meiprate81w

2022-03-18

A state is considering legislation to establisha law to help homeless individuals. A sample of 309 individuals regarding their opinions on the potential new law. Is there a relationship between sex and support for this law? Calculate the expected frequencies and compute a chi-square test for a statistically significant relationship between sex and support for the new law. Be sure to round to the second decimal.

### Answer & Explanation

Talan Kent

The given information is about the sample of 309 individuals regarding their opinions on the potential new law.
Hypotheses:
${H}_{0}$: There is no significant relationship between sex and support for the new law.
${H}_{1}$: There is a significant relationship between sex and support for the new law.
Observed Frequencies:

Calculate the expected frequencies:

Chi-square test statistic:
${\chi }^{2}=\sum _{\left\{i=1\right\}}^{n}\frac{{\left({O}_{i}-{E}_{i}\right)}^{2}}{{E}_{i}}$
Calculate the chi-square test statistic value is as follows:
${\chi }^{2}=\frac{{\left(98-81.79\right)}^{2}}{81.79}+\frac{{\left(101-117.21\right)}^{2}}{117.21}+\frac{{\left(29-45.21\right)}^{2}}{45.21}+\frac{{\left(81-64.79\right)}^{2}}{64.79}$
$=3.2127+2.2418+5.8121+4.0556$
$=15.3222$
$\sim 15.32$
Therefore, the value of chi-square test statistic is 15.32.
P-value:
Degrees of freedom:
(r – 1)(c – 1) = (2 – 1)(2 – 1) = 1.
Therefore, the degrees of freedom is 1.
Calculate the p-value using Excel is as follows:
Therefore, the p-value is 0.0001.
Decision rule:
Let assume level of significance is 0.05.
If p-value < level if significance, then reject the null hypothesis.
Here, p-value (0.0001) < level of significance (0.05), then reject the null hypothesis.
Conclusion:
Therefore, there is an evidence to conclude that there is a significant relationship between sex and support for the new law.

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