Question

A 20-minute consumer survey mailed to 400 adults aged 25-34 included a $5 Starbucks gift certificate. The same survey was mailed to 400 adults aged 25-34 without the gift certificate. There were 57 responses from the first group and 31 from the second group.

(a) Perform a two tailed test comparing the response rate (proportions) at a = .05
(b) Form a confidence interval for the difference of proportions, without pooling the sample. Does it includes zero?

Answer 1

a)
`z=\frac{0.1425-0.0775}{\sqrt{0.11(1-0.11)((1)/(400)+(1)/(400))}}=2.938`

`p_v =2*P(Z>2.938)=0.0033`

If we compare the p value and the significance level given
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the two proportions are different at 5% of significance. .

b)
`(0.1425-0.0775) - 1.96 \sqrt{(0.1425(1-0.1425))/(400) +(0.0775(1-0.0775))/(400)}=0.02187`
`(0.1425-0.0775) + 1.96 \sqrt{(0.1425(1-0.1425))/(400) +(0.0775(1-0.0775))/(400)}=0.10813`

And the 95% confidence interval would be given (0.02187;0.10813).

And as we can see the confidence interval not contains the 0.

Explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

`p_A` represent the real population proportion for the gift

`\hat p_A =(57)/(400)=0.1425` represent the estimated proportion for the gift

`n_A=400` is the sample size required for the gift

`p_B` represent the real population proportion for no gift

`\hat p_B =(31)/(400)=0.0775` represent the estimated proportion for no gift

`n_B=400` is the sample size required for no gift

`z` represent the critical value for the margin of error

The population proportion have the following distribution

`p \sim N(p,\sqrt{(p(1-p))/(n)})`

Part a

We need to conduct a hypothesis in order to check if the proportions are equal , the system of hypothesis would be:

Null hypothesis:
`p_A = p_B`

Alternative hypothesis:
`p_(A) \\eq p_(B)`

We need to apply a z test to compare proportions, and the statistic is given by:

`z=\frac{p_(A)-p_(B)}{\sqrt{\hat p (1-\hat p)((1)/(n_(A))+(1)/(n_(B)))}}` (1)

Where
`\hat p=(X_(A)+X_(B))/(n_(A)+n_(B))=(51+37)/(400+400)=0.11`

Calculate the statistic

Replacing in formula (1) the values obtained we got this:

`z=\frac{0.1425-0.0775}{\sqrt{0.11(1-0.11)((1)/(400)+(1)/(400))}}=2.938`

Statistical decision

We have a significance level provided
`\alpha=0.05`, and now we can calculate the p value for this test.

Since is a two tailed test the p value would be:

`p_v =2*P(Z>2.938)=0.0033`

If we compare the p value and the significance level given
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the two proportions are different at 5% of significance. .

Part b

The confidence interval for the difference of two proportions would be given by this formula

`(\hat p_A -\hat p_B) \pm z_(\alpha/2) \sqrt{(\hat p_A(1-\hat p_A))/(n_A) +(\hat p_B (1-\hat p_B))/(n_B)}`

For the 95% confidence interval the value of and , with that value we can find the quantile required for the interval in the normal standard distribution.

`z_(\alpha/2)=1.96`

And replacing into the confidence interval formula we got:
`(0.1425-0.0775) - 1.96 \sqrt{(0.1425(1-0.1425))/(400) +(0.0775(1-0.0775))/(400)}=0.02187`
`(0.1425-0.0775) + 1.96 \sqrt{(0.1425(1-0.1425))/(400) +(0.0775(1-0.0775))/(400)}=0.10813`And the 95% confidence interval would be given (0.02187;0.10813).

We are confident at 95% that the difference between the two proportions is between
`0.02187\leq p_A -p_b \leq 0.10813`

And as we can see the interval not contains the 0.