Math / Data & Statistics

QuestionQUESTION 12 Provide an appropriate response. Find the value of $E$ , the margin of error, for $c=0.90, n=10$ and $s=3.7$ . 0.68 2.14 1.62 2.12

Studdy Solution

STEP 1

What is this asking? We need to find the margin of error E for a 90% confidence level with a sample size of 10 and a sample standard deviation of 3.7. Watch out! Don't forget to use the t-distribution since the sample size is small and we're using the sample standard deviation!

STEP 2

1. Find the critical value.
2. Calculate the margin of error.

STEP 3

Alright, so we're dealing with a 90% confidence level, which means we want to find the t-value that leaves 5% in each tail of the distribution.
Why each tail?
Because $100\% - 90\% = 10\%$ , and we split that $10\%$ evenly between the two tails.

STEP 4

Our sample size is $n = 10$, so our degrees of freedom are $n - 1 = 10 - 1 = 9$.
We'll use this and our tail percentage to look up our critical t-value.

STEP 5

Checking our t-table for 9 degrees of freedom and a 0.05 tail probability, we find our **critical t-value** is $t_{0.05,9} = 1.833$.
Boom!

STEP 6

Now, it's time to plug everything into the margin of error formula for a t-distribution: $E = t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}$ Where E is the margin of error, $t_{\alpha/2}$ is our critical t-value, s is the sample standard deviation, and n is the sample size.

STEP 7

Let's substitute our values: $t_{\alpha/2} = 1.833$ , $s = 3.7$ , and $n = 10$ . $E = 1.833 \cdot \frac{3.7}{\sqrt{10}}$

STEP 8

Time to crunch the numbers: $E = 1.833 \cdot \frac{3.7}{3.162}$ $E = 1.833 \cdot 1.169$ $E \approx 2.14$

STEP 9

Our margin of error, E, is approximately 2.14.
So the correct answer choice is 2.14.

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QuestionQUESTION 12 Provide an appropriate response. Find the value of $E$ , the margin of error, for $c=0.90, n=10$ and $s=3.7$ . 0.68 2.14 1.62 2.12

Studdy Solution

STEP 1

What is this asking? We need to find the margin of error E for a 90% confidence level with a sample size of 10 and a sample standard deviation of 3.7. Watch out! Don't forget to use the t-distribution since the sample size is small and we're using the sample standard deviation!

STEP 2

1. Find the critical value.
2. Calculate the margin of error.

STEP 3

Alright, so we're dealing with a 90% confidence level, which means we want to find the t-value that leaves 5% in each tail of the distribution.
Why each tail?
Because $100\% - 90\% = 10\%$ , and we split that $10\%$ evenly between the two tails.

STEP 4

Our sample size is \(n = 10\), so our degrees of freedom are \(n - 1 = 10 - 1 = 9\).
We'll use this and our tail percentage to look up our critical t-value.

STEP 5

Checking our t-table for 9 degrees of freedom and a 0.05 tail probability, we find our **critical t-value** is \(t_{0.05,9} = 1.833\).
Boom!

STEP 6

Now, it's time to plug everything into the margin of error formula for a t-distribution: $E = t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}$ Where E is the margin of error, $t_{\alpha/2}$ is our critical t-value, s is the sample standard deviation, and n is the sample size.

STEP 7

Let's substitute our values: $t_{\alpha/2} = 1.833$ , $s = 3.7$ , and $n = 10$ . $E = 1.833 \cdot \frac{3.7}{\sqrt{10}}$

STEP 8

Time to crunch the numbers: $E = 1.833 \cdot \frac{3.7}{3.162}$ $E = 1.833 \cdot 1.169$ $E \approx 2.14$

STEP 9

Our margin of error, E, is approximately 2.14.
So the correct answer choice is 2.14.

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QuestionQUESTION 12 Provide an appropriate response. Find the value of EEE, the margin of error, for c=0.90,n=10c=0.90, n=10c=0.90,n=10 and s=3.7s=3.7s=3.7. 0.68 2.14 1.62 2.12

Studdy Solution

STEP 1

What is this asking? We need to find the margin of error *E* for a 90% confidence level with a sample size of 10 and a sample standard deviation of 3.7. Watch out! Don't forget to use the *t*-distribution since the sample size is small and we're using the sample standard deviation!

STEP 2

1. Find the critical value.2. Calculate the margin of error.

STEP 3

Alright, so we're dealing with a **90% confidence level**, which means we want to find the *t*-value that leaves **5%** in each tail of the distribution.Why each tail?Because 100%−90%=10%100\% - 90\% = 10\%100%−90%=10%, and we split that 10%10\%10% evenly between the two tails.

STEP 4

Our **sample size** is \(n = 10\), so our **degrees of freedom** are \(n - 1 = 10 - 1 = 9\).We'll use this and our tail percentage to look up our critical *t*-value.

STEP 5

Checking our *t*-table for 9 degrees of freedom and a 0.05 tail probability, we find our **critical *t*-value** is \(t_{0.05,9} = 1.833\).Boom!

STEP 6

STEP 7

Let's substitute our **values**: tα/2=1.833t_{\alpha/2} = 1.833tα/2​=1.833, s=3.7s = 3.7s=3.7, and n=10n = 10n=10. E=1.833⋅3.710E = 1.833 \cdot \frac{3.7}{\sqrt{10}}E=1.833⋅10​3.7​

STEP 8

Time to **crunch the numbers**: E=1.833⋅3.73.162E = 1.833 \cdot \frac{3.7}{3.162}E=1.833⋅3.1623.7​ E=1.833⋅1.169E = 1.833 \cdot 1.169E=1.833⋅1.169E≈2.14E \approx 2.14E≈2.14

STEP 9

Our margin of error, *E*, is approximately **2.14**.So the correct answer choice is 2.14.

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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.

QuestionQUESTION 12 Provide an appropriate response. Find the value of $E$ , the margin of error, for $c=0.90, n=10$ and $s=3.7$ . 0.68 2.14 1.62 2.12

What is this asking? We need to find the margin of error E for a 90% confidence level with a sample size of 10 and a sample standard deviation of 3.7. Watch out! Don't forget to use the t-distribution since the sample size is small and we're using the sample standard deviation!

1. Find the critical value.
2. Calculate the margin of error.

Alright, so we're dealing with a 90% confidence level, which means we want to find the t-value that leaves 5% in each tail of the distribution.
Why each tail?
Because $100\% - 90\% = 10\%$ , and we split that $10\%$ evenly between the two tails.

Our sample size is \(n = 10\), so our degrees of freedom are \(n - 1 = 10 - 1 = 9\).
We'll use this and our tail percentage to look up our critical t-value.

Checking our t-table for 9 degrees of freedom and a 0.05 tail probability, we find our **critical t-value** is \(t_{0.05,9} = 1.833\).
Boom!

Let's substitute our values: $t_{\alpha/2} = 1.833$ , $s = 3.7$ , and $n = 10$ . $E = 1.833 \cdot \frac{3.7}{\sqrt{10}}$

Time to crunch the numbers: $E = 1.833 \cdot \frac{3.7}{3.162}$ $E = 1.833 \cdot 1.169$ $E \approx 2.14$

Our margin of error, E, is approximately 2.14.
So the correct answer choice is 2.14.