Question

# A study of young children was designed to increase their intake of whole-grain, rather than regular-grain, snacks. At the end of the study, the 79 chi

Study design
A study of young children was designed to increase their intake of whole-grain, rather than regular-grain, snacks. At the end of the study, the 79 children who participated in the study were presented with a choice between a regular-grain snack and a whole-grain alternative. The whole-grain alternative was chosen by 55 children. You want to examine the possibility that the children are equally likely to choose each type of snack.
$$\displaystyle{H}_{{0}}:{p}={0.5}$$
$$\displaystyle{H}_{{a}}:{p}\ne{0.5}$$
Perform the significance test. (Use $$\displaystyle\alpha={0.01}$$. Round your test statistic to two decimal places and your P-value to four decimal places.)

2021-02-25

Step 1
Given Information
No of children participated $$(n) = 79$$
No of children chosen $$(x) = 55$$
$$\displaystyle\hat{{{p}}}=\frac{{x}}{{n}}=\frac{{55}}{{79}}={0.696}$$
Null and alternative Hypothesis
Null Hypothesis: $$p = 0.5$$
Alternative Hypothesis: $$\displaystyle{p}\ne{0.5}$$
Test statistic
$$\displaystyle{z}=\frac{{\hat{{{p}}}-{p}}}{{\sqrt{{\frac{{{p}{\left({1}-{p}\right)}}}{{n}}}}=\frac{{{0.696}-{0.5}}}{{\sqrt{{\frac{{{0.5}\times{\left({1}-{0.5}\right)}}}{{79}}}}}}={3.48}}}$$
Step 2
p-value
$$\displaystyle{p}-{v}{a}{l}{u}{e}={2}{x}{P}{\left({Z}{>}{3.48}\right)}={2}{x}{\left({1}–{P}{\left({Z}{<}{3.48}\right)}\right)}={2}{x}{\left({1}–{0.999748}\right)}={0.000501}$$
$$P(Z<3.48)=0.999748\ \text{(Using Excel function=NORM.S.DIST(B 20, TRUE)}$$
$$p-value < 0.01$$,
So, reject null hypothesis and conclude that children are not equally likely to choose each type of snack.