A chi-square homogeneity test is to be conducted to decide whether four populations are nonhomogeneous with respect to a variable that has eight possible values. What are the degrees of freedom for the x^2-statistic?

ringearV 2020-10-19 Answered
A chi-square homogeneity test is to be conducted to decide whether four populations are nonhomogeneous with respect to a variable that has eight possible values. What are the degrees of freedom for the \(\displaystyle{x}^{{2}}\)-statistic?

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svartmaleJ
Answered 2020-10-20 Author has 7739 answers
Step 1
The test for homogeneity determines if two or more populations (or subgroups of a population) have the same distribution of a single categorical variable.
Step 2
For chi-square tests based on two-way tables (both the test of independence and the test of homogeneity), the degrees of freedom are calculated using the following formula: df = (r-1)(c-1) where r is the number of rows and c is the number of columns.
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asked 2021-01-17
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(a) What is the level of significance?
State the null and alternate hypotheses.
\(H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}>?_{2}^{2}H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}\neq?_{2}^{2}H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}?_{2}^{2},H1:?_{1}^{2}=?_{2}^{2}\)
(b) Find the value of the sample F statistic. (Round your answer to two decimal places.)
What are the degrees of freedom?
\(df_{N} = ?\)
\(df_{D} = ?\)
What assumptions are you making about the original distribution?
The populations follow independent normal distributions. We have random samples from each population.The populations follow dependent normal distributions. We have random samples from each population.The populations follow independent normal distributions.The populations follow independent chi-square distributions. We have random samples from each population.
(c) Find or estimate the P-value of the sample test statistic. (Round your answer to four decimal places.)
(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?
At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant. At the ? = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.
(e) Interpret your conclusion in the context of the application.
Reject the null hypothesis, there is sufficient evidence that the population variance is larger in the old thermostat temperature readings.Fail to reject the null hypothesis, there is sufficient evidence that the population variance is larger in the old thermostat temperature readings. Fail to reject the null hypothesis, there is insufficient evidence that the population variance is larger in the old thermostat temperature readings.Reject the null hypothesis, there is insufficient evidence that the population variance is larger in the old thermostat temperature readings.
asked 2021-03-05
For each of the following situations, state whether you’d use a chi-square goodness-of-fit test, a chi-square test of homogeneity, a chi-square test of independence, or some other statistical test:
a) Is the quality of a car affected by what day it was built? A car manufacturer examines a random sample of the warranty claims filed over the past two years to test whether defects are randomly distributed across days of the work week.
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\(\begin{array}{cc}\hline & \text{Afraid to walk at night?} \\ \hline & \text{Yes} & \text{No} & \text{Total} \\ \hline \text{Male} & 173 & 598 & 771 \\ \hline \text{Female} & 393 & 540 & 933 \\ \hline \text{Total} & 566 & 1138 & 1704 \\ \hline \text{Source:}2014\ GSS \end{array}\)
If the chi-square \((\chi 2)\) test statistic \(=73.7\) what is the p-value you would report? Use Table C.Remember to calculate the df first.
Group of answer choices
\(P>0.250\)
\(P=0.01\)
\(P<0.001\)
\(P<0.002\)
image
\(\begin{array}{|c|cc|}\hline & \text{Right-Tail Probability} \\ \hline df & 0.250 & 0.100 & 0.050 & 0.025 & 0.010 & 0.005 & 0.001 \\ \hline 1 & 1.32 & 2.71 & 3.84 & 5.02 & 6.63 & 7.88 & 10.83 \\ 2 & 2.77 & 4.61 & 5.99 & 7.38 & 9.21 & 10.60 & 13.82 \\ 3 & 4.11 & 6.25 & 7.81 & 9.35 & 11.34 & 12.84 & 16.27 \\ 4 & 5.39 & 7.78 & 9.49 & 11.14 & 13.28 & 14.86 & 18.47 \\ 5 & 6.63 & 9.24 & 11.07 & 12.83 & 15.09 & 16.75 & 20.52 \\ 6&7.84&10.64&12.59&14.45&16.81&18.55&22.46 \\ 7&9.04&12.02&14.07&16.01&18.48&20.28&24.32\\ 8&10.22&13.36&15.51&17.53&20.09&21.96&26.12 \\ 9&11.39&14.68&16.92&19.02&21.67&23.59&27.88 \\ 10&12.55&15.99&18.31&20.48&23.21&25.19&29.59 \\ 11&13.70&17.28&19.68&21.92&24.72&26.76&31.26 \\ 12&14.85&18.55&21.03&23.34&26.22&28.30&32.91 \\ 13&15.98&19.81&22.36 & 24.74 & 27.69 & 29.82 & 34.53 \\ 14 & 17.12 & 21.06 & 23.68 & 26.12 & 29.14 & 31.32 & 36.12 \\15 & 18.25 & 22.31 & 25.00 & 27.49 & 30.58 & 32.80 & 37.70 \\ 16 & 19.37 & 32.54 & 26.30 & 28.85 & 32.00 & 34.27 & 39.25 \\ 17 & 20.49 & 24.77 & 27.59 & 30.19 & 33.41 & 35.72 & 40.79 \\ 18 & 21.60 & 25.99 & 28.87 & 31.53 & 34.81 & 37.16 & 42.31 \\ 19 & 22.72 & 27.20 & 30.14 & 32.85 & 36.19 & 38.58 & 43.82 \\ 20 & 23.83 & 28.41 & 31.41 & 34.17 & 37.57 & 40.00 & 45.32 \\ \hline \end{array}\)

asked 2021-02-12
For the following situations, identify the test you would run to analyze the data:
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