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78 . For a chi-square distribution with five degrees of freedom, the curve is ______________.
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 $\frac{{(O-E)}^{2}}{z}$ 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?
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?
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?
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?
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; $EBM=\text{}{z}_{\frac{\alpha}{2}}\left(\frac{\sigma}{\sqrt{n}}\right)=\left(1.96\right)\left(0.73\right)=\text{}1.4308$
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:
$\frac{\sigma}{\sqrt{n}}=\text{}\frac{4}{\sqrt{50}}=0.57$
$EBM=\text{}{z}_{\frac{\alpha}{2}}\left(\frac{\sigma}{\sqrt{n}}\right)=\left(1.96\right)\left(0.57\right)=\text{}1.12$
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.
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