the observed distribution of X may be fitted by the geometric distribution g(x; ½), x = 1, 2, 3, ....

90. In an experiment to study the dependence of hypertension on smoking habits, the following data were taken on 180 individuals:

Non-Smokers Moderate Smokers Heavy Smokers

Hypertension 21 36 30

No Hypertension 48 26 19

Test the hypothesis that the presence or absence of hypertension is independent of smoking habits. Use a 0.05 level of significance.

92. A random sample of 200 married men, all retired, were classified according to education and number of children:

NUMBER OF CHILDREN

EDUCATION 0-1 2-3 Over 3

Elementary 14 37 32

Secondary 19 42 17

College 12 17 10

Test the hypothesis, at the 0.05 level of significance, that the size of a family is independent of the level of education attained by the father.

Please see attached file for full problem description.

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##### Solution Summary

This problem set has 6 statistics problems involving the following:

-- estimating the variance of a population,

-- estimating the ratio of two variances/determining if two populations have the same variance,

-- using the chi-square goodness-of-fit test, and

-- using the chi-square test for independence.

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9.13 One-and Two-Sample Estimation Problems: Two Samples: Estimating the Ratio of Two Variances

72. A random sample of 20 students obtained a mean of x = 72 and a variance of s2 = 16 on a college placement test in mathematics. Assuming the scores to be normally distributed, construct a 98% confidence interval for ơ2.

The formula for the confidence interval for a variance is:

The values that we need to plug into the formula are:

n = 20, s2 = 16, and the chi-square critical values:

the left one will have a significance level of 0.01 (1 - 0.98 = 0.02; 0.02/2 = 0.01) and 19 degrees of freedom: X2 = 36.191

the right one will have a significance level of 0.99 and 19 degrees of freedom: X2 = 7.363

Using these values, the 98% CI for the variance is:

19(16) < σ2 < 19(16)

36.191 7.363

8.3999 < σ2 < 41.2875

78. Construct a 90% confidence interval for ơ21 / ơ22 in Exercise 43. Were we justified in assuming that ơ21 does not equal ơ22 when we constructed our confidence interval for μ1 - μ2?

Exercise 43: A taxi company is trying to decide whether to purchase brand A or brand B tires for its fleet of taxis. To estimate the difference in the two brands, an experiment is conducted using 12 of each brand. The tires are run until they wear out. The results are

Brand A: x1 = 36,300 kilometers,

S1 = 5,000 kilometers.

Brand B: x2 = 38, 100 kilometers

S2 = 6, 100 kilometers.

Compute a 95% confidence interval for μA - μB assuming the populations to be approximately normally distributed. You may not assume that the variances are equal.

Answer: -6,536 < μ1 - μ2 < 2,936

This is similar to the previous question, but when we have a ratio of variances, we need to use the F statistic instead of the chi-squared statistic. The formula for the confidence interval is:

The values that we need to plug into the formula are:

n1 = n2 = 12, ...

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