Math / Data & Statistics

QuestionQuestion 33 A sample of 30 incoming students was taken and whether or not they are left-handed was recorded. A $95 \%$ confidence interval for $p$ is $(9.5 \%, 11.1 \%)$ . a.) The individual object in the study was a randomly selected $\square$ .
This is computer-graded so use exact wording from the problem above. Remember an individual is ONE of something. b.) What was the variable information recorded for each object in the study?
This is computer-graded so use exact wording from the problem above. The variable information was whether or not they $\square$ c.) State the statistical interpretation of the confidence interval in the context of this problem.
Select an answer the $\square$ Select an answer of $\square$ $?$ incoming students that are left-handed is between $\square$ \% and $\square$ \% d.) What is the symbol and value of the point estimate for $p$ ? $?+=$ $\square$ \% e.) What is the margin of error for the given interval? $\qquad$ \% f.) Fill in the boxes below to show the relation on the number line between the numeric values of the point estimate and the interval estimate for $p$ . $\qquad$ $\qquad$ B $\qquad$ $A=$ $\square$ \% B = $\square$ \% C = $\square$ \%

Studdy Solution

STEP 1

What is this asking? We're looking at a study of left-handed students and figuring out what the confidence interval really means, along with the point estimate and margin of error. Watch out! Don't mix up the point estimate and the confidence interval!
The interval is a range of plausible values, while the point estimate is a single best guess.

STEP 2

1. Identify the individual
2. Determine the variable
3. Interpret the confidence interval
4. Calculate the point estimate
5. Calculate the margin of error
6. Relate the values on a number line

STEP 3

The problem says "a sample of 30 incoming students was taken".
Each individual in the study is one incoming student!

STEP 4

The problem states that the information recorded was "whether or not they are left-handed".

STEP 5

The problem gives us a 95% confidence interval of (9.5%, 11.1%).

STEP 6

This means we are 95% confident that the true percentage of left-handed incoming students is somewhere between 9.5% and 11.1%.

STEP 7

The point estimate is the middle of the confidence interval.
It's our best single guess for the true percentage.

STEP 8

To find the middle, we add the endpoints and divide by two: $\frac{9.5\% + 11.1\%}{2} = \frac{20.6\%}{2} = 10.3\%$ So, our point estimate is 10.3%.

STEP 9

The margin of error is the distance from the point estimate to either end of the interval.

STEP 10

We can find this by subtracting the lower endpoint from the point estimate (or vice-versa): $11.1\% - 10.3\% = 0.8\%$ Our margin of error is 0.8%.

STEP 11

We have our lower bound (A), point estimate (B), and upper bound (C).

STEP 12

From the confidence interval, A = 9.5% and C = 11.1%.

STEP 13

We calculated our point estimate B = 10.3%.
Notice that B is right in between A and C!

STEP 14

a) incoming student b) are left-handed c) We are 95% confident that the true proportion of incoming students that are left-handed is between 9.5% and 11.1% d) $\hat{p} = 10.3\%$ e) 0.8% f) A = 9.5%, B = 10.3%, C = 11.1%

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Studdy Solution

STEP 1

STEP 2

1. Identify the individual
2. Determine the variable
3. Interpret the confidence interval
4. Calculate the point estimate
5. Calculate the margin of error
6. Relate the values on a number line

STEP 3

The problem says "a sample of 30 incoming students was taken".
Each individual in the study is one incoming student!

STEP 4

The problem states that the information recorded was "whether or not they are left-handed".

STEP 5

The problem gives us a 95% confidence interval of (9.5%, 11.1%).

STEP 6

This means we are 95% confident that the true percentage of left-handed incoming students is somewhere between 9.5% and 11.1%.

STEP 7

The point estimate is the middle of the confidence interval.
It's our best single guess for the true percentage.

STEP 8

To find the middle, we add the endpoints and divide by two: $\frac{9.5\% + 11.1\%}{2} = \frac{20.6\%}{2} = 10.3\%$ So, our point estimate is 10.3%.

STEP 9

The margin of error is the distance from the point estimate to either end of the interval.

STEP 10

We can find this by subtracting the lower endpoint from the point estimate (or vice-versa): $11.1\% - 10.3\% = 0.8\%$ Our margin of error is 0.8%.

STEP 11

We have our lower bound (A), point estimate (B), and upper bound (C).

STEP 12

From the confidence interval, A = 9.5% and C = 11.1%.

STEP 13

We calculated our point estimate B = 10.3%.
Notice that B is right in between A and C!

STEP 14

a) incoming student b) are left-handed c) We are 95% confident that the true proportion of incoming students that are left-handed is between 9.5% and 11.1% d) $\hat{p} = 10.3\%$ e) 0.8% f) A = 9.5%, B = 10.3%, C = 11.1%

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Studdy Solution

STEP 1

STEP 2

1. Identify the individual2. Determine the variable3. Interpret the confidence interval4. Calculate the point estimate5. Calculate the margin of error6. Relate the values on a number line

STEP 3

The problem says "a sample of **30 incoming students** was taken".Each individual in the study is one incoming student!

STEP 4

The problem states that the information recorded was "**whether or not they are left-handed**".

STEP 5

The problem gives us a **95%** confidence interval of **(9.5%, 11.1%)**.

STEP 6

This means we are 95% confident that the *true* percentage of left-handed incoming students is somewhere between **9.5%** and **11.1%**.

STEP 7

The point estimate is the middle of the confidence interval.It's our best single guess for the true percentage.

STEP 8

To find the middle, we add the endpoints and divide by two: 9.5%+11.1%2=20.6%2=10.3% \frac{9.5\% + 11.1\%}{2} = \frac{20.6\%}{2} = 10.3\% 29.5%+11.1%​=220.6%​=10.3% So, our point estimate is **10.3%**.

STEP 9

The margin of error is the distance from the point estimate to either end of the interval.

STEP 10

We can find this by subtracting the lower endpoint from the point estimate (or vice-versa): 11.1%−10.3%=0.8% 11.1\% - 10.3\% = 0.8\% 11.1%−10.3%=0.8% Our margin of error is **0.8%**.

STEP 11

We have our lower bound (A), point estimate (B), and upper bound (C).

STEP 12

From the confidence interval, A = **9.5%** and C = **11.1%**.

STEP 13

We calculated our point estimate B = **10.3%**.Notice that B is right in between A and C!

STEP 14

a) incoming student b) are left-handed c) We are 95% confident that the true proportion of incoming students that are left-handed is between 9.5% and 11.1% d) p^=10.3%\hat{p} = 10.3\%p^​=10.3% e) 0.8% f) A = 9.5%, B = 10.3%, C = 11.1%

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

1. Identify the individual
2. Determine the variable
3. Interpret the confidence interval
4. Calculate the point estimate
5. Calculate the margin of error
6. Relate the values on a number line

The problem says "a sample of 30 incoming students was taken".
Each individual in the study is one incoming student!

The problem states that the information recorded was "whether or not they are left-handed".

The problem gives us a 95% confidence interval of (9.5%, 11.1%).

This means we are 95% confident that the true percentage of left-handed incoming students is somewhere between 9.5% and 11.1%.

The point estimate is the middle of the confidence interval.
It's our best single guess for the true percentage.

To find the middle, we add the endpoints and divide by two: $\frac{9.5\% + 11.1\%}{2} = \frac{20.6\%}{2} = 10.3\%$ So, our point estimate is 10.3%.

We can find this by subtracting the lower endpoint from the point estimate (or vice-versa): $11.1\% - 10.3\% = 0.8\%$ Our margin of error is 0.8%.

From the confidence interval, A = 9.5% and C = 11.1%.

We calculated our point estimate B = 10.3%.
Notice that B is right in between A and C!

a) incoming student b) are left-handed c) We are 95% confident that the true proportion of incoming students that are left-handed is between 9.5% and 11.1% d) $\hat{p} = 10.3\%$ e) 0.8% f) A = 9.5%, B = 10.3%, C = 11.1%