Mathematics, 19.02.2020 01:54, honeytrail429

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?

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answered: ddcope1030

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

p_v =2*P(Z2.938)=0.0033

If we compare the p value and the significance level given $\alpha=0.05$ we see that p_v 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) $(0.1425-0.0775) - 1.96 \sqrt{\frac{0.1425(1-0.1425)}{400} +\frac{0.0775(1-0.0775)}{400}}=0.02187$ $(0.1425-0.0775) + 1.96 \sqrt{\frac{0.1425(1-0.1425)}{400} +\frac{0.0775(1-0.0775)}{400}}=0.10813$

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

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

Step-by-step 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

$\hat p_A =\frac{57}{400}=0.1425$ 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

$\hat p_B =\frac{31}{400}=0.0775$ represent the estimated proportion for no gift

n_B=400 is the sample size required for no gift

represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

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: $p_{A} \neq p_{B}$

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

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

Where $\hat p=\frac{X_{A}+X_{B}}{n_{A}+n_{B}}=\frac{51+37}{400+400}=0.11$

Calculate the statistic

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

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

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(Z2.938)=0.0033

Part b

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

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

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: $(0.1425-0.0775) - 1.96 \sqrt{\frac{0.1425(1-0.1425)}{400} +\frac{0.0775(1-0.0775)}{400}}=0.02187$ $(0.1425-0.0775) + 1.96 \sqrt{\frac{0.1425(1-0.1425)}{400} +\frac{0.0775(1-0.0775)}{400}}=0.10813$ 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 $0.02187\leq p_A -p_b \leq 0.10813$