Chapter 11.4, Problem 9E

To determine

(a)

To find:

The null and alternative hypotheses for given scenario.

Answer to Problem 9E

Solution:

The alternate hypothesis is $p_2-p_1<0$ and the null hypothesis is $p_2-p_1\ge 0$.

Explanation of Solution

Given information:

There is an old wives’ tale that women who eat chocolate during pregnancy are more likely to have happy babies. A pregnancy magazine wants to test this claim, and it gathers 100 randomly selected pregnant women for its study. Half of the women sampled agree to cat chocolate at least once a day, while the other half agree to forego chocolate for the duration of their pregnancies. A year later, the ladies complete a survey regarding the overall happiness of their babies. The results are given in the following table. At the 0.01 level of significance, test the claim of the old wives’ tale.

Number of babies
	Happy babies	Unhappy Babies
With chocolate	24	26
Without chocolate	22	28

Concept:

The null hypothesis is a statement of no difference, that there is no significant difference between the two phenomena. It is considered to be true until it is nullified by statistical evidence for an alternative hypothesis.

An alternative hypothesis is a contradicting statement to the null hypothesis and states a significant difference between the two phenomena It is accepted when the null hypothesis is false.

Assume, be the estimate proportion for the happy babies whose mothers ate chocolate during their pregnancies.

And be the estimate proportion for the happy babies whose mothers did not eat chocolate during their pregnancies.

From the given information, the claim is made that the pregnant women gives happy child after having chocolates, that is,

$\begin{array}{c}{p}_2<p_1\\ p_2-p_1<0\end{array}$

The required alternate hypothesis is $p_2-p_1<0$.

It is known that the null hypothesis for the given proportion is $H_0:p_2-p_1\ge 0$. The required null hypothesis is $p_2-p_1\ge 0$.

To determine

(b)

To find:

The distribution is to be used for the statistical test and its level of significance.

Answer to Problem 9E

Solution:

The standard normal variate z-test will be applied for given samples at the level of significance is $0.01$ or $1\%$.

Explanation of Solution

Given information:

There is an old wives’ tale that women who eat chocolate during pregnancy are more likely to have happy babies. A pregnancy magazine wants to test this claim, and it gathers 100 randomly selected pregnant women for its study. Half of the women sampled agree to cat chocolate at least once a day, while the other half agree to forego chocolate for the duration of their pregnancies. A year later, the ladies complete a survey regarding the overall happiness of their babies. The results are given in the following table. At the 0.01 level of significance, test the claim of the old wives’ tale.

Number of babies
	Happy babies	Unhappy Babies
With chocolate	24	26
Without chocolate	22	28

Since the both sample size is greater than 30, normal distribution will be followed and the standard normal variate z-test will be applied.

According to the null hypothesis, left-tailed test is to be applied.

From the given information, the level of significance is $0.01$ or $1\%$.

To determine

(c)

To find:

The test statistic for a hypothesis test for two population means.

Answer to Problem 9E

Solution:

The test statistic value for given population is $0.401286$.

Explanation of Solution

Given information:

There is an old wives’ tale that women who eat chocolate during pregnancy are more likely to have happy babies. A pregnancy magazine wants to test this claim, and it gathers 100 randomly selected pregnant women for its study. Half of the women sampled agree to cat chocolate at least once a day, while the other half agree to forego chocolate for the duration of their pregnancies. A year later, the ladies complete a survey regarding the overall happiness of their babies. The results are given in the following table. At the 0.01 level of significance, test the claim of the old wives’ tale.

Number of babies
	Happy babies	Unhappy Babies
With chocolate	24	26
Without chocolate	22	28

Formula used:

The formula used to perform a statistic test for a hypothesis test for two population proportions is,

$z=\frac{\left({\widehat{p}}_1-{\widehat{p}}_2\right)-\left(p_1-p_2\right)}{\sqrt{\overline{p}\left(1-\overline{p}\right)\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}$

Where, ${\widehat{p}}_1$ and ${\widehat{p}}_2$ are the estimate proportion for samples 1 and 2 respectively, that be calculated as ${\widehat{p}}_1=\frac{x_1}{n_1}$ and ${\widehat{p}}_2=\frac{x_2}{n_2}$ respectively. And $p_1$ and $p_2$ are the estimate proportion for samples 1 and 2 respectively.

The weighted estimated proportion for the population is $\overline{p}$ and calculated by. $n_1$$\overline{p}=\frac{x_1+x_2}{n_1+n_2}$

Also $x_1$ and $x_2$ are the successes that happens in samples 1 and 2 respectively, The sample size of samples 1 and 2 are $n_1$ and $n_2$ respectively.

Calculation:

From the problem, the sample successes are $x_1=24$ and $x_2=22$. The sample sizes are $n_1=24+26=50$ and $n_2=22+28=50$. The null hypothesis for the population is $H_0:p_2-p_1\ge 0$ at $\alpha =1\%$.

It is already assumed that $p_1$ be the estimate proportion for the happy babies whose mothers ate chocolate during their pregnancies. And $p_2$ be the estimate proportion for the happy babies whose mothers did not eat chocolate during their pregnancies.

The sample success and size for sample 1 is $x_1=24$ and $n_1=50$ respectively. The estimate proportion for samples 1 is,

${\widehat{p}}_1=\frac{x_1}{n_1}$

Substitute $24$ for $x_1$ and for $n_1$

in above equation.

$\begin{array}{rll}{\widehat{p}}_1& =& \frac{24}{50}\\ & =& 0.48\end{array}$---(1)

The sample success and size for sample 2 is $x_2=22$ and $n_2=50$ respectively. The estimate proportion for samples 2 is

${\widehat{p}}_2=\frac{x_2}{n_2}$

Substitute $22$ for $x_2$ and $50$ for $n_2$ in above equation.

$\begin{array}{rll}{\widehat{p}}_2& =& \frac{22}{50}\\ & =& 0.44\end{array}$---(2)

The weighted estimated proportion for the population is given as,

$\overline{p}=\frac{x_1+x_2}{n_1+n_2}$

Substitute $24$ for , $22$ for , for and $50$ for in above equation.

$\begin{array}{rll}\overline{p}& =& \frac{24+22}{50+50}\\ & =& \frac{46}{100}\\ & =& 0.46\end{array}$---(3)

The statistical test for two population proportions is given as

$z=\frac{\left({\widehat{p}}_1-{\widehat{p}}_2\right)-\left(p_1-p_2\right)}{\sqrt{\overline{p}\left(1-\overline{p}\right)\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}$

From equations (1),(2) and (3), substitute $0.46$ for $\overline{p}$, $0.48$ for ${\widehat{p}}_1$, $0.44$ for ${\widehat{p}}_2$, for $n_1$ and $50$ for $n_2$ in above equation.

$z=\frac{\left(0.48-0.44\right)-\left(p_1-p_2\right)}{\sqrt{\left(0.46\right)\left(1-0.46\right)\left(\frac{1}{50}+\frac{1}{50}\right)}}$

It is given that the null hypothesis for the given proportion is $H_0:p_2-p_1\ge 0$. Substitute 0 for $p_2-p_1$.

$\begin{array}{rll}z& =& \frac{\left(0.04\right)-\left(0\right)}{\sqrt{\left(0.2484\right)\left(0.04\right)}}\\ & =& 0.401286\end{array}$

Thus, the test statistic value is $z=0.401286$.

To determine

(d)

To find:

The value and conclude the decision at given level of significance.

Answer to Problem 9E

Solution:

The claim is incorrectly made that the pregnant women gives happy child after having chocolates. There will no difference between babies by eating chocolates.

Explanation of Solution

Given information:

There is an old wives’ tale that women who eat chocolate during pregnancy are more likely to have happy babies. A pregnancy magazine wants to test this claim, and it gathers 100 randomly selected pregnant women for its study. Half of the women sampled agree to cat chocolate at least once a day, while the other half agree to forego chocolate for the duration of their pregnancies. A year later, the ladies complete a survey regarding the overall happiness of their babies. The results are given in the following table. At the 0.01 level of significance, test the claim of the old wives’ tale.

Number of babies
	Happy babies	Unhappy Babies
With chocolate	24	26
Without chocolate	22	28

Formula Used:

The $p$-value of given population for calculated $z=k$ is,

$p=P(z\le k)$

When $p>\alpha$, then null hypothesis is accepted at the level of significance $\alpha$.

It is already computed that $z_{\text{calc}}=0.401286$ for given samples. Its value is,

$\begin{array}{rll}p& =& \text{P}\left(z\le 0.401286\right)\\ & =& \varphi \left(0.401286\right)\\ & =& 0.344136\\ & \approx & 0.3441\end{array}$

And the level of significance is $0.01$, that is $\alpha =0.01$.

$\begin{array}{c}0.01<0.3441\\ \alpha <p\end{array}$

The null hypothesis is accepted as $p_2-p_1\ge 0$ and the alternative hypothesis is rejected as $p_2-p_1<0$.

Thus, the claim is incorrectly made that the pregnant women gives happy child after having chocolates. There will no difference between babies by eating chocolates.