Question # 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}

Comparing two groups
ANSWERED $$\begin{array}{|c|c|} \hline & Housework Hours \\ \hline Gender & Sample\ Size & Mean & Standard\ Deviation \\ \hline Women & 473473 & 33.133.1 & 14.214.2 \\ \hline Men & 488488 & 18.618.6 & 15.715.7 \\ \end{array}$$

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 ​$$\displaystyle{\left(\mu_{{W}}-\mu_{{M}}​\right)}$$ 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 $$\displaystyle{Z}={\frac{{{x}_{{1}}-{x}_{{2}}}}{{\sqrt{{\frac{{{s}_{{1}}^{{2}}}}{{n}_{{1}}}+\frac{{{s}_{{2}}^{{2}}}}{{n}_{{2}}}}}}}}$$

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)