Chapter 11, Problem 1CS

To determine

To find:

The P-value for given test.

Answer to Problem 1CS

The $p-$ value for this test is $0.264,0.005,0.079,0.025$ .

Explanation of Solution

Given Information:

The results for five factors are presented in the following table.

	Class	Controls
Proportion with high income	$0.09$	$0.13$
Proportion owning a home	$0.72$	$0.76$
Proportion with underground wiring	$0.14$	$0.07$
Proportion with father white non-Hispanic	$0.43$	$0.36$
Proportion with father Hispanic	$0.38$	$0.47$

Calculation:

The z-test statistics is.

By using MINITAB, find z-test statistics with the help of following steps:

1. Import the data.
2. Select $2$ Proportions from Basic Statistics from Stat menu.
3. Enter Summarized data.
4. Click option button choose level of significance and alternative hypothesis
5. Click 0k.

Test and $\text{CI}$ for Two Proportion

  $\begin{array}{l}\text{Sample X N Sample p}\\ 1\text{ 216 300 0}\text{.720000}\\ \text{2 228 300 0}\text{.760000}\end{array}$

Difference $=p\left(1\right)-p\left(2\right)$

Estimate for difference: $-0.04$

  $95\%\text{ CI}$ For difference: $\left(-0.110122,0.0301218\right)$

Test for difference $=0\left(\text{vs not}=0\right):Z=-1.12\text{ P-Value}=0.264$

  $-1.12$ From the MINITAB output, the z-test statistics is,

The $p-$ value for this test is,

  $0.264$ From the MINITAB output, the $p-$ value for this test is,

To test the hypothesis is that the two population proportions of cases and control are differs with respect to owning a home at $5\%$ significance leve.

The null and alternative hypothesis is,

  $\begin{array}{l}{H}_0:p_1=p_2\\ H_a:p_1\ne p_2\end{array}$

The z-test statistics is.

By using MINITAB, find z-test statistics with the help of following steps:

1. Import the data.
2. >Select $2$ Proportions from Basic Statistics from Stat menu.
3. Enter Summarized data.
4. Click option button choose level of significance and alternative hypothesis
5. Click Ok.

Test and $\text{CI}$ for Two Proportion

  $\begin{array}{l}\text{Sample X N Sample p}\\ 1\text{ 42 300 0}\text{.140000}\\ \text{2 21 300 0}\text{.070000}\end{array}$

Difference $=p\left(1\right)-p\left(2\right)$

Estimate for difference: $0.07$

  $95\%\text{ CI}$ For difference: $\left(0.0212629,0.118737\right)$

Test for difference $=0\left(\text{vs not}=0\right):Z=2.82\text{ P-Value}=0.005$

  $2.82$ From the MINITAB output, the z-test statistics is,

The $p-$ value for this test is,

  $0.005$ From the MINITAB output, the $p-$ value for this test is,

To test the hypothesis is that the two population proportions of cases and control are differs with respect to father white non-Hispanic at $5\%$ significance leve.

The null and alternative hypothesis is,

  $\begin{array}{l}{H}_0:p_1=p_2\\ H_a:p_1\ne p_2\end{array}$

The z-test statistics is.

By using MINITAB, find z-test statistics with the help of following steps:

1. Import the data.
2. Select $2$ Proportions from Basic Statistics from Stat menu.
3. Enter Summarized data.
4. Click option button choose level of significance and alternative hypothesis
5. Click Ok.

Test and $\text{CI}$ for Two Proportion

  $\begin{array}{l}\text{Sample X N Sample p}\\ 1\text{ 129 300 0}\text{.430000}\\ \text{2 108 300 0}\text{.360000}\end{array}$

Difference $=p\left(1\right)-p\left(2\right)$

Estimate for difference: $0.07$

  $95\%\text{ CI}$ For difference: $\left(-0.00803020,0.148030\right)$

Test for difference $=0\left(\text{vs not}=0\right):Z=1.76\text{ P-Value}=0.079$

  $1.76$ From the MINITAB output, the z-test statistics is,

The $p-$ value for this test is,

  $0.079$ From the MINITAB output, the $p-$ value for this test is,

To test the hypothesis is that the two population proportions of cases and control are differs with respect to father Hispanic at $5\%$ significance leve.

The null and alternative hypothesis is,

  $\begin{array}{l}{H}_0:p_1=p_2\\ H_a:p_1\ne p_2\end{array}$

The z-test statistics is.

By using MINITAB, find z-test statistics with the help of following steps:

1. Import the data.
2. Select $2$ Proportions from Basic Statistics from Stat menu.
3. Enter Summarized data.
4. Click option button choose level of significance and alternative hypothesis
5. Click Ok.

Test and $\text{CI}$ for Two Proportion

  $\begin{array}{l}\text{Sample X N Sample p}\\ 1\text{ 114 300 0}\text{.380000}\\ \text{2 141 }\text{300 0}\text{.470000}\end{array}$

Difference $=p\left(1\right)-p\left(2\right)$

Estimate for difference: $-0.09$

  $95\%\text{ CI}$ For difference: $\left(-0.168781,-0.0112185\right)$

Test for difference $=0\left(\text{vs not}=0\right):Z=-2.24\text{ P-Value}=0.025$

  $-2.24$ From the MINITAB output, the z-test statistics is,

The $p-$ value for this test is,

  $0.025$ From the MINITAB output, the $p-$ value for this test is,