Math / Data & Statistics

Question2. $[-/ 2$ Points]
DETAILS MY NOTES MENDSTAT15 8.2.010. ASK YOUR TEACHER
PRACTICE ANOTHER Calculate the $95 \%$ margin of error in estimating a population mean $\mu$ for the following values. (Round your answer to three decimal places.) $n=7,000, s^{2}=64$ $\qquad$ Consider that for $s^{2}=64$ and sample sizes of 50,100, and 500 the margins of error are 2.217, 1.568, and 0.701 respectively. Comment on how an increased sample size affects the margin of error. As the sample size increases the margin of error decreases. As the sample size increases the margin of error remains relatively constant. As the sample size increases the margin of error also increases. You may need to use the appropriate appendix table or technology to answer this question. Need Help? Read it Submit Answer
3. [-/1 Points $]$
DETAILS MY NOTES MENDSTAT15 8.3.003.S. ASK YOUR TEACHER PRACTICE ANOTHER

Studdy Solution

STEP 1

What is this asking? We need to find the 95% margin of error when estimating the average of a population, given a sample size of 7000 and a variance of 64, and then see how changing the sample size affects the margin of error. Watch out! Don't mix up variance and standard deviation!
Also, make sure to use the correct t-distribution value for our sample size and confidence level.
Since our sample size is large, we can use the z-score for a 95% confidence interval.

STEP 2

1. Calculate the standard deviation.
2. Find the critical value.
3. Calculate the margin of error.
4. Analyze the effect of sample size.

STEP 3

We're given the variance, $\$s^2 = 64$ , but we need the standard deviation, $\$s$ , for the margin of error formula.
The standard deviation is the square root of the variance!

STEP 4

So, we calculate $\$s = \sqrt{s^2} = \sqrt{64} = 8$ .
Our standard deviation is $\$s = 8$ .

STEP 5

We want a 95% confidence interval.
This means we need to find the z-score that leaves 2.5% in each tail of the normal distribution (since $(100\% - 95\%) / 2 = 2.5\%$ ).

STEP 6

Using a z-table or calculator, we find that the critical value is approximately $\$z = 1.96$ .
This means that 95% of the data falls within 1.96 standard deviations of the mean.

STEP 7

The margin of error formula is given by $\text{Margin of Error} = z \cdot \frac{s}{\sqrt{n}}$ , where $\$z$ is the critical value, $\$s$ is the standard deviation, and $\$n$ is the sample size.

STEP 8

We plug in our values: $\$n = 7000$ , $\$s = 8$ , and $\$z = 1.96$ .

STEP 9

So, the margin of error is $1.96 \cdot \frac{8}{\sqrt{7000}} \approx 1.96 \cdot \frac{8}{83.666} \approx 1.96 \cdot 0.0956 \approx 0.187$ .
Rounded to three decimal places, our margin of error is $\$0.187$ .

STEP 10

We're told that for sample sizes of 50, 100, and 500 (with the same variance), the margins of error are 2.217, 1.568, and 0.701, respectively.

STEP 11

Notice that as the sample size increases, the margin of error decreases.
This makes sense!
A larger sample size gives us more information about the population, so our estimate of the mean becomes more precise, and our margin of error shrinks!

STEP 12

The 95% margin of error for a sample size of 7000 and variance of 64 is approximately 0.187.
As the sample size increases, the margin of error decreases.

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Studdy Solution

STEP 1

STEP 2

1. Calculate the standard deviation.
2. Find the critical value.
3. Calculate the margin of error.
4. Analyze the effect of sample size.

STEP 3

We're given the variance, $\$s^2 = 64$ , but we need the standard deviation, $\$s$ , for the margin of error formula.
The standard deviation is the square root of the variance!

STEP 4

So, we calculate $\$s = \sqrt{s^2} = \sqrt{64} = 8$ .
Our standard deviation is $\$s = 8$ .

STEP 5

We want a 95% confidence interval.
This means we need to find the z-score that leaves 2.5% in each tail of the normal distribution (since $(100\% - 95\%) / 2 = 2.5\%$ ).

STEP 6

Using a z-table or calculator, we find that the critical value is approximately $\$z = 1.96$ .
This means that 95% of the data falls within 1.96 standard deviations of the mean.

STEP 7

The margin of error formula is given by $\text{Margin of Error} = z \cdot \frac{s}{\sqrt{n}}$ , where $\$z$ is the critical value, $\$s$ is the standard deviation, and $\$n$ is the sample size.

STEP 8

We plug in our values: $\$n = 7000$ , $\$s = 8$ , and $\$z = 1.96$ .

STEP 9

So, the margin of error is $1.96 \cdot \frac{8}{\sqrt{7000}} \approx 1.96 \cdot \frac{8}{83.666} \approx 1.96 \cdot 0.0956 \approx 0.187$ .
Rounded to three decimal places, our margin of error is $\$0.187$ .

STEP 10

We're told that for sample sizes of 50, 100, and 500 (with the same variance), the margins of error are 2.217, 1.568, and 0.701, respectively.

STEP 11

Notice that as the sample size increases, the margin of error decreases.
This makes sense!
A larger sample size gives us more information about the population, so our estimate of the mean becomes more precise, and our margin of error shrinks!

STEP 12

The 95% margin of error for a sample size of 7000 and variance of 64 is approximately 0.187.
As the sample size increases, the margin of error decreases.

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Studdy Solution

STEP 1

STEP 2

1. Calculate the standard deviation.2. Find the critical value.3. Calculate the margin of error.4. Analyze the effect of sample size.

STEP 3

We're given the **variance**, $s2=64\$s^2 = 64$s2=64, but we need the **standard deviation**, $s\$s$s, for the margin of error formula.The standard deviation is the square root of the variance!

STEP 4

So, we **calculate** $s=s2=64=8\$s = \sqrt{s^2} = \sqrt{64} = 8$s=s2​=64​=8.Our **standard deviation** is $s=8\$s = 8$s=8.

STEP 5

We want a **95% confidence interval**.This means we need to find the *z*-score that leaves 2.5% in each tail of the normal distribution (since (100%−95%)/2=2.5%(100\% - 95\%) / 2 = 2.5\%(100%−95%)/2=2.5%).

STEP 6

Using a *z*-table or calculator, we find that the **critical value** is approximately $z=1.96\$z = 1.96$z=1.96.This means that 95% of the data falls within 1.96 standard deviations of the mean.

STEP 7

The **margin of error** formula is given by Margin of Error=z⋅sn \text{Margin of Error} = z \cdot \frac{s}{\sqrt{n}} Margin of Error=z⋅n​s​, where $z\$z$z is the **critical value**, $s\$s$s is the **standard deviation**, and $n\$n$n is the **sample size**.

STEP 8

We **plug in** our values: $n=7000\$n = 7000$n=7000, $s=8\$s = 8$s=8, and $z=1.96\$z = 1.96$z=1.96.

STEP 9

STEP 10

We're told that for sample sizes of 50, 100, and 500 (with the same variance), the margins of error are 2.217, 1.568, and 0.701, respectively.

STEP 11

Notice that as the **sample size increases**, the **margin of error decreases**.This makes sense!A larger sample size gives us more information about the population, so our estimate of the mean becomes more precise, and our margin of error shrinks!

STEP 12

The **95% margin of error** for a sample size of 7000 and variance of 64 is approximately **0.187**.As the sample size increases, the margin of error decreases.

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1. Calculate the standard deviation.
2. Find the critical value.
3. Calculate the margin of error.
4. Analyze the effect of sample size.

We're given the variance, $\$s^2 = 64$ , but we need the standard deviation, $\$s$ , for the margin of error formula.
The standard deviation is the square root of the variance!

So, we calculate $\$s = \sqrt{s^2} = \sqrt{64} = 8$ .
Our standard deviation is $\$s = 8$ .

We want a 95% confidence interval.
This means we need to find the z-score that leaves 2.5% in each tail of the normal distribution (since $(100\% - 95\%) / 2 = 2.5\%$ ).

Using a z-table or calculator, we find that the critical value is approximately $\$z = 1.96$ .
This means that 95% of the data falls within 1.96 standard deviations of the mean.

The margin of error formula is given by $\text{Margin of Error} = z \cdot \frac{s}{\sqrt{n}}$ , where $\$z$ is the critical value, $\$s$ is the standard deviation, and $\$n$ is the sample size.

We plug in our values: $\$n = 7000$ , $\$s = 8$ , and $\$z = 1.96$ .

Notice that as the sample size increases, the margin of error decreases.
This makes sense!
A larger sample size gives us more information about the population, so our estimate of the mean becomes more precise, and our margin of error shrinks!

The 95% margin of error for a sample size of 7000 and variance of 64 is approximately 0.187.
As the sample size increases, the margin of error decreases.