Question

1) A survey compared adult gamers with teen gamers with respect to the devices on which they play. There were 1063 adult gamers, of whom 54% played on game consoles. There were 1064 teens, of whom 89% played on game consoles. Find a 95% confidence interval for the difference in the two (population) proportions.

2) Using the data from #1, test the null hypothesis that the two proportions are equal.

Answer 1

Part 1

Confidence interval for difference between two population proportions:

Confidence interval = (P1 – P2) ± Z*sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]

Where, P1 and P2 are sample proportions for first and second groups respectively.

Let first group as the group of 1064 teens and second group as the group of 1063 adult gamers.

We are given

P1 = 0.89

P2 = 0.54

N1 = 1064

N2 = 1063

P1 – P2 = 0.89 – 0.54 = 0.35

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

Confidence interval = (P1 – P2) ± Z*sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]

Confidence interval = (0.89 – 0.54) ± 1.96*sqrt[(0.89*(1 – 0.89)/1064) + (0.54*(1 – 0.54)/1063)]

Confidence interval = 0.35 ± 1.96*0.0180

Confidence interval = 0.35 ± 0.0354

Lower limit = 0.35 - 0.0354 = 0.3146

Upper limit = 0.35 + 0.0354 = 0.3854

Confidence interval = (0.3146, 0.3854)

Part 2

Here, we have to use z test for difference between population proportions.

H₀: p₁ = p₂ versus H_a: p₁ ≠ p₂

This is a two tailed test.

We assume level of significance = α = 0.05

Test statistic for z test for two population proportions is given as below:

Z = (P₁ – P₂) / sqrt[(P₁*(1 – P₁)/n₁) + (P₂*(1 – P₂)/n₂)]

Z = 0.35 / 0.0180

(by using values from part 1)

Z = 19.44444444

P-value = 0.00

(by using z-table or excel)

P-value < α = 0.05

So, we reject the null hypothesis

There is insufficient evidence to conclude that the two proportions are equal.