Math / Data & Statistics

QuestionThe Gallup organization recently conducted a survey of 1,015 randomly selected U.S. adults about "Black Friday" shopping. They asked the following question: "As you know, the Friday after Thanksgiving is one of the biggest shopping days of the year.
Looking ahead, do you personally plan on shopping on the Friday after Thanksgiving, or not?"
Of the 515 men who responded, $16 \%$ said "Yes." Of the 500 women who responded, 20\% said "Yes."
Construct a 95\% confidence interval for the difference between the proportion of men and women who planned to shop on the Friday after Thanksgiving. Use three decimal places when computing the margin of error.

Studdy Solution

STEP 1

What is this asking? We want to find a range where we're 95% sure the true difference between men and women Black Friday shoppers falls. Watch out! Don't mix up the number of men and women surveyed!
Also, remember we're looking at the difference between the groups.

STEP 2

1. Calculate the proportions.
2. Find the standard error.
3. Compute the margin of error.
4. Build the confidence interval.

STEP 3

Let's dive in!
We've got 515 men, and 16% said they'd shop.
So, the number of men shopping is $0.16 \cdot 515 = 82.4$ .
Since we can't have parts of people, let's keep the proportion as $\frac{82.4}{515} \approx 0.16$ .
This is our $\hat{p}_1$ , the sample proportion of men who plan to shop.

STEP 4

For the ladies, we have 500 women, and 20% plan to shop.
That's $0.20 \cdot 500 = 100$ women.
The sample proportion of women who plan to shop, $\hat{p}_2$ , is $\frac{100}{500} = 0.20$ .

STEP 5

The standard error measures how much our sample proportions might vary.
The formula is a little scary, but we've got this! $SE = \sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1} + \frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}}$ Where $n_1$ and $n_2$ are the number of men and women.

STEP 6

Let's plug in our values: $SE = \sqrt{\frac{0.16(1 - 0.16)}{515} + \frac{0.20(1 - 0.20)}{500}}$ $SE = \sqrt{\frac{0.16 \cdot 0.84}{515} + \frac{0.20 \cdot 0.80}{500}}$ $SE = \sqrt{\frac{0.1344}{515} + \frac{0.16}{500}}$ $SE = \sqrt{0.0002607767 + 0.00032}$ $SE = \sqrt{0.0005807767} \approx 0.024$ So, our standard error is approximately 0.024!

STEP 7

For a 95% confidence interval, we use a z-score of 1.96.
The margin of error is the z-score times the standard error. $ME = 1.96 \cdot SE$

STEP 8

Plugging in our standard error: $ME = 1.96 \cdot 0.024 \approx 0.047$

STEP 9

The confidence interval is the difference in proportions, plus or minus the margin of error. The difference in proportions is $\hat{p}_1 - \hat{p}_2 = 0.16 - 0.20 = -0.04$ .

STEP 10

So, our 95% confidence interval is: $-0.04 \pm 0.047$ That's $(-0.04 - 0.047, -0.04 + 0.047)$ , which simplifies to $(-0.087, 0.007)$ .

STEP 11

Our 95% confidence interval for the difference in proportions is $(-0.087, 0.007)$ .
This means we're 95% confident that the true difference between the proportion of men and women who plan to shop on Black Friday is somewhere between -0.087 and 0.007.

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Studdy Solution

STEP 1

What is this asking? We want to find a range where we're 95% sure the true difference between men and women Black Friday shoppers falls. Watch out! Don't mix up the number of men and women surveyed!
Also, remember we're looking at the difference between the groups.

STEP 2

1. Calculate the proportions.
2. Find the standard error.
3. Compute the margin of error.
4. Build the confidence interval.

STEP 3

Let's dive in!
We've got 515 men, and 16% said they'd shop.
So, the number of men shopping is $0.16 \cdot 515 = 82.4$ .
Since we can't have parts of people, let's keep the proportion as $\frac{82.4}{515} \approx 0.16$ .
This is our $\hat{p}_1$ , the sample proportion of men who plan to shop.

STEP 4

For the ladies, we have 500 women, and 20% plan to shop.
That's $0.20 \cdot 500 = 100$ women.
The sample proportion of women who plan to shop, $\hat{p}_2$ , is $\frac{100}{500} = 0.20$ .

STEP 5

The standard error measures how much our sample proportions might vary.
The formula is a little scary, but we've got this! $SE = \sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1} + \frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}}$ Where $n_1$ and $n_2$ are the number of men and women.

STEP 6

STEP 7

For a 95% confidence interval, we use a z-score of 1.96.
The margin of error is the z-score times the standard error. $ME = 1.96 \cdot SE$

STEP 8

Plugging in our standard error: $ME = 1.96 \cdot 0.024 \approx 0.047$

STEP 9

The confidence interval is the difference in proportions, plus or minus the margin of error. The difference in proportions is $\hat{p}_1 - \hat{p}_2 = 0.16 - 0.20 = -0.04$ .

STEP 10

So, our 95% confidence interval is: $-0.04 \pm 0.047$ That's $(-0.04 - 0.047, -0.04 + 0.047)$ , which simplifies to $(-0.087, 0.007)$ .

STEP 11

Our 95% confidence interval for the difference in proportions is $(-0.087, 0.007)$ .
This means we're 95% confident that the true difference between the proportion of men and women who plan to shop on Black Friday is somewhere between -0.087 and 0.007.

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Studdy Solution

STEP 1

What is this asking? We want to find a range where we're 95% sure the *true* difference between men and women Black Friday shoppers falls. Watch out! Don't mix up the number of men and women surveyed!Also, remember we're looking at the *difference* between the groups.

STEP 2

1. Calculate the proportions.2. Find the standard error.3. Compute the margin of error.4. Build the confidence interval.

STEP 3

STEP 4

For the ladies, we have 500 women, and 20% plan to shop.That's 0.20⋅500=1000.20 \cdot 500 = 1000.20⋅500=100 women.The **sample proportion of women** who plan to shop, p^2\hat{p}_2p^​2​, is 100500=0.20\frac{100}{500} = 0.20500100​=0.20.

STEP 5

STEP 6

STEP 7

For a 95% confidence interval, we use a **z-score** of 1.96.The **margin of error** is the z-score times the standard error. ME=1.96⋅SEME = 1.96 \cdot SEME=1.96⋅SE

STEP 8

**Plugging in** our standard error: ME=1.96⋅0.024≈0.047ME = 1.96 \cdot 0.024 \approx 0.047ME=1.96⋅0.024≈0.047

STEP 9

The confidence interval is the **difference in proportions**, plus or minus the **margin of error**. The difference in proportions is p^1−p^2=0.16−0.20=−0.04\hat{p}_1 - \hat{p}_2 = 0.16 - 0.20 = -0.04p^​1​−p^​2​=0.16−0.20=−0.04.

STEP 10

So, our 95% confidence interval is: −0.04±0.047-0.04 \pm 0.047−0.04±0.047 That's (−0.04−0.047,−0.04+0.047)(-0.04 - 0.047, -0.04 + 0.047)(−0.04−0.047,−0.04+0.047), which simplifies to (−0.087,0.007)(-0.087, 0.007)(−0.087,0.007).

STEP 11

Our 95% confidence interval for the difference in proportions is (−0.087,0.007)(-0.087, 0.007)(−0.087,0.007).This means we're 95% confident that the *true* difference between the proportion of men and women who plan to shop on Black Friday is somewhere between -0.087 and 0.007.

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Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

What is this asking? We want to find a range where we're 95% sure the true difference between men and women Black Friday shoppers falls. Watch out! Don't mix up the number of men and women surveyed!
Also, remember we're looking at the difference between the groups.

1. Calculate the proportions.
2. Find the standard error.
3. Compute the margin of error.
4. Build the confidence interval.

For the ladies, we have 500 women, and 20% plan to shop.
That's $0.20 \cdot 500 = 100$ women.
The sample proportion of women who plan to shop, $\hat{p}_2$ , is $\frac{100}{500} = 0.20$ .

For a 95% confidence interval, we use a z-score of 1.96.
The margin of error is the z-score times the standard error. $ME = 1.96 \cdot SE$

Plugging in our standard error: $ME = 1.96 \cdot 0.024 \approx 0.047$

The confidence interval is the difference in proportions, plus or minus the margin of error. The difference in proportions is $\hat{p}_1 - \hat{p}_2 = 0.16 - 0.20 = -0.04$ .

So, our 95% confidence interval is: $-0.04 \pm 0.047$ That's $(-0.04 - 0.047, -0.04 + 0.047)$ , which simplifies to $(-0.087, 0.007)$ .

Our 95% confidence interval for the difference in proportions is $(-0.087, 0.007)$ .
This means we're 95% confident that the true difference between the proportion of men and women who plan to shop on Black Friday is somewhere between -0.087 and 0.007.