Chapter 10.1, Problem 19E

(a)

To determine

To explain why the sample results give evidence for the alternative hypothesis.

Explanation of Solution

This problem refers to the previous question, that is,

It is given in the question that the researchers want to know that if a greater proportion of $6\text{ to }7$ year olds can sort correctly than $4\text{ to }5$ year olds or not.

So, the given claim that: proportion is greater for the $6\text{ to }7$ year olds.

Now, we have to find out the appropriate hypotheses for performing a significance test.

Thus, the claim is either the null hypothesis or the alternative hypothesis. The null hypothesis states that the population proportions are equal. If the null hypothesis is the claim then the alternative hypothesis states the opposite of the null hypothesis.

Therefore, the appropriate hypotheses for this is:

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

Where we have,

  $p_1=$ the proportion of $4\text{ to }5$ year olds who can sort correctly.

  $p_2=$ the proportion of $6\text{ to }7$ year olds who can sort correctly.

Now, the sample proportion is the number of successes divided by the sample size, that is:

  $\begin{array}{l}{\widehat{p}}_1=\frac{x_1}{n_1}\\ =\frac{10}{50}\\ =0.20\\ {\widehat{p}}_2=\frac{x_2}{n_2}\\ =\frac{28}{53}\\ =0.5283\end{array}$

From this we conclude that the sample proportion for the second sample is larger than the sample proportion for the first sample which is same as the alternative hypothesis i.e. $p_1<p_2$ and thus this implies that there is some evidence for the alternative hypothesis. This states that the sample proportion of $4\text{ to }5$ year olds who can sort correctly is smaller than the the sample proportion of $6\text{ to }7$ year olds who can sort correctly.

(b)

To determine

To calculate the standardized test statistics and the P -value.

Answer to Problem 19E

The P-value is $0.0003$ and the value of test statistics value is $-3.45$ .

Explanation of Solution

Now referring to the part (a), we know that:

  $\begin{array}{l}{x}_1=10\\ x_2=28\\ n_1=50\\ n_2=53\end{array}$

And the appropriate hypotheses for this is:

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

Where we have,

  $p_1=$ the proportion of $4\text{ to }5$ year olds who can sort correctly.

  $p_2=$ the proportion of $6\text{ to }7$ year olds who can sort correctly.

And the sample proportion is the number of successes divided by the sample size, that is:

  $\begin{array}{l}{\widehat{p}}_1=\frac{x_1}{n_1}\\ =\frac{10}{50}\\ =0.20\\ {\widehat{p}}_2=\frac{x_2}{n_2}\\ =\frac{28}{53}\\ =0.5283\end{array}$

  $\begin{array}{l}{\widehat{p}}_p=\frac{x_1+x_2}{n_1+n_2}\\ =\frac{10+28}{50+53}\\ =\frac{38}{103}\\ =0.3689\end{array}$

Now, we will calculate the value of test statistics as:

  $\begin{array}{l}z=\frac{{\widehat{p}}_1-{\widehat{p}}_2-(p_1-p_2)}{\sqrt{{\widehat{p}}_p(1-{\widehat{p}}_p)}\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\\ =\frac{0.20-0.5283-0}{\sqrt{0.3689(1-0.3689)}\sqrt{\frac{1}{50}+\frac{1}{53}}}\\ \approx -3.45\end{array}$

The P-value is the probability of obtaining the value of the test statistics or a value more extreme assuming that the null hypothesis is true. Thus, we have,

  $\begin{array}{l}P=P(Z<-3.45)\\ =0.0003\end{array}$

Therefore, the P-value is $0.0003$ and the value of test statistics value is $-3.45$ .

(c)

To determine

To explain what conclusion would you make.

Answer to Problem 19E

There is a convincing evidence that there is a greater proportion of $6\text{ to }7$ year olds who can sort correctly than the proportion of $4\text{ to }5$ year olds who can sort correctly.

Explanation of Solution

It is given that:

  $\begin{array}{l}{x}_1=10\\ x_2=28\\ n_1=50\\ n_2=53\end{array}$

And we calculated in part (a) and part (b), that the P-value is $0.0003$ and the value of test statistics value is $-3.45$ .

And the hypotheses are as:

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

Thus, if the P-value is smaller than the significance level, then we will reject the null hypothesis, thus, we have,

  $P<0.05\Rightarrow \text{Reject }{H}_0$

Thus, we conclude that there is a convincing evidence that there is a greater proportion of $6\text{ to }7$ year olds who can sort correctly than the proportion of $4\text{ to }5$ year olds who can sort correctly.