Math / Data & Statistics

QuestionPart 2 of 4 Points: 0 of 1
In a simple random sample of 1600 people age 20 and over in a certain country, the proportion with a certain disease was found to be 0.090 (or $9.0 \%$ ). Complete parts (a) through (d) below. a. What is the standard error of the estimate of the proportion of all people in the country age 20 and over with the disease? $S E_{\text {est }}=0.0072$ (Round to four decimal places as needed.) b. Find the margin of error, using a $95 \%$ confidence level, for estimating this proportion. $\mathrm{m}=$ $\square$ (Round to three decimal places as needed.)

Studdy Solution

STEP 1

What is this asking? We're looking at a survey and trying to figure out how accurate the results are, specifically how far off our estimate of the disease proportion might be from the real value in the whole country. Watch out! Don't mix up the standard error with the margin of error!
The standard error tells us about the spread of our estimates, while the margin of error gives us a range where the true value probably lies.

STEP 2

1. Calculate the margin of error

STEP 3

The margin of error ( $m$ ) is calculated using the formula: $m = z^* \cdot SE_{est}$ where $z^*$ is the critical value corresponding to the desired confidence level, and $SE_{est}$ is the standard error of the estimate.
Remember, the margin of error tells us how much "wiggle room" we have around our sample proportion.
It's like giving our estimate a little buffer zone!

STEP 4

We're given a 95% confidence level.
This means we want a $z^$ value that captures the middle 95% of the normal distribution.
This leaves 2.5% in each tail (because $100\% - 95\% = 5\%$ , and $\frac{5\%}{2} = 2.5\%$ ).
Looking up this value in a z-table or using a calculator, we find that $z^$ for a 95% confidence level is approximately 1.96.

STEP 5

We're given the standard error $SE_{est} =$ 0.0072.
Now, we can plug everything into our formula: $m = 1.96 \cdot 0.0072$ $m = 0.014112$

STEP 6

The problem asks us to round to three decimal places.
So, our margin of error is approximately 0.014.

STEP 7

The margin of error for estimating the proportion of people age 20 and over with the disease is approximately 0.014.

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Studdy Solution

STEP 1

STEP 2

1. Calculate the margin of error

STEP 3

STEP 4

STEP 5

We're given the standard error $SE_{est} =$ 0.0072.
Now, we can plug everything into our formula: $m = 1.96 \cdot 0.0072$ $m = 0.014112$

STEP 6

The problem asks us to round to three decimal places.
So, our margin of error is approximately 0.014.

STEP 7

The margin of error for estimating the proportion of people age 20 and over with the disease is approximately 0.014.

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Studdy Solution

STEP 1

STEP 2

1. Calculate the margin of error

STEP 3

STEP 4

STEP 5

We're given the **standard error** SEest=SE_{est} = SEest​= **0.0072**.Now, we can plug everything into our formula: m=1.96⋅0.0072 m = 1.96 \cdot 0.0072 m=1.96⋅0.0072 m=0.014112 m = 0.014112 m=0.014112

STEP 6

The problem asks us to round to three decimal places.So, our **margin of error** is approximately **0.014**.

STEP 7

The margin of error for estimating the proportion of people age 20 and over with the disease is approximately **0.014**.

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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.

We're given the standard error $SE_{est} =$ 0.0072.
Now, we can plug everything into our formula: $m = 1.96 \cdot 0.0072$ $m = 0.014112$

The problem asks us to round to three decimal places.
So, our margin of error is approximately 0.014.

The margin of error for estimating the proportion of people age 20 and over with the disease is approximately 0.014.