Children make choices Refer to Exercise 15.

a. Explain why the sample results give some evidence for the alternative hypothesis.

b. Calculate the standardized test statistic and $P$-value.

c. What conclusion would you make?

It is given that researchers want to know that if a greater proportion of $6$to $7$year old can sort correctly than $4$to $5$year old or not.

$x_1=10$

$x_2=28$

$n_1=50$

$n_2=53$

Claim is population is greater for $6-7$years old.

The hypothesis is:

Null: $H_0:p_1=p_2$

Alternative: $H_a:p_1<p_2$

$p_1$is the proportion of $4-5$year old who can sort correctly.

$p_2$is the proportion of $6-7$year old who can sort correctly.

Sample proportion is ${\hat{p}}_1=\frac{x_1}{n_1}=\frac{10}{50}=0.20$

${\hat{p}}_2=\frac{x_2}{n_2}=\frac{28}{53}=0.5283$

$p_1<p_2$, there is some evidence for alternative hypothesis.

From above part, ${\hat{p}}_1=0.20$and ${\hat{p}}_2==0.5283$

and ${\hat{p}}_p=\frac{x_1+x_2}{n_1+n_2}=\frac{10+28}{50+53}=\frac{38}{103}=0.3689$

Value of test statistics is:

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

$=\frac{0.20-0.5283-0}{\sqrt{0.3689(1-0.3689)}\sqrt{\frac{1}{50}+\frac{1}{53}}}$

$\approx -3.45$

Thus probability is $P=P(Z<-3.45)=0.0003$

From above:

$H_0:p_1=p_2$

$H_a:p_1<p_2$

As $P$value is less than significance level $P<0.05\Rightarrow Reject{H}_0$
There is evidence present that there is greater proportion of$6-7$years old who can sort correctly than proportion of$4-5$years old who can sort correctly.