# A random sample of 2,500 people was selected, and the people were asked to give their favorite season. Their responses, along with their age group, are summarized in the two-way table below. begin{array}{c|cccc|c} & text {Winter} &text{Spring}& text {Summer } & text {Fall}& text {Total} hline text {Children} & 30 & 0 & 170&0&200 text{Teens} & 150 & 75 & 250&25&500 text {Adults } & 250 & 250 & 250&250&1000 text {Seniors} & 300 & 150 & 50&300&800 hline text {Total} & 730 & 475 & 720 &575&2500 end{array} Among those whose favorite season is spring, what proportion are adults? a) frac{250}{1000} b) frac{250}{2500} c) frac{475}{2500} d) frac{250}{475} e) frac{225}{475}

Question
Two-way tables
A random sample of 2,500 people was selected, and the people were asked to give their favorite season. Their responses, along with their age group, are summarized in the two-way table below.
$$\begin{array}{c|cccc|c} & \text {Winter} &\text{Spring}& \text {Summer } & \text {Fall}& \text {Total}\\ \hline \text {Children} & 30 & 0 & 170&0&200 \\ \text{Teens} & 150 & 75 & 250&25&500 \\ \text {Adults } & 250 & 250 & 250&250&1000 \\ \text {Seniors} & 300 & 150 & 50&300&800 \\ \hline \text {Total} & 730 & 475 & 720 &575&2500 \end{array}$$
Among those whose favorite season is spring, what proportion are adults?
$$a) \frac{250}{1000}$$
$$b) \frac{250}{2500}$$
$$c) \frac{475}{2500}$$
$$d) \frac{250}{475}$$
$$e) \frac{225}{475}$$

2020-11-27

Step 1
Obtain the proportion of adults whose favourite season is spring:
The value of probability is obtained by:
$$P(E)=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$$
$$P(\text{Adults whose favorite season is spring})=\frac{\text{Number of adults whose favorite season is spring}}{\text{Total number of outcomes}}$$
$$=\frac{250}{2500}$$
Step 2
Therefore, the correct answer is “Option B.) $$\frac{250}{2500}$$.

### Relevant Questions

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.
1950 randomly selected adults were asked if they think they are financially better off than their parents. The following table gives the two-way classification of the responses based on the education levels of the persons included in the survey and whether they are financially better off, the same as, or worse off than their parents
$$\begin{array}{|c|c|c|}\hline &\text{Less Than High School}&\text{High School}&\text{More Than High School}\\\hline \text{Better off} &140&440&430\\ \hline \text{Same as}&60&230&110\\ \hline \text{Worse off}&180&280&80\\ \hline\end{array}\\$$
Suppose one adult is selected at random from these 1950 adults. Find the following probablity.
Round your answer to three decimal places.
$$P(\text{more than high school or worse off})=?$$
A group of 125 truck owners were asked what brand of truck they owned and whether or not the truck has four-wheel drive. The results are summarized in the two-way table below. Suppose we randomly select one of these truck owners.
$$\begin{array}{c|cc} & \text { Four-wheel drive} & \text { No four-wheel drive } \\\hline \text { Ford } & 28 & 17 \\ \text { Chevy } & 32 & 18 \\ \text { Dodge } & 20 & 10 \end{array}$$
What is the probability that the person owns a Dodge or has four-wheel drive?
$$(a) \frac{20}{80}$$
$$(b) \frac{20}{125}$$
$$(c) \frac{80}{125}$$
$$(d) \frac{90}{125}$$
$$(e) \frac{110}{125}$$
Men and women were surveyed regarding their favorite leisure sport, as shown below. All questions pertain to this two-way frequency table.
$$\begin{array}{|c|c|c|}\hline\text{Leisure Sport}&\text{Golf}&\text{Tennis}&\text{Skiing}&\text{Total}\\\hline\text{Men} &32 & 21 & 30&83\\ \hline\text{Women}& 35 & 30&26&91\\\hline \ \text{Total}&67&51&56&174 \\ \hline \end{array}$$
$$\text{Find P(men)} \cdot \text{P(skiing)}.$$
(Choose a numbered choice from the list below.)
$$1) \frac{83}{174}$$
$$2) \frac{56}{174}4$$
$$3) \frac{4648}{174}$$
$$4) \frac{4648}{174^2}$$
In this exercise , a two-way table is shown for two groups , 1 and 2 , and two possible outcomes , A nad B $$\begin{array}{|c|c|c|}\hline &\text{Outcome A}&\text{Outcome B}&\text{Total}\\\hline \text{Group 1} &30&20&50\\ \hline \text{Group 2}&40&110&150\\ \hline \text{Total}&70&130&200\\ \hline \end{array}\\$$
a) What proportion of all cases had Outcome A?
b) What proportion of all cases are in Group 1?
c) What proportion of cases in group 1 had Outcome B?
d) What proportion of cases who had Outcome A were in group 2?
Use the two-way table below to answer the following question. Round the answer to the nearest whole percent. There are 300 seniors at Bellmere Senior High School who are enrolled in elective science classes as shown below.
$$\begin{array}{c|cc|c} &\text{Physics}&\text{Chemistry}&\text{Total}\\ \hline \text{Males}&100&68&168\\ \text{Females}&71&61&132\\ \hline \text{Total}&171&129&300 \end{array}\$$
What is the probability that a senior student chosen at random is enrolled in physics given that student is female?

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
$$\mu_1 - \mu_2$$.
(Round your answers to two decimal places.)
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.
$$\begin{array}{|c|c|c|c|}\hline&\text{Intramural Basketball}&\text{Chess Club}&\text{Jazz Band}&\text{Not Involved}&\text{Total}\\\hline\text{Females} &20 & 10&10&20&60 \\ \hline\text{Males}& 20 & 2&8&10&40\\\hline\text{Total}&40&12&18&30&100\\\hline \end{array}$$ $$a)\frac{2}{10}=0.20$$
$$b)\frac{2}{40}=0.05$$
$$c)\frac{2}{12}\approx0.167$$
$$d)\frac{2}{100}=0.02$$
$$\begin{array}{c|c}&Yes&No\\\hline\text{Children}&0.15&0.25\\\hline\text{Adults}&0.1&0.6\end{array}$$
$$\begin{array}{c|ccc|c} & 18-34 & 35-54 & 55+ & \text { Total } \\ \hline \text { iPhone } & 169 & 171 & 127 & 467 \\ \text { Androod } & 214 & 189 & 100 & 503 \\ \text { Other } & 134 & 277 & 643 & 1054 \\ \hline \text { Total } & 517 & 637 & 870 & 2024 \end{array}$$