# begin{array}{|c|c|} hline & Housework Hours hline Gender & Sample Size & Mean & Standard Deviation hline Women & 473473 & 33.133.1 & 14.214.2 hline Men & 488488 & 18.618.6 & 15.715.7 end{array} a. Based on this​ study, calculate how many more hours per​ week, on the​ average, women spend on housework than men. b. Find the standard error for comparing the means. What factor causes the standard error to be small compared to the sample standard deviations for the two​ groups? The cause the standard error to be small compared to the sample standard deviations for the two groups. c. Calculate the​ 95% confidence interval comparing the population means for women Interpret the result including the relevance of 0 being within the interval or not. The​ 95% confidence interval for ​(mu_W - mu_M​

Question
Comparing two groups
$$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}&{H}{o}{u}{s}{e}{w}{\quad\text{or}\quad}{k}{H}{o}{u}{r}{s}\backslash{h}{l}\in{e}{G}{e}{n}{d}{e}{r}&{S}{a}\mp\le\ {S}{i}{z}{e}&{M}{e}{a}{n}&{S}{\tan{{d}}}{a}{r}{d}\ {D}{e}{v}{i}{a}{t}{i}{o}{n}\backslash{h}{l}\in{e}{W}{o}{m}{e}{n}&{473473}&{33.133}{.1}&{14.214}{.2}\backslash{h}{l}\in{e}{M}{e}{n}&{488488}&{18.618}{.6}&{15.715}{.7}\backslash{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$ a. Based on this​ study, calculate how many more hours per​ week, on the​ average, women spend on housework than men. b. Find the standard error for comparing the means. What factor causes the standard error to be small compared to the sample standard deviations for the two​ groups? The cause the standard error to be small compared to the sample standard deviations for the two groups. c. Calculate the​ 95% confidence interval comparing the population means for women Interpret the result including the relevance of 0 being within the interval or not. The​ 95% confidence interval for ​$$\displaystyle{\left(\mu_{{W}}-\mu_{{M}}​\right)}$$ is: (Round to two decimal places as​ needed.) The values in the​ 95% confidence interval are less than 0, are greater than 0, include 0, which implies that the population mean for women could be the same as is less than is greater than the population mean for men. d. State the assumptions upon which the interval in part c is based. Upon which assumptions below is the interval​ based? Select all that apply. A.The standard deviations of the two populations are approximately equal. B.The population distribution for each group is approximately normal. C.The samples from the two groups are independent. D.The samples from the two groups are random.

2021-01-28
Given data for women mean(x1) = 33133.1 std(s1) = 14214.1 population(n1) = 473473 for men, mean(x1) = 18618.6 std(s1) = 15.715.7 population(n1) = 488488 a) how many more hours per​ week, on the​ average, women spend on housework than men. = 488488 - 473473 = 15015, actually less b) We use Z test denominator is standard error Z = \frac{x1 - x2}{\sqrt{s1^2/n1 + s2^2/n2}} standard error= 30.53 c) Z = 491.7 which is far above critical value at Z = 0.95 i.e 1.645 d) The samples are independent because of large difference (C)

### Relevant Questions

Iron is very important for babies' growth. A common belief is that breastfeeding will help the baby to get more iron than formula feeding. To justify the belief, a study followed 2 groups of babies from born to 6 months. With one group babies are breast fed, and the other group are formula fed without iron supplements. Data below shows iron levels of those two groups of babies. $$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}{G}{r}{o}{u}{p}&{S}{a}\mp\le\ {s}{i}{z}{e}&{m}{e}{a}{n}&{S}{\tan{{d}}}{a}{r}{d}\ {d}{e}{v}{i}{a}{t}{i}{o}{n}\backslash{h}{l}\in{e}{B}{r}{e}\ast-{f}{e}{d}&{23}&{13.3}&{1.7}\backslash{h}{l}\in{e}{F}{\quad\text{or}\quad}\mu{l}{a}-{f}{e}{d}&{23}&{12.4}&{1.8}\backslash{h}{l}\in{e}{D}{I}{F}{F}={B}{r}{e}\ast-{F}{\quad\text{or}\quad}\mu{l}{a}&{23}&{0.9}&{1.4}\backslash{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$ (1) There are two groups we need to compare for the study: Breast-Fed and Formula- Fed. Are those two groups dependent or independent? Based on your answer, what inference procedure should we apply for this research? (2) Please perform the inference you decided in (1), and make sure to follow the 5-step procedure for any hypothesis test. (3) Based on your conclusion in (2), what kind of error could you make? Explain the type of error using the context words for this research
A two-sample inference deals with dependent and independent inferences. In a two-sample hypothesis testing problem, underlying parameters of two different populations are compared. In a longitudinal (or follow-up) study, the same group of people is followed over time. Two samples are said to be paired when each data point in the first sample is matched and related to a unique data point in the second sample.
This problem demonstrates inference from two dependent (follow-up) samples using the data from the hypothetical study of new cases of tuberculosis (TB) before and after the vaccination was done in several geographical areas in a country in sub-Saharan Africa. Conclusion about the null hypothesis is to note the difference between samples.
The problem that demonstrates inference from two dependent samples uses hypothetical data from the TB vaccinations and the number of new cases before and after vaccination. PSK\begin{array}{|c|c|} \hline Geographical\ regions & Before\ vaccination & After\ vaccination\\ \hline 1 & 85 & 11\\ \hline 2 & 77 & 5\\ \hline 3 & 110 & 14\\ \hline 4 & 65 & 12\\ \hline 5 & 81 & 10\\\hline 6 & 70 & 7\\ \hline 7 & 74 & 8\\ \hline 8 & 84 & 11\\ \hline 9 & 90 & 9\\ \hline 10 & 95 & 8\\ \hline \end{array}ZSK
Using the Minitab statistical analysis program to enter the data and perform the analysis, complete the following: Construct a one-sided $$\displaystyle{95}\%$$ confidence interval for the true difference in population means. Test the null hypothesis that the population means are identical at the 0.05 level of significance.
A new thermostat has been engineered for the frozen food cases in large supermarkets. Both the old and new thermostats hold temperatures at an average of $$25^{\circ}F$$. However, it is hoped that the new thermostat might be more dependable in the sense that it will hold temperatures closer to $$25^{\circ}F$$. One frozen food case was equipped with the new thermostat, and a random sample of 21 temperature readings gave a sample variance of 5.1. Another similar frozen food case was equipped with the old thermostat, and a random sample of 19 temperature readings gave a sample variance of 12.8. Test the claim that the population variance of the old thermostat temperature readings is larger than that for the new thermostat. Use a $$5\%$$ level of significance. How could your test conclusion relate to the question regarding the dependability of the temperature readings? (Let population 1 refer to data from the old thermostat.)
(a) What is the level of significance?
State the null and alternate hypotheses.
$$H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}>?_{2}^{2}H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}\neq?_{2}^{2}H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}?_{2}^{2},H1:?_{1}^{2}=?_{2}^{2}$$
(b) Find the value of the sample F statistic. (Round your answer to two decimal places.)
What are the degrees of freedom?
$$df_{N} = ?$$
$$df_{D} = ?$$
What assumptions are you making about the original distribution?
The populations follow independent normal distributions. We have random samples from each population.The populations follow dependent normal distributions. We have random samples from each population.The populations follow independent normal distributions.The populations follow independent chi-square distributions. We have random samples from each population.
(c) Find or estimate the P-value of the sample test statistic. (Round your answer to four decimal places.)
(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?
At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant. At the ? = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.
(e) Interpret your conclusion in the context of the application.
Reject the null hypothesis, there is sufficient evidence that the population variance is larger in the old thermostat temperature readings.Fail to reject the null hypothesis, there is sufficient evidence that the population variance is larger in the old thermostat temperature readings. Fail to reject the null hypothesis, there is insufficient evidence that the population variance is larger in the old thermostat temperature readings.Reject the null hypothesis, there is insufficient evidence that the population variance is larger in the old thermostat temperature readings.
Would you rather spend more federal taxes on art? Of a random sample of $$n_{1} = 86$$ politically conservative voters, $$r_{1} = 18$$ responded yes. Another random sample of $$n_{2} = 85$$ politically moderate voters showed that $$r_{2} = 21$$ responded yes. Does this information indicate that the population proportion of conservative voters inclined to spend more federal tax money on funding the arts is less than the proportion of moderate voters so inclined? Use $$\alpha = 0.05.$$ (a) State the null and alternate hypotheses. $$H_0:p_{1} = p_{2}, H_{1}:p_{1} > p_2$$
$$H_0:p_{1} = p_{2}, H_{1}:p_{1} < p_2$$
$$H_0:p_{1} = p_{2}, H_{1}:p_{1} \neq p_2$$
$$H_{0}:p_{1} < p_{2}, H_{1}:p_{1} = p_{2}$$ (b) What sampling distribution will you use? What assumptions are you making? The Student's t. The number of trials is sufficiently large. The standard normal. The number of trials is sufficiently large.The standard normal. We assume the population distributions are approximately normal. The Student's t. We assume the population distributions are approximately normal. (c)What is the value of the sample test statistic? (Test the difference $$p_{1} - p_{2}$$. Do not use rounded values. Round your final answer to two decimal places.) (d) Find (or estimate) the P-value. (Round your answer to four decimal places.) (e) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level alpha? At the $$\alpha = 0.05$$ level, we reject the null hypothesis and conclude the data are statistically significant. At the $$\alpha = 0.05$$ level, we fail to reject the null hypothesis and conclude the data are statistically significant. At the $$\alpha = 0.05$$ level, we fail to reject the null hypothesis and conclude the data are not statistically significant. At the $$\alpha = 0.05$$ 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 sufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters. Fail to reject the null hypothesis, there is sufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters. Fail to reject the null hypothesis, there is insufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters. Reject the null hypothesis, there is insufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters.
1. A researcher is interested in finding a 98% confidence interval for the mean number of times per day that college students text. The study included 144 students who averaged 44.7 texts per day. The standard deviation was 16.5 texts. a. To compute the confidence interval use a ? z t distribution. b. With 98% confidence the population mean number of texts per day is between and texts. c. If many groups of 144 randomly selected members are studied, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population number of texts per day and about percent will not contain the true population mean number of texts per day. 2. You want to obtain a sample to estimate how much parents spend on their kids birthday parties. Based on previous study, you believe the population standard deviation is approximately $$\displaystyle\sigma={40.4}$$ dollars. You would like to be 90% confident that your estimate is within 1.5 dollar(s) of average spending on the birthday parties. How many parents do you have to sample? n = 3. You want to obtain a sample to estimate a population mean. Based on previous evidence, you believe the population standard deviation is approximately $$\displaystyle\sigma={57.5}$$. You would like to be 95% confident that your estimate is within 0.1 of the true population mean. How large of a sample size is required?
The table below shows the number of people for three different race groups who were shot by police that were either armed or unarmed. These values are very close to the exact numbers. They have been changed slightly for each student to get a unique problem.
Suspect was Armed:
Black - 543
White - 1176
Hispanic - 378
Total - 2097
Suspect was unarmed:
Black - 60
White - 67
Hispanic - 38
Total - 165
Total:
Black - 603
White - 1243
Hispanic - 416
Total - 2262
Give your answer as a decimal to at least three decimal places.
a) What percent are Black?
b) What percent are Unarmed?
c) In order for two variables to be Independent of each other, the P $$(A and B) = P(A) \cdot P(B) P(A and B) = P(A) \cdot P(B).$$
This just means that the percentage of times that both things happen equals the individual percentages multiplied together (Only if they are Independent of each other).
Therefore, if a person's race is independent of whether they were killed being unarmed then the percentage of black people that are killed while being unarmed should equal the percentage of blacks times the percentage of Unarmed. Let's check this. Multiply your answer to part a (percentage of blacks) by your answer to part b (percentage of unarmed).
Remember, the previous answer is only correct if the variables are Independent.
d) Now let's get the real percent that are Black and Unarmed by using the table?
If answer c is "significantly different" than answer d, then that means that there could be a different percentage of unarmed people being shot based on race. We will check this out later in the course.
Let's compare the percentage of unarmed shot for each race.
e) What percent are White and Unarmed?
f) What percent are Hispanic and Unarmed?
If you compare answers d, e and f it shows the highest percentage of unarmed people being shot is most likely white.
Why is that?
This is because there are more white people in the United States than any other race and therefore there are likely to be more white people in the table. Since there are more white people in the table, there most likely would be more white and unarmed people shot by police than any other race. This pulls the percentage of white and unarmed up. In addition, there most likely would be more white and armed shot by police. All the percentages for white people would be higher, because there are more white people. For example, the table contains very few Hispanic people, and the percentage of people in the table that were Hispanic and unarmed is the lowest percentage.
Think of it this way. If you went to a college that was 90% female and 10% male, then females would most likely have the highest percentage of A grades. They would also most likely have the highest percentage of B, C, D and F grades
The correct way to compare is "conditional probability". Conditional probability is getting the probability of something happening, given we are dealing with just the people in a particular group.
g) What percent of blacks shot and killed by police were unarmed?
h) What percent of whites shot and killed by police were unarmed?
i) What percent of Hispanics shot and killed by police were unarmed?
You can see by the answers to part g and h, that the percentage of blacks that were unarmed and killed by police is approximately twice that of whites that were unarmed and killed by police.
j) Why do you believe this is happening?
Do a search on the internet for reasons why blacks are more likely to be killed by police. Read a few articles on the topic. Write your response using the articles as references. Give the websites used in your response. Your answer should be several sentences long with at least one website listed. This part of this problem will be graded after the due date.
A random sample of $$\displaystyle{n}_{{1}}={16}$$ communities in western Kansas gave the following information for people under 25 years of age.
$$\displaystyle{X}_{{1}}:$$ Rate of hay fever per 1000 population for people under 25
$$\begin{array}{|c|c|} \hline 97 & 91 & 121 & 129 & 94 & 123 & 112 &93\\ \hline 125 & 95 & 125 & 117 & 97 & 122 & 127 & 88 \\ \hline \end{array}$$
A random sample of $$\displaystyle{n}_{{2}}={14}$$ regions in western Kansas gave the following information for people over 50 years old.
$$\displaystyle{X}_{{2}}:$$ Rate of hay fever per 1000 population for people over 50
$$\begin{array}{|c|c|} \hline 94 & 109 & 99 & 95 & 113 & 88 & 110\\ \hline 79 & 115 & 100 & 89 & 114 & 85 & 96\\ \hline \end{array}$$
(i) Use a calculator to calculate $$\displaystyle\overline{{x}}_{{1}},{s}_{{1}},\overline{{x}}_{{2}},{\quad\text{and}\quad}{s}_{{2}}.$$ (Round your answers to two decimal places.)
(ii) Assume that the hay fever rate in each age group has an approximately normal distribution. Do the data indicate that the age group over 50 has a lower rate of hay fever? Use $$\displaystyle\alpha={0.05}.$$
(a) What is the level of significance?
State the null and alternate hypotheses.
$$\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}<\mu_{{2}}$$
$$\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}>\mu_{{2}}$$
$$\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}\ne\mu_{{2}}$$
$$\displaystyle{H}_{{0}}:\mu_{{1}}>\mu_{{2}},{H}_{{1}}:\mu_{{1}}=\mu_{{12}}$$
(b) What sampling distribution will you use? What assumptions are you making?
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,
The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations,
The Student's t. We assume that both population distributions are approximately normal with known standard deviations,
What is the value of the sample test statistic? (Test the difference $$\displaystyle\mu_{{1}}-\mu_{{2}}$$. Round your answer to three decimalplaces.)
What is the value of the sample test statistic? (Test the difference $$\displaystyle\mu_{{1}}-\mu_{{2}}$$. Round your answer to three decimal places.)
(c) Find (or estimate) the P-value.
P-value $$\displaystyle>{0.250}$$
$$\displaystyle{0.125}<{P}-\text{value}<{0},{250}$$
$$\displaystyle{0},{050}<{P}-\text{value}<{0},{125}$$
$$\displaystyle{0},{025}<{P}-\text{value}<{0},{050}$$
$$\displaystyle{0},{005}<{P}-\text{value}<{0},{025}$$
P-value $$\displaystyle<{0.005}$$
Sketch the sampling distribution and show the area corresponding to the P-value.
P.vaiue Pevgiue
P-value f P-value
Effect of alcoholic parents A study1 compared personality characteristics between 49 children of alcoholics and a control group of 49 children of nonalcoholics who were matched on age and gender. On a measure of well being, the 49 children of alcoholics had a mean of $$\displaystyle{26.1}{\left({s}={7.2}\right)}$$ and the 49 subjects in the control group had a mean of $$\displaystyle{\left[{28.8}{\left({s}={6.4}\right)}\right]}$$. The difference scores between the matched subjects from the two groups had a mean of $$\displaystyle{2.7}{\left({s}={9.7}\right)}$$.