Mathematics: A 20-minute consumer survey mailed to 500 adults aged 25-34 included a $5 Starbucks gift certiﬁcate. The same surveywas mailed to 500 adults aged 25-34 without the gift certificate. There were 65 responses from the first group and 45 fromthe second group. Form a 95 percent confidence interval for the difference of proportions. Does it include zero?

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A 20-minute consumer survey mailed to 500 adults

Ответ:

esme72

18.01.2022

Mathematics

$(0.13-0.09) - 1.96 \sqrt{\frac{0.13(1-0.13)}{500} +\frac{0.09(1-0.09)}{500}}=0.00129$

$(0.13-0.09) + 1.96 \sqrt{\frac{0.13(1-0.13)}{500} +\frac{0.09(1-0.09)}{500}}=0.0787$

And the 95% confidence interval would be given (0.00129;0.0787).

We are confident at 95% that the difference between the two proportions is between $0.00129 \leq p_B -p_A \leq 0.0787$

And as we can see the confidence interval for the difference on this case 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".

Solution to the problem

p_A represent the real population proportion for the gift certificate

$\hat p_A =\frac{65}{500}=0.13$ represent the estimated proportion for the gift certificate

n_A=500 is the sample size required for the gift certificate

p_B represent the real population proportion for without the gift certificate

$\hat p_B =\frac{45}{500}=0.09$ represent the estimated proportion for without the gift certificate

n_B=500 is the sample size required for Brand B

represent the critical value for the margin of error

The population proportion have the following distribution

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

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 $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , 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:

$(0.13-0.09) - 1.96 \sqrt{\frac{0.13(1-0.13)}{500} +\frac{0.09(1-0.09)}{500}}=0.00129$

$(0.13-0.09) + 1.96 \sqrt{\frac{0.13(1-0.13)}{500} +\frac{0.09(1-0.09)}{500}}=0.0787$

And the 95% confidence interval would be given (0.00129;0.0787).

We are confident at 95% that the difference between the two proportions is between $0.00129 \leq p_B -p_A \leq 0.0787$

And as we can see the confidence interval for the difference on this case not contains the 0.

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