Question

A large snack company claims that children enjoy their whole wheat snack equally as much as normal snacks. A study testing this claim on a SRS of 86 children were given both snack alternatives and then each child was asked which snack they preferred. The whole wheat snack was chosen by 48 children.

a) Using the 68-95-99.7 rule, if the snack company claims were true, you would expect ?̂ to fall between what two percent about 95% of the time?

b) Using your answer from a, is the snack company's claim correct?

c) Perform a significance test for α = 0.5

Answer 1

Part a

As per 68-95-99.7 rule, we would expect the p-hat to fall within 2 standard deviations from p-hat.

If the snack company’s claim is true, then population proportion = p = 0.5, q= 1 – p = 0.5

We are given n=86, X=48,

p-hat = X/n = 48/86 = 0.558139535

q-hat = 1 – p-hat = 1 - 0.558139535 = 0.44186

Estimate for standard deviation = SE = sqrt(p-hat*q-hat/n) = sqrt(0.558139535*0.44186/86) = 0.053551

p-hat - 2*SD = 0.558139535- 2*0.053551= 0.451038

p-hat + 2*SD = 0.558139535+ 2*0.053551= 0.665242

Required percentages = 45.10% and 66.52%

Part b

Using above answer, the snack company’s claim is correct because the hypothesized value 0.5 of proportion is lies within above interval.

Part c

Here, we have to use z test for population proportion.

H₀: p=0.5 versus H_a: p ≠0.5

We are given α = 0.5

n=86, X=48,

p-hat = X/n = 48/86 = 0.558139535

Z = (P-hat – p)/sqrt(p*(1-p)/n)

Z = (0.558139535 – 0.5) / sqrt(0.5*0.5/86)

Z = 1.0783

Critical value = ±1.96 (by using z-table)

P-value = 0.2809 (by using z-table)

P-value > α = 0.5

So, we do not reject the null hypothesis

There is sufficient evidence to conclude that children enjoy their whole wheat snack equally as much as normal snacks.