Math / Data & Statistics

QuestionYou measure 36 randomly selected textbooks' weights, and find they have a mean weight of 76 ounces. Assume the population standard deviation is 10 ounces. Based on this, construct a $95 \%$ confidence interval for the true population mean textbook weight.
Give your answers as decimals, to two places

Studdy Solution

STEP 1

What is this asking? We're trying to estimate the average weight of all textbooks, using the average weight from our sample of 36 textbooks, and we want to be 95% confident that the true average falls within our estimated range. Watch out! Don't mix up the sample standard deviation and the population standard deviation!
We're given the population standard deviation here, which simplifies things a bit.

STEP 2

1. Find the critical value.
2. Calculate the margin of error.
3. Construct the confidence interval.

STEP 3

Since we're aiming for a 95% confidence interval, this means there's $5\%$ left over for the tails of our normal distribution.
Because the normal distribution is symmetric, we split that $5\%$ evenly between the two tails, leaving $\frac{5\%}{2} = 2.5\%$ in each tail.

STEP 4

We need to find the z-score that corresponds to the area to the left of it being $95\% + 2.5\% = 97.5\%$ or $0.975$ .
This special z-score is called the critical value, and we can look it up in a z-table or use a calculator.
For a 95% confidence level, the critical value is $z = 1.96$ .

STEP 5

The margin of error tells us how much "wiggle room" we have around our sample mean.
The formula for the margin of error when we know the population standard deviation is: $\text{Margin of Error} = z \cdot \frac{\sigma}{\sqrt{n}}$ where $z$ is our critical value, $\sigma$ is the population standard deviation, and $n$ is our sample size.

STEP 6

Let's plug in our values.
We found $z = 1.96$ , we were given $\sigma = 10$ , and our sample size is $n = 36$ . $\text{Margin of Error} = 1.96 \cdot \frac{10}{\sqrt{36}}$ $\text{Margin of Error} = 1.96 \cdot \frac{10}{6}$ $\text{Margin of Error} = 1.96 \cdot 1.6666\dots$ $\text{Margin of Error} \approx 3.27$ So our margin of error is approximately 3.27 ounces.

STEP 7

The confidence interval is built around our sample mean, which is $\bar{x} = 76$ ounces.
We take the sample mean and add/subtract the margin of error to get our interval: $\text{Confidence Interval} = \bar{x} \pm \text{Margin of Error}$

STEP 8

Plugging in our values: $\text{Confidence Interval} = 76 \pm 3.27$ This gives us a lower bound of $76 - 3.27 = 72.73$ ounces and an upper bound of $76 + 3.27 = 79.27$ ounces.

STEP 9

We are 95% confident that the true population mean textbook weight is between 72.73 ounces and 79.27 ounces.

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Studdy Solution

STEP 1

STEP 2

1. Find the critical value.
2. Calculate the margin of error.
3. Construct the confidence interval.

STEP 3

Since we're aiming for a 95% confidence interval, this means there's $5\%$ left over for the tails of our normal distribution.
Because the normal distribution is symmetric, we split that $5\%$ evenly between the two tails, leaving $\frac{5\%}{2} = 2.5\%$ in each tail.

STEP 4

We need to find the z-score that corresponds to the area to the left of it being $95\% + 2.5\% = 97.5\%$ or $0.975$ .
This special z-score is called the critical value, and we can look it up in a z-table or use a calculator.
For a 95% confidence level, the critical value is $z = 1.96$ .

STEP 5

STEP 6

STEP 7

The confidence interval is built around our sample mean, which is $\bar{x} = 76$ ounces.
We take the sample mean and add/subtract the margin of error to get our interval: $\text{Confidence Interval} = \bar{x} \pm \text{Margin of Error}$

STEP 8

Plugging in our values: $\text{Confidence Interval} = 76 \pm 3.27$ This gives us a lower bound of $76 - 3.27 = 72.73$ ounces and an upper bound of $76 + 3.27 = 79.27$ ounces.

STEP 9

We are 95% confident that the true population mean textbook weight is between 72.73 ounces and 79.27 ounces.

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Studdy Solution

STEP 1

STEP 2

1. Find the critical value.2. Calculate the margin of error.3. Construct the confidence interval.

STEP 3

Since we're aiming for a **95% confidence interval**, this means there's 5%5\%5% left over for the tails of our normal distribution.Because the normal distribution is symmetric, we split that 5%5\%5% evenly between the two tails, leaving 5%2=2.5%\frac{5\%}{2} = 2.5\%25%​=2.5% in each tail.

STEP 4

STEP 5

STEP 6

STEP 7

STEP 8

Plugging in our values: Confidence Interval=76±3.27 \text{Confidence Interval} = 76 \pm 3.27 Confidence Interval=76±3.27 This gives us a lower bound of 76−3.27=72.7376 - 3.27 = 72.7376−3.27=72.73 ounces and an upper bound of 76+3.27=79.2776 + 3.27 = 79.2776+3.27=79.27 ounces.

STEP 9

We are 95% confident that the true population mean textbook weight is between **72.73 ounces** and **79.27 ounces**.

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Studdy solves anything!

Start learning now

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

1. Find the critical value.
2. Calculate the margin of error.
3. Construct the confidence interval.

Since we're aiming for a 95% confidence interval, this means there's $5\%$ left over for the tails of our normal distribution.
Because the normal distribution is symmetric, we split that $5\%$ evenly between the two tails, leaving $\frac{5\%}{2} = 2.5\%$ in each tail.

Plugging in our values: $\text{Confidence Interval} = 76 \pm 3.27$ This gives us a lower bound of $76 - 3.27 = 72.73$ ounces and an upper bound of $76 + 3.27 = 79.27$ ounces.

We are 95% confident that the true population mean textbook weight is between 72.73 ounces and 79.27 ounces.