a social service organization reports that th...
a social service organization reports that the level of educational attainment of mothers receiving food stamps is uniformly distributed. to test this claim, you randomly select 96 mothers who currently receive food stamps and record the educational attainment of each. the results are shown in the table on the right. at \\(\\alpha = 0.05\\), can you reject the claim that the distribution is uniform? complete parts (a) through (d) below.\n(a) state \\(h_0\\) and \\(h_a\\) and identify the claim.\n\\(h_0\\): the distribution of educational - attainment responses is uniform\n\\(h_a\\): the distribution of educational - attainment responses is not uniform\nwhich hypothesis is the claim?\n\\(h_0\\)\n(b) determine the critical value, \\(\\chi_0^2\\), and the rejection region.\n\\(\\chi_0^2=7.378\\) (round to three decimal places as needed.)\nchoose the correct rejection region below.\na. \\(\\chi^2\\leq\\chi_0^2\\)\nb. \\(\\chi^2<\\chi_0^2\\)\nc. \\(\\chi^2\\geq\\chi_0^2\\)\nd. \\(\\chi^2>\\chi_0^2\\)\n(c) calculate the test statistic.\n\\(\\chi^2 = 7.378\\) (round to three decimal places as needed.)
Answer
# Explanation:
## Step1: Recall chi - square goodness - of - fit test
For a goodness - of - fit test to check if a distribution is uniform, we use the chi - square test statistic $\chi^{2}=\sum\frac{(O - E)^{2}}{E}$, where $O$ is the observed frequency and $E$ is the expected frequency.
## Step2: Calculate expected frequency
There are $n=96$ observations and $k = 3$ categories (Not a high school graduate, High school graduate, College (1 year or more)). For a uniform distribution, the expected frequency $E=\frac{n}{k}=\frac{96}{3}=32$ for each category.
## Step3: Calculate the test statistic
Let $O_1 = 35$, $O_2=39$, $O_3 = 22$.
\[
\begin{align*}
\chi^{2}&=\frac{(O_1 - E)^{2}}{E}+\frac{(O_2 - E)^{2}}{E}+\frac{(O_3 - E)^{2}}{E}\\
&=\frac{(35 - 32)^{2}}{32}+\frac{(39 - 32)^{2}}{32}+\frac{(22 - 32)^{2}}{32}\\
&=\frac{3^{2}}{32}+\frac{7^{2}}{32}+\frac{(- 10)^{2}}{32}\\
&=\frac{9 + 49+100}{32}\\
&=\frac{158}{32}\\
& = 4.9375\approx4.938
\end{align*}
\]
## Step4: Make a decision
The critical value $\chi_{0}^{2}=7.378$ with degrees of freedom $df=k - 1=3 - 1 = 2$ and $\alpha=0.05$. Since the test statistic $\chi^{2}=4.938<\chi_{0}^{2}=7.378$, we fail to reject the null hypothesis $H_0$.
# Answer:
We fail to reject the null hypothesis $H_0$. So, there is not enough evidence at the $\alpha = 0.05$ level of significance to reject the claim that the distribution of educational - attainment responses is uniform.