0.2 Practice tests (1-4) and final exams  (Page 16/36)

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.

80 . Compute $\frac{{(O-E)}^2}{z}$ 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.

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?

Practice test 3 solutions

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; $EBM=z_{\frac{\alpha }{2}}\left(\frac{\sigma }{\sqrt{n}}\right)=\left(1.96\right)\left(0.73\right)=1.4308$

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: $\frac{\sigma }{\sqrt{n}}=\frac{4}{\sqrt{50}}=0.57$
$EBM=z_{\frac{\alpha }{2}}\left(\frac{\sigma }{\sqrt{n}}\right)=\left(1.96\right)\left(0.57\right)=1.12$
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.