Ch8 A.

3.

Let W = women’s height and M = men’s height. From the information given in the question, we know the following relationship

\[\begin{align} W = 0.92\cdot M \end{align}\]

If we plot this equation (relationship), we will have the following figure:

## [1] The linear correlation coefficient is 1

We can see all the data points could be connected by a straight line, so there is a perfect and positive linear correlation between women’s height and men’s height, and the linear correlation coefficient, r=1

Ch27 B.

3.

We first want to test if there is any significant difference between the percentage in 1970 and in 2005:

\[\begin{align} H_0: p_{1970} - p_{2005} = 0 \\ H_a: p_{1970} - p_{2005} \ne 0 \end{align}\]

We were also given

\(\hat{p}_{1970} = 59%\), \(\hat{p}_{2005} = 35%\), and \(n_{1975}=n_{2005}=1000\)

As the result, we know \(\hat{p}_C = \frac{1000\cdot 0.59 + 1000\cdot 0.35}{1000+1000}=0.47\)

Because two samples are independent and \(n_{1970}\cdot\hat{p}_{1970}\), \(n_{1970}\cdot\left(1-\hat{p}_{1970}\right)\), \(n_{2005}\cdot\hat{p}_{2005}\), and \(n_{2005}\cdot\left(1-\hat{p}_{2005}\right)\) are all at least \(10\), we can perform a 2-sample proportion z test to test our hypotheses stated above.

\[\begin{align} z &= \frac{\hat{p}_{1970}-\hat{p}_{2005}-0}{\sqrt{\hat{p}_C \cdot \left(1-\hat{p}_C\right)\cdot\left(\frac{1}{n_{1975}} + \frac{1}{n_{2005}}\right)}} \\ &= \frac{.59-.35-0}{\sqrt{0.47\cdot(1-0.47)\cdot\left(\frac{1}{1000}+\frac{1}{1000}\right)}} \\ &= 10.7525 \end{align}\]

Since \(z=10.7525\) is way greater than \(3\), which implies the p value is \(< 0.001\), so the difference is hard to be explained by chance.

## $p_common
## [1] 0.47
## 
## $Diff_Percentage
## [1] 0.24
## 
## $SE
## [1] 0.02232039
## 
## $z_score
## [1] 10.7525

Also because the z score is large, so it also implies practical significance in the difference.

Ch 27 D.

3

From the question, we were given the total number of form A and form B are:

\[\begin{align} n_A = 112 + 84 = 196 \\ n_B = 84 + 17 = 101 \\ \end{align}\]

Then for form A and form B, the percentage favored radiation are:

\[\begin{align} \hat{p}_A = \frac{84}{196} \\ \hat{p}_B = \frac{17}{101} \\ \end{align}\]

As the result, we know \(\hat{p}_C = \frac{84 + 17}{196+101}=\frac{101}{297}\)

Now we want to test

\[\begin{align} H_0: p_{A} - p_{B} = 0 \\ H_a: p_{A} - p_{B} \ne 0 \end{align}\]

Because two samples are independent and \(n_{A}\cdot\hat{p}_{A}\), \(n_{A}\cdot\left(1-\hat{p}_{A}\right)\), \(n_{B}\cdot\hat{p}_{B}\), and \(n_{B}\cdot\left(1-\hat{p}_{B}\right)\) are all at least \(10\), we can perform a 2-sample proportion z test to test our hypotheses stated above.

\[\begin{align} z &= \frac{\hat{p}_{A}-\hat{p}_{B}-0}{\sqrt{\hat{p}_C \cdot \left(1-\hat{p}_C\right)\cdot\left(\frac{1}{n_{A}} + \frac{1}{n_{B}}\right)}} \\ &= \frac{\frac{84}{196}-\frac{17}{101}-0}{\sqrt{\frac{101}{297}\cdot(1-\frac{101}{297})\cdot\left(\frac{1}{196}+\frac{1}{101}\right)}} \\ &= 4.485147 \end{align}\]

Since \(z=4.485147\) is way greater than \(3\), which implies the p value is \(< 0.01\), so the difference is hard to be explained by chance.

## $p_common
## [1] 0.3400673
## 
## $Diff_Percentage
## [1] 0.2602546
## 
## $SE
## [1] 0.05802589
## 
## $z_score
## [1] 4.485147

Note, the solution states that \(z=5\) because the percentages was rounded before calculation of z

Note 2, also in the solution, the percentages were who favored surgery, but the hint in the question implies that we should use the percentage favored radiation