Math / Data & Statistics

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Changing jobs: A sociologist sampled 214 people who work in computer-related jobs, and found that 44 of them have changed jobs in the past 6 months. Part 1 of 2 (a) Construct a $99.8 \%$ confidence interval for those who work in computer-related jobs who have changed jobs in the past 6 months. Round the answer to at least three decimal places.
A $99.8 \%$ confidence interval for the proportion of those who work in computer-related jobs who have changed jobs in the past 6 months is 0.1203 $\square$
Part: $1 / 2$
Part 2 of 2 (b) Among the 214 people, 128 of them are under the age of 35 . These constitute a simple random sample of workers under the age of 35. If this sample were used to construct a $99.8 \%$ confidence interval for the proportion of workers under the age of 35 who have changed jobs in the past 6 months, is it likely that the margin of error would be larger, smaller, or about the same as the one in Part (a)?
The margin of error would be (Choose one) $\square$ , because the size of the sample is $\square$ (Choose one) . Skip Part Check Save For Later Submit Assignment

Studdy Solution

STEP 1

What is this asking? We need to find a confidence interval for the proportion of computer workers who changed jobs recently, and then compare the margin of error if we only looked at younger workers. Watch out! Don't mix up the sample sizes when calculating the confidence intervals!
Also, remember that a larger sample size leads to a smaller margin of error.

STEP 2

1. Calculate the Sample Proportion
2. Calculate the Margin of Error
3. Construct the Confidence Interval
4. Compare Sample Sizes and Margin of Error

STEP 3

Alright, let's dive in!
We've got 44 people out of 214 who switched jobs.
Let's call this our sample proportion, often denoted by $\hat{p}$ .

STEP 4

To calculate $\hat{p}$ , we simply divide the number of job changers by the total number of people surveyed.
So, $\hat{p} = \frac{44}{214} \approx \textbf{0.2056}$ .
This means about 20.56% of our sample changed jobs.

STEP 5

Now for the margin of error!
This tells us how much our sample proportion might differ from the true proportion of all computer workers.
We need a z-score for our 99.8% confidence level.
This z-score is approximately 3.09.
You can find this using a z-table or calculator.

STEP 6

The margin of error formula is $z \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ , where $n$ is our sample size.
Plugging in our values, we get $3.09 \cdot \sqrt{\frac{\textbf{0.2056}(1-\textbf{0.2056})}{\textbf{214}}} \approx 3.09 \cdot \sqrt{\frac{0.2056 \cdot 0.7944}{214}} \approx 3.09 \cdot \sqrt{\frac{0.1632}{214}} \approx 3.09 \cdot 0.0275 \approx \textbf{0.0850}$ .

STEP 7

The confidence interval is our sample proportion plus or minus the margin of error.
So, it's $\hat{p} \pm \text{Margin of Error}$ .

STEP 8

Let's calculate: $\textbf{0.2056} \pm \textbf{0.0850}$ .
This gives us an interval of $(0.2056 - 0.0850, 0.2056 + 0.0850)$ , which simplifies to $(0.1206, 0.2906)$ .
Rounded to three decimal places, we get $(\textbf{0.121}, \textbf{0.291})$ .

STEP 9

Part (b) asks about the margin of error if we only looked at the 128 people under 35.
Remember, a smaller sample size means a larger margin of error.

STEP 10

Since 128 is smaller than 214, the margin of error would be larger.
We don't need to calculate it, just understand the relationship between sample size and margin of error!

STEP 11

(a) A 99.8% confidence interval for the proportion of computer workers who changed jobs is (0.121, 0.291).
(b) The margin of error would be larger, because the size of the sample is smaller.

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Studdy Solution

STEP 1

STEP 2

1. Calculate the Sample Proportion
2. Calculate the Margin of Error
3. Construct the Confidence Interval
4. Compare Sample Sizes and Margin of Error

STEP 3

Alright, let's dive in!
We've got 44 people out of 214 who switched jobs.
Let's call this our sample proportion, often denoted by $\hat{p}$ .

STEP 4

To calculate $\hat{p}$ , we simply divide the number of job changers by the total number of people surveyed.
So, $\hat{p} = \frac{44}{214} \approx \textbf{0.2056}$ .
This means about 20.56% of our sample changed jobs.

STEP 5

Now for the margin of error!
This tells us how much our sample proportion might differ from the true proportion of all computer workers.
We need a z-score for our 99.8% confidence level.
This z-score is approximately 3.09.
You can find this using a z-table or calculator.

STEP 6

STEP 7

The confidence interval is our sample proportion plus or minus the margin of error.
So, it's $\hat{p} \pm \text{Margin of Error}$ .

STEP 8

Let's calculate: $\textbf{0.2056} \pm \textbf{0.0850}$ .
This gives us an interval of $(0.2056 - 0.0850, 0.2056 + 0.0850)$ , which simplifies to $(0.1206, 0.2906)$ .
Rounded to three decimal places, we get $(\textbf{0.121}, \textbf{0.291})$ .

STEP 9

Part (b) asks about the margin of error if we only looked at the 128 people under 35.
Remember, a smaller sample size means a larger margin of error.

STEP 10

Since 128 is smaller than 214, the margin of error would be larger.
We don't need to calculate it, just understand the relationship between sample size and margin of error!

STEP 11

(a) A 99.8% confidence interval for the proportion of computer workers who changed jobs is (0.121, 0.291).
(b) The margin of error would be larger, because the size of the sample is smaller.

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Studdy Solution

STEP 1

STEP 2

1. Calculate the Sample Proportion2. Calculate the Margin of Error3. Construct the Confidence Interval4. Compare Sample Sizes and Margin of Error

STEP 3

Alright, let's **dive in**!We've got **44** people out of **214** who switched jobs.Let's call this our **sample proportion**, often denoted by p^\hat{p}p^​.

STEP 4

To calculate p^\hat{p}p^​, we simply divide the **number of job changers** by the **total number of people surveyed**.So, p^=44214≈0.2056\hat{p} = \frac{44}{214} \approx \textbf{0.2056}p^​=21444​≈0.2056.This means about **20.56%** of our sample changed jobs.

STEP 5

Now for the **margin of error**!This tells us how much our sample proportion might differ from the *true* proportion of *all* computer workers.We need a **z-score** for our **99.8% confidence level**.This z-score is approximately **3.09**.You can find this using a z-table or calculator.

STEP 6

STEP 7

The **confidence interval** is our **sample proportion** plus or minus the **margin of error**.So, it's p^±Margin of Error\hat{p} \pm \text{Margin of Error}p^​±Margin of Error.

STEP 8

STEP 9

Part (b) asks about the **margin of error** if we only looked at the **128** people under 35.Remember, a *smaller* sample size means a *larger* margin of error.

STEP 10

Since **128** is *smaller* than **214**, the margin of error would be *larger*.We don't need to calculate it, just understand the relationship between sample size and margin of error!

STEP 11

(a) A 99.8% confidence interval for the proportion of computer workers who changed jobs is (0.121, 0.291).(b) The margin of error would be *larger*, because the size of the sample is *smaller*.

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Start learning now

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

1. Calculate the Sample Proportion
2. Calculate the Margin of Error
3. Construct the Confidence Interval
4. Compare Sample Sizes and Margin of Error

Alright, let's dive in!
We've got 44 people out of 214 who switched jobs.
Let's call this our sample proportion, often denoted by $\hat{p}$ .

To calculate $\hat{p}$ , we simply divide the number of job changers by the total number of people surveyed.
So, $\hat{p} = \frac{44}{214} \approx \textbf{0.2056}$ .
This means about 20.56% of our sample changed jobs.

Now for the margin of error!
This tells us how much our sample proportion might differ from the true proportion of all computer workers.
We need a z-score for our 99.8% confidence level.
This z-score is approximately 3.09.
You can find this using a z-table or calculator.

The confidence interval is our sample proportion plus or minus the margin of error.
So, it's $\hat{p} \pm \text{Margin of Error}$ .

Part (b) asks about the margin of error if we only looked at the 128 people under 35.
Remember, a smaller sample size means a larger margin of error.

Since 128 is smaller than 214, the margin of error would be larger.
We don't need to calculate it, just understand the relationship between sample size and margin of error!

(a) A 99.8% confidence interval for the proportion of computer workers who changed jobs is (0.121, 0.291).
(b) The margin of error would be larger, because the size of the sample is smaller.