# The following is a two-way table showing preferences for an award (A, B, C) by gender for the students sampled in survey. Test whether the data indicate there is some association between gender and preferred award. begin{array}{|c|c|c|}hline &text{A}&text{B}&text{C}&text{Total}hline text{Female} &20&76&73&169 hline text{Male}&11&73&109&193 hline text{Total}&31&149&182&360 hline end{array} Chi-square statistic=? p-value=? Conclusion: (reject or do not reject H_0) Does the test indicate an association between gender and preferred award? (yes/no)

Question
Two-way tables
The following is a two-way table showing preferences for an award (A, B, C) by gender for the students sampled in survey. Test whether the data indicate there is some association between gender and preferred award.
$$\begin{array}{|c|c|c|}\hline &\text{A}&\text{B}&\text{C}&\text{Total}\\\hline \text{Female} &20&76&73&169\\ \hline \text{Male}&11&73&109&193 \\ \hline \text{Total}&31&149&182&360 \\ \hline \end{array}\\$$
Chi-square statistic=?
p-value=?
Conclusion: (reject or do not reject $$H_0$$)
Does the test indicate an association between gender and preferred award? (yes/no)

2021-01-20
Step 1
From the provided information,
The hypotheses can be constructed as:
$$H_0$$: There is no association between gender and preferred award.
$$H_1$$: There is association between gender and preferred award.
Assume the level of significance $$(\alpha) = 0.05$$
Step 2
The expected frequencies can be obtained as:
$$E_11=\frac{\text{Row total} \times \text{Column total}}{\text{Total}} =\frac{169 \times 31}{362}$$
$$=14.47$$
$$E_12=\frac{169 \times 149}{362}$$
$$=69.56$$
$$E_13=\frac{169 \times182}{362}$$
$$=84.97$$
$$E_21=\frac{193 \times 31}{362}$$
$$=16.53$$
$$E_22=\frac{193 \times 149}{362}$$
$$=79.44$$
$$E_23=\frac{193 \times 182}{362}$$
$$=97.03$$ Step 3 The value of chi square test statistic can be obtained as: $$x^2=\sum \frac{(O-E)^2}{E}$$
$$=\left[ \frac{(20-14.47)^2}{14.47}+\frac{(76-69.56)^2}{69.56}+\frac{(73-84.97)^2}{84.97}+\frac{(11-16.53)^2}{16.53}+\frac{(73-79.44)^2}{79.44}+\frac{(109-97.03)^2}{97.03}\right]$$
the chi square test statistic is 8.245
Step 4
The degree of freedom $$=(r-1) \cdot (c-1) = (2-1) \cdot (3-1) = 2$$ The p value of the test statistic with 2 degree of freedom from the chi square table is 0.0162 which is less than level of significance, therefore, the null hypothesis would reject and it can be concluded that there is sufficient evidence to support the claim that there is some association between gender and preferred award.
Yes, the test indicates an association between gender and preferred award.

### Relevant Questions

A random sample of $$n_1 = 14$$ winter days in Denver gave a sample mean pollution index $$x_1 = 43$$.
Previous studies show that $$\sigma_1 = 19$$.
For Englewood (a suburb of Denver), a random sample of $$n_2 = 12$$ winter days gave a sample mean pollution index of $$x_2 = 37$$.
Previous studies show that $$\sigma_2 = 13$$.
Assume the pollution index is normally distributed in both Englewood and Denver.
(a) State the null and alternate hypotheses.
$$H_0:\mu_1=\mu_2.\mu_1>\mu_2$$
$$H_0:\mu_1<\mu_2.\mu_1=\mu_2$$
$$H_0:\mu_1=\mu_2.\mu_1<\mu_2$$
$$H_0:\mu_1=\mu_2.\mu_1\neq\mu_2$$
(b) What sampling distribution will you use? What assumptions are you making? NKS The Student's t. We assume that both population distributions are approximately normal with known standard deviations.
The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations.
The standard normal. We assume that both population distributions are approximately normal with known standard deviations.
The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations.
(c) What is the value of the sample test statistic? Compute the corresponding z or t value as appropriate.
(Test the difference $$\mu_1 - \mu_2$$. Round your answer to two decimal places.) NKS (d) Find (or estimate) the P-value. (Round your answer to four decimal places.)
(e) Based on your answers in parts (i)−(iii), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level \alpha?
At the $$\alpha = 0.01$$ level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
At the $$\alpha = 0.01$$ level, we reject the null hypothesis and conclude the data are statistically significant.
At the $$\alpha = 0.01$$ level, we fail to reject the null hypothesis and conclude the data are statistically significant.
At the $$\alpha = 0.01$$ level, we reject the null hypothesis and conclude the data are not statistically significant.
(f) Interpret your conclusion in the context of the application.
Reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver. (g) Find a 99% confidence interval for
$$\mu_1 - \mu_2$$.
lower limit
upper limit
(h) Explain the meaning of the confidence interval in the context of the problem.
Because the interval contains only positive numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, we can not say that the mean population pollution index for Englewood is different than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains only negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is less than that of Denver.
How are the smoking habits of students related to their parents' smoking? Here is a two-way table from a survey of student s in eight Arizona high schools:
$$\begin{array}{c|c}&\text{Student smokes}&\text{Student does not smoke}&\text{Total}\\\hline\text{Both parents smoke}&400&1380&400+1380=1780\\\hline\text{One parent smokes}&416&1823&416+1823=2239\\\hline\text{Neither parent smokes}&188&1168&188+1168=1356\\\hline\text{Total}&400+416+188=1004&1380+1823+1168=4371&1004+4371=5375\end{array}$$
(a) Write the null and alternative hypotheses for the question of interest.
(b) Find the expected cell counts. Write a sentence that explains in simple language what "expected counts" are.
(c) Find the chi-square statistic, its degrees of freedom, and the P-value.
The following table gives a two-way classification of all basketball players at a state university who began their college careers between 2004 and 2008, based on gender and whether or not they graduated.
$$\begin{array}{|c|c|c|}\hline &\text{Graduated}&\text{Did not Graduate}\\\hline \text{Male} &129&51\\ \hline \text{Female}&134&36 \\ \hline \end{array}\\$$
If one of these players is selected at random, find the following probability.
$$P(\text{graduated or male})=$$ Enter your answer in accordance to the question statement
You randomly survey shoppers at a supermarket about whether they use reusable bags. Of 60 male shoppers 15 use reusable bags. Of 110 female shoppers, 60 use reusable bags. Organize your results in a two-way table. Include the marginal frequencies.
In a General Social Survey of Americans in 1991, two variables, gender and finding life exciting or dull, were measured on 980 individuals. The two-way table below summarizes the results.
Let A = randomly chosen person is female
Let B = randomly chosen person finds life exciting
(a) Find P(A | B)
(b) Are the events A & B independent?
$$\begin{array}{ccc}\text{Original Counts}&\text{Exciting}&\text{Routine}&\text{Dull}&\text{Total}\\\hline \text{Male} &213&200&12&425\\ \text{Female}&221&305&29&555\\ \text{Female}&434&505&41&980 \end{array}$$
Find the expected count and the contribution to the chi-square statistic for the (Group 1, Yes) cell in the two-way table below.
$$\begin{array}{|c|c|c|}\hline&\text{Yes}&\text{No}&\text{Total}\\\hline\text{Group 1} &710 & 277 & 987\\ \hline\text{Group 2}& 1175 & 323&1498\\\hline \ \text{Total}&1885&600&2485 \\ \hline \end{array}$$
Round your answer for the excepted count to one decimal place, and your answer for the contribution to the chi-square statistic to three decimal places.
Expected count=?
contribution to the chi-square statistic=?
A survey of 4826 randomly selected young adults (aged 19 to 25 ) asked, "What do you think are the chances you will have much more than a middle-class income at age 30? The two-way table summarizes the responses.
$$\begin{array} {c|cc|c} & \text { Female } & \text { Male } & \text { Total } \\ \hline \text { Almost no chance } & 96 & 98 & 194 \\ \hline \text { Some chance but probably not } & 426 & 286 & 712 \\ \hline \text { A 50-50 chance } & 696 & 720 & 1416 \\ \hline \text { A good chance } & 663 & 758 & 1421 \\ \hline \text { Almost certain } & 486 & 597 & 1083 \\ \hline \text { Total } & 2367 & 2459 & 4826 \end{array}$$
Choose a survey respondent at random. Define events G: a good chance, M: male, and N: almost no chance. Given that the chosen student didn't say "almost no chance," what's the probability that this person is female? Write your answer as a probability statement using correct symbols for the events.
$$\begin{array}{|c|c|c|} \hline &\text{Yes}&\text{No}&\text{Total}\\ \hline \text{Group 1} & 56&44&100\\ \hline \text{Group 2}&132&68&200\\ \hline \text{Group 3}&72&28&100\\ \hline \text{Total}&260&140&400\\ \hline \end{array}$$
$$\begin{array} {lc} & \text{Class} \ \text {Survived } & \begin{array}{c|c|c|c} & \text { First } & \text { Second } & \text { Third } \\ \hline \text { Yes } & 197 & 94 & 151 \\ \hline \text { No } & 122 & 167 & 476 \end{array}\ \end{array}$$
$$\begin{array}{c|ccc|c} &\text { Green } & \text { Blue }& \text { Yellow }& \text { Total }\\ \hline \text{ Male } & 40&24&16&80\\ \text{ Female} &30&60&30&120\\ \hline \text{Total}&70&84&46&200 \end{array}\$$