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.