# 0.2 Practice tests (1-4) and final exams  (Page 21/36)

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78 . For a chi-square distribution with five degrees of freedom, the curve is ______________.

## 11.3: test of independence

Use the following information to answer the next four exercises. You are considering conducting a chi-square test of independence for the data in this table, which displays data about cell phone ownership for freshman and seniors at a high school. Your null hypothesis is that cell phone ownership is independent of class standing.

79 . Compute the expected values for the cells.

Cell = Yes Cell = No
Freshman 100 150
Senior 200 50

80 . Compute $\frac{{\left(O-E\right)}^{2}}{z}$ for each cell, where O = observed and E = expected.

81 . What is the chi-square statistic and degrees of freedom for this study?

82 . At the α = 0.5 significance level, what is your decision regarding the null hypothesis?

## 11.4: test of homogeneity

83 . You conduct a chi-square test of homogeneity for data in a five by two table. What is the degrees of freedom for this test?

## 11.5: comparison summary of the chi-square tests: goodness-of-fit, independence and homogeneity

84 . A 2013 poll in the State of California surveyed people about taxing sugar-sweetened beverages. The results are presented in the following table, and are classified by ethnic group and response type. Are the poll responses independent of the participants’ ethnic group? Conduct a hypothesis test at the 5% significance level.

Ethnic Group \ Response Type Favor Oppose No Opinion Row Total
White / Non-Hispanic 234 433 43 710
Latino 147 106 19 272
African American 24 41 6 71
Asian American 54 48 16 118
Column Total 459 628 84 1171

85 . In a test of homogeneity, what must be true about the expected value of each cell?

86 . Stated in general terms, what are the null and alternative hypotheses for the chi-square test of independence?

87 . Stated in general terms, what are the null and alternative hypotheses for the chi-square test of homogeneity?

## 11.6: test of a single variance

88 . A lab test claims to have a variance of no more than five. You believe the variance is greater. What are the null and alternative hypothesis to test this?

## 8.1: confidence interval, single population mean, population standard deviation known, normal

1 . $\frac{\sigma }{\sqrt{n}}=\frac{4}{\sqrt{30}}=0.73$

2 . normal

3 . 0.025 or 2.5%; A 95% confidence interval contains 95% of the probability, and excludes five percent, and the five percent excluded is split evenly between the upper and lower tails of the distribution.

4 . z -score = 1.96;

5 . 41 ± 1.43 = (39.57, 42.43); Using the calculator function Zinterval, answer is (40.74, 41.26. Answers differ due to rounding.

6 . The z -value for a 90% confidence interval is 1.645, so EBM = 1.645(0.73) = 1.20085.
The 90% confidence interval is 41 ± 1.20 = (39.80, 42.20).
The calculator function Zinterval answer is (40.78, 41.23). Answers differ due to rounding.

7 . The standard error of measurement is:

The 95% confidence interval is 41 ± 1.12 = (39.88, 42.12).
The calculator function Zinterval answer is (40.84, 41.16). Answers differ due to rounding.

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Extraordinary observation (too distant, high, low etc)
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