# 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|}hlinetext{Leisure Sport}&text{Golf}&text{Tennis}&text{Skiing}&text{Total}hlinetext{Men} &32 & 21 & 30&83 hlinetext{Women}& 35 & 30&26&91hline 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}

Question
Two-way tables
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.)
$$1) \frac{83}{174}$$
$$2) \frac{56}{174}4$$
$$3) \frac{4648}{174}$$
$$4) \frac{4648}{174^2}$$

2021-02-14

Step 1
From the table, we see that there are total 83 men.
There are total 174 persons.
So the probability of getting a man $$\text{P(men)}=\frac{83}{174}$$ From the table we see that total 56 people are skiing. So the probability of skiing is
$$\text{P(Skiing)}=\frac{56}{174}$$
Step 2
Plugging the probabilities in the given expression we get:
$$\text{P(men)}\cdot\text{P(skiing)}$$
$$=\frac{83}{174} \times \frac{56}{174}$$
$$=\frac{4648}{174^2}$$

Answer: $$4)\frac{4648}{174^2}$$

### Relevant Questions

1. Who seems to have more variability in their shoe sizes, men or women?
a) Men
b) Women
c) Neither group show variability
d) Flag this Question
2. In general, why use the estimate of $$n-1$$ rather than n in the computation of the standard deviation and variance?
a) The estimate n-1 is better because it is used for calculating the population variance and standard deviation
b) The estimate n-1 is never used to calculate the sample variance and standard deviation
c) $$n-1$$ provides an unbiased estimate of the population and allows more variability when using a sample and gives a better mathematical estimate of the population
d) The estimate n-1 is better because it is use for calculation of both the population and sample variance as well as standard deviation.
$$\begin{array}{|c|c|}\hline \text{Shoe Size (in cm)} & \text{Gender (M of F)} \\ \hline 25.7 & M \\ \hline 25.4 & F \\ \hline 23.8 & F \\ \hline 25.4 & F \\ \hline 26.7 & M \\ \hline 23.8 & F \\ \hline 25.4 & F \\ \hline 25.4 & F \\ \hline 25.7 & M \\ \hline 25.7 & F \\ \hline 23.5 & F \\ \hline 23.1 & F \\ \hline 26 & M \\ \hline 23.5 & F \\ \hline 26.7 & F \\ \hline 26 & M \\ \hline 23.1 & F \\ \hline 25.1 & F \\ \hline 27 & M \\ \hline 25.4 & F \\ \hline 23.5 & F \\ \hline 23.8 & F \\ \hline 27 & M \\ \hline 25.7 & F \\ \hline \end{array}$$
$$\begin{array}{|c|c|}\hline \text{Shoe Size (in cm)} & \text{Gender (M of F)} \\ \hline 27.6 & M \\ \hline 26.9 & F \\ \hline 26 & F \\ \hline 28.4 & M \\ \hline 23.5 & F \\ \hline 27 & F \\ \hline 25.1 & F \\ \hline 28.4 & M \\ \hline 23.1 & F \\ \hline 23.8 & F \\ \hline 26 & F \\ \hline 25.4 & M \\ \hline 23.8 & F \\ \hline 24.8 & M \\ \hline 25.1 & F \\ \hline 24.8 & F \\ \hline 26 & M \\ \hline 25.4 & F \\ \hline 26 & M \\ \hline 27 & M \\ \hline 25.7 & F \\ \hline 27 & M \\ \hline 23.5 & F \\ \hline 29 & F \\ \hline \end{array}$$
$$\begin{array}{cc}\hline & \text{Afraid to walk at night?} \\ \hline & \text{Yes} & \text{No} & \text{Total} \\ \hline \text{Male} & 173 & 598 & 771 \\ \hline \text{Female} & 393 & 540 & 933 \\ \hline \text{Total} & 566 & 1138 & 1704 \\ \hline \text{Source:}2014\ GSS \end{array}$$
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.
$$P>0.250$$
$$P=0.01$$
$$P<0.001$$
$$P<0.002$$

$$\begin{array}{|c|cc|}\hline & \text{Right-Tail Probability} \\ \hline df & 0.250 & 0.100 & 0.050 & 0.025 & 0.010 & 0.005 & 0.001 \\ \hline 1 & 1.32 & 2.71 & 3.84 & 5.02 & 6.63 & 7.88 & 10.83 \\ 2 & 2.77 & 4.61 & 5.99 & 7.38 & 9.21 & 10.60 & 13.82 \\ 3 & 4.11 & 6.25 & 7.81 & 9.35 & 11.34 & 12.84 & 16.27 \\ 4 & 5.39 & 7.78 & 9.49 & 11.14 & 13.28 & 14.86 & 18.47 \\ 5 & 6.63 & 9.24 & 11.07 & 12.83 & 15.09 & 16.75 & 20.52 \\ 6&7.84&10.64&12.59&14.45&16.81&18.55&22.46 \\ 7&9.04&12.02&14.07&16.01&18.48&20.28&24.32\\ 8&10.22&13.36&15.51&17.53&20.09&21.96&26.12 \\ 9&11.39&14.68&16.92&19.02&21.67&23.59&27.88 \\ 10&12.55&15.99&18.31&20.48&23.21&25.19&29.59 \\ 11&13.70&17.28&19.68&21.92&24.72&26.76&31.26 \\ 12&14.85&18.55&21.03&23.34&26.22&28.30&32.91 \\ 13&15.98&19.81&22.36 & 24.74 & 27.69 & 29.82 & 34.53 \\ 14 & 17.12 & 21.06 & 23.68 & 26.12 & 29.14 & 31.32 & 36.12 \\15 & 18.25 & 22.31 & 25.00 & 27.49 & 30.58 & 32.80 & 37.70 \\ 16 & 19.37 & 32.54 & 26.30 & 28.85 & 32.00 & 34.27 & 39.25 \\ 17 & 20.49 & 24.77 & 27.59 & 30.19 & 33.41 & 35.72 & 40.79 \\ 18 & 21.60 & 25.99 & 28.87 & 31.53 & 34.81 & 37.16 & 42.31 \\ 19 & 22.72 & 27.20 & 30.14 & 32.85 & 36.19 & 38.58 & 43.82 \\ 20 & 23.83 & 28.41 & 31.41 & 34.17 & 37.57 & 40.00 & 45.32 \\ \hline \end{array}$$
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?
$$a) \frac{250}{1000}$$
$$b) \frac{250}{2500}$$
$$c) \frac{475}{2500}$$
$$d) \frac{250}{475}$$
$$e) \frac{225}{475}$$
Consider the accompanying data on flexural strength (MPa) for concrete beams of a certain type.
$$\begin{array}{|c|c|}\hline 11.8 & 7.7 & 6.5 & 6 .8& 9.7 & 6.8 & 7.3 \\ \hline 7.9 & 9.7 & 8.7 & 8.1 & 8.5 & 6.3 & 7.0 \\ \hline 7.3 & 7.4 & 5.3 & 9.0 & 8.1 & 11.3 & 6.3 \\ \hline 7.2 & 7.7 & 7.8 & 11.6 & 10.7 & 7.0 \\ \hline \end{array}$$
a) Calculate a point estimate of the mean value of strength for the conceptual population of all beams manufactured in this fashion. $$[Hint.\ ?x_{j}=219.5.]$$ (Round your answer to three decimal places.)
MPa
State which estimator you used.
$$x$$
$$p?$$
$$\frac{s}{x}$$
$$s$$
$$\tilde{\chi}$$
b) Calculate a point estimate of the strength value that separates the weakest $$50\%$$ of all such beams from the strongest $$50\%$$.
MPa
State which estimator you used.
$$s$$
$$x$$
$$p?$$
$$\tilde{\chi}$$
$$\frac{s}{x}$$
c) Calculate a point estimate of the population standard deviation ?. $$[Hint:\ ?x_{i}2 = 1859.53.]$$ (Round your answer to three decimal places.)
MPa
Interpret this point estimate.
This estimate describes the linearity of the data.
This estimate describes the bias of the data.
This estimate describes the spread of the data.
This estimate describes the center of the data.
Which estimator did you use?
$$\tilde{\chi}$$
$$x$$
$$s$$
$$\frac{s}{x}$$
$$p?$$
d) Calculate a point estimate of the proportion of all such beams whose flexural strength exceeds 10 MPa. [Hint: Think of an observation as a "success" if it exceeds 10.] (Round your answer to three decimal places.)
e) Calculate a point estimate of the population coefficient of variation $$\frac{?}{?}$$. (Round your answer to four decimal places.)
State which estimator you used.
$$p?$$
$$\tilde{\chi}$$
$$s$$
$$\frac{s}{x}$$
$$x$$
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=?
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?
A survey of 120 students about which sport , baseball , basketball , football ,hockey , or other , they prefer to watch on TV yielded the following two-way frequency table . What is the conditional relative frequency that a student prefers to watch baseball , given that the student is a girl? Round the answer to two decimal places as needed
$$\begin{array}{|c|c|c|}\hline &\text{Baseball}&\text{Basketball}&\text{Football}&\text{Hockey}&\text{Other}&\text{Total}\\\hline \text{Boys} &18&14&20&6&2&60\\ \hline \text{Girls}&14&16&13&5&12&60\\ \hline \text{Total}&32&30&33&11&14&120\\ \hline \end{array}\\$$
a) 11.67%
b) 23.33%
c) 43.75%
d) 53.33%
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?
$$\begin{array}{ccc}\text{Original Counts}&\text{Exciting}&\text{Routine}&\text{Dull}&\text{Total}\\\hline \text{Male} &213&200&12&425\\ \text{Female}&221&305&29&555\\ \text{Female}&434&505&41&980 \end{array}$$
$$\begin{array}{|c|c|c|c|}\hline&\text{Intramural Basketball}&\text{Chess Club}&\text{Jazz Band}&\text{Not Involved}&\text{Total}\\\hline\text{Females} &20 & 10&10&20&60 \\ \hline\text{Males}& 20 & 2&8&10&40\\\hline\text{Total}&40&12&18&30&100\\\hline \end{array}$$ $$a)\frac{2}{10}=0.20$$
$$b)\frac{2}{40}=0.05$$
$$c)\frac{2}{12}\approx0.167$$
$$d)\frac{2}{100}=0.02$$
$$\begin{array}{|c|c|c|}\hline&\text{morning}&\text{afternoon}&\text{total}\\\hline\text{grade 6} &15 & 8 & 23\\ \hline\text{grade 8}& 18 & 21&39\\\hline\text{grade 10}& 12 & 26&38\\ \hline \text{total}&45&55&100 \\ \hline \end{array}$$