Math / Data & Statistics

QuestionEat your cereal: Boxes of cereal are labeled as containing 14 ounces. Following are the weights, in ounces, of a sample of 12 boxes. It is reasonable to assume that the population is approximately normal. \begin{tabular}{llllll} \hline 13.01 & 14.96 & 13.10 & 13.11 & 13.09 & 13.01 \\ 13.14 & 14.96 & 13.04 & 13.03 & 13.10 & 13.11 \\ \hline \end{tabular} Send data to Excel
Part: $0 / 2$
Part 1 of 2 (a) Construct a $90 \%$ confidence interval for the mean weight. Round the answers to at least three decimal places.
A $90 \%$ confidence interval for the mean weight is $12.818^{\otimes}<\mu<13.458{ }^{\otimes}$ .

Studdy Solution

STEP 1

What is this asking? We need to find a range where we're 90% sure the true average weight of a cereal box lies, based on weighing 12 boxes. Watch out! Don't mix up the different formulas for confidence intervals – we need the one for the mean weight, and make sure to use the correct t-score since our sample size is small!

STEP 2

1. Calculate the sample mean.
2. Calculate the sample standard deviation.
3. Find the critical t-value.
4. Calculate the margin of error.
5. Construct the confidence interval.

STEP 3

Let's add up all the weights and divide by the number of boxes to get the sample mean, often written as $\bar{x}$ .
This tells us the average weight of the boxes in our sample.

STEP 4

$\bar{x} = \frac{13.01 + 14.96 + 13.10 + 13.11 + 13.09 + 13.01 + 13.14 + 14.96 + 13.04 + 13.03 + 13.10 + 13.11}{12}$

STEP 5

$\bar{x} = \frac{156.66}{12} \approx 13.055$ So, our sample mean is approximately $13.055$ ounces!

STEP 6

Now, let's see how spread out our weights are.
We'll use the sample standard deviation, usually shown as $s$ , to measure this spread.

STEP 7

First, we find the variance, $s^2$ , by calculating the average of the squared differences between each weight and the sample mean: $s^2 = \frac{\sum_{i=1}^{12}(x_i - \bar{x})^2}{12-1}$

STEP 8

Plugging in our values, we get: $s^2 = \frac{(13.01-13.055)^2 + (14.96-13.055)^2 + \dots + (13.11-13.055)^2}{11} \approx 0.602$

STEP 9

The sample standard deviation, $s$ , is the square root of the variance: $s = \sqrt{s^2} = \sqrt{0.602} \approx 0.776$ So, our sample standard deviation is about $0.776$ ounces.

STEP 10

Since our sample size is small (12 boxes), we'll use a t-distribution.
We want a 90% confidence interval, so we need to find the critical t-value, $t_{\alpha/2}$ , that leaves 5% in each tail of the distribution (since $100\% - 90\% = 10\%$ , and $10\% / 2 = 5\%$ ).

STEP 11

We have $12 - 1 = 11$ degrees of freedom.
Looking up the t-value for a 90% confidence interval with 11 degrees of freedom in a t-table (or using a calculator), we find $t_{\alpha/2} \approx 1.796$ .

STEP 12

The margin of error, $E$ , tells us how much our sample mean might differ from the true population mean.
It's calculated as: $E = t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}$ where $s$ is the sample standard deviation and $n$ is the sample size.

STEP 13

Plugging in our values, we get: $E = 1.796 \cdot \frac{0.776}{\sqrt{12}} \approx 0.403$ Our margin of error is approximately $0.403$ ounces.

STEP 14

Finally, we can build our confidence interval!
It's centered around our sample mean and extends out by the margin of error on either side.

STEP 15

The formula is: $\bar{x} - E < \mu < \bar{x} + E$

STEP 16

Plugging in our values: $13.055 - 0.403 < \mu < 13.055 + 0.403$ $12.652 < \mu < 13.458$

STEP 17

We are 90% confident that the true mean weight of the cereal boxes lies between $\bf{12.652}$ ounces and $\bf{13.458}$ ounces.

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Studdy Solution

STEP 1

STEP 2

1. Calculate the sample mean.
2. Calculate the sample standard deviation.
3. Find the critical t-value.
4. Calculate the margin of error.
5. Construct the confidence interval.

STEP 3

Let's add up all the weights and divide by the number of boxes to get the sample mean, often written as $\bar{x}$ .
This tells us the average weight of the boxes in our sample.

STEP 4

$\bar{x} = \frac{13.01 + 14.96 + 13.10 + 13.11 + 13.09 + 13.01 + 13.14 + 14.96 + 13.04 + 13.03 + 13.10 + 13.11}{12}$

STEP 5

$\bar{x} = \frac{156.66}{12} \approx 13.055$ So, our sample mean is approximately $13.055$ ounces!

STEP 6

Now, let's see how spread out our weights are.
We'll use the sample standard deviation, usually shown as $s$ , to measure this spread.

STEP 7

First, we find the variance, $s^2$ , by calculating the average of the squared differences between each weight and the sample mean: $s^2 = \frac{\sum_{i=1}^{12}(x_i - \bar{x})^2}{12-1}$

STEP 8

Plugging in our values, we get: $s^2 = \frac{(13.01-13.055)^2 + (14.96-13.055)^2 + \dots + (13.11-13.055)^2}{11} \approx 0.602$

STEP 9

The sample standard deviation, $s$ , is the square root of the variance: $s = \sqrt{s^2} = \sqrt{0.602} \approx 0.776$ So, our sample standard deviation is about $0.776$ ounces.

STEP 10

Since our sample size is small (12 boxes), we'll use a t-distribution.
We want a 90% confidence interval, so we need to find the critical t-value, $t_{\alpha/2}$ , that leaves 5% in each tail of the distribution (since $100\% - 90\% = 10\%$ , and $10\% / 2 = 5\%$ ).

STEP 11

We have $12 - 1 = 11$ degrees of freedom.
Looking up the t-value for a 90% confidence interval with 11 degrees of freedom in a t-table (or using a calculator), we find $t_{\alpha/2} \approx 1.796$ .

STEP 12

The margin of error, $E$ , tells us how much our sample mean might differ from the true population mean.
It's calculated as: $E = t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}$ where $s$ is the sample standard deviation and $n$ is the sample size.

STEP 13

Plugging in our values, we get: $E = 1.796 \cdot \frac{0.776}{\sqrt{12}} \approx 0.403$ Our margin of error is approximately $0.403$ ounces.

STEP 14

Finally, we can build our confidence interval!
It's centered around our sample mean and extends out by the margin of error on either side.

STEP 15

The formula is: $\bar{x} - E < \mu < \bar{x} + E$

STEP 16

Plugging in our values: $13.055 - 0.403 < \mu < 13.055 + 0.403$ $12.652 < \mu < 13.458$

STEP 17

We are 90% confident that the true mean weight of the cereal boxes lies between $\bf{12.652}$ ounces and $\bf{13.458}$ ounces.

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Studdy Solution

STEP 1

STEP 2

1. Calculate the sample mean.2. Calculate the sample standard deviation.3. Find the critical t-value.4. Calculate the margin of error.5. Construct the confidence interval.

STEP 3

Let's **add up** all the weights and **divide** by the number of boxes to get the **sample mean**, often written as xˉ\bar{x}xˉ.This tells us the average weight of the boxes in our sample.

STEP 4

xˉ=13.01+14.96+13.10+13.11+13.09+13.01+13.14+14.96+13.04+13.03+13.10+13.1112 \bar{x} = \frac{13.01 + 14.96 + 13.10 + 13.11 + 13.09 + 13.01 + 13.14 + 14.96 + 13.04 + 13.03 + 13.10 + 13.11}{12} xˉ=1213.01+14.96+13.10+13.11+13.09+13.01+13.14+14.96+13.04+13.03+13.10+13.11​

STEP 5

xˉ=156.6612≈13.055 \bar{x} = \frac{156.66}{12} \approx 13.055 xˉ=12156.66​≈13.055 So, our **sample mean** is approximately 13.05513.05513.055 ounces!

STEP 6

Now, let's see how spread out our weights are.We'll use the **sample standard deviation**, usually shown as sss, to measure this spread.

STEP 7

First, we find the **variance**, s2s^2s2, by calculating the average of the squared differences between each weight and the sample mean: s2=∑i=112(xi−xˉ)212−1 s^2 = \frac{\sum_{i=1}^{12}(x_i - \bar{x})^2}{12-1} s2=12−1∑i=112​(xi​−xˉ)2​

STEP 8

Plugging in our values, we get: s2=(13.01−13.055)2+(14.96−13.055)2+⋯+(13.11−13.055)211≈0.602 s^2 = \frac{(13.01-13.055)^2 + (14.96-13.055)^2 + \dots + (13.11-13.055)^2}{11} \approx 0.602 s2=11(13.01−13.055)2+(14.96−13.055)2+⋯+(13.11−13.055)2​≈0.602

STEP 9

The **sample standard deviation**, sss, is the square root of the variance: s=s2=0.602≈0.776 s = \sqrt{s^2} = \sqrt{0.602} \approx 0.776 s=s2​=0.602​≈0.776 So, our **sample standard deviation** is about 0.7760.7760.776 ounces.

STEP 10

STEP 11

We have 12−1=1112 - 1 = 1112−1=11 degrees of freedom.Looking up the t-value for a 90% confidence interval with 11 degrees of freedom in a t-table (or using a calculator), we find tα/2≈1.796t_{\alpha/2} \approx 1.796tα/2​≈1.796.

STEP 12

The **margin of error**, EEE, tells us how much our sample mean might differ from the true population mean.It's calculated as: E=tα/2⋅sn E = t_{\alpha/2} \cdot \frac{s}{\sqrt{n}} E=tα/2​⋅n​s​ where sss is the sample standard deviation and nnn is the sample size.

STEP 13

Plugging in our values, we get: E=1.796⋅0.77612≈0.403 E = 1.796 \cdot \frac{0.776}{\sqrt{12}} \approx 0.403 E=1.796⋅12​0.776​≈0.403 Our **margin of error** is approximately 0.4030.4030.403 ounces.

STEP 14

Finally, we can build our **confidence interval**!It's centered around our sample mean and extends out by the margin of error on either side.

STEP 15

The formula is: xˉ−E<μ<xˉ+E \bar{x} - E < \mu < \bar{x} + E xˉ−E<μ<xˉ+E

STEP 16

Plugging in our values: 13.055−0.403<μ<13.055+0.403 13.055 - 0.403 < \mu < 13.055 + 0.403 13.055−0.403<μ<13.055+0.403 12.652<μ<13.458 12.652 < \mu < 13.458 12.652<μ<13.458

STEP 17

We are 90% confident that the true mean weight of the cereal boxes lies between 12.652\bf{12.652}12.652 ounces and 13.458\bf{13.458}13.458 ounces.

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Studdy solves anything!

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1. Calculate the sample mean.
2. Calculate the sample standard deviation.
3. Find the critical t-value.
4. Calculate the margin of error.
5. Construct the confidence interval.

Let's add up all the weights and divide by the number of boxes to get the sample mean, often written as $\bar{x}$ .
This tells us the average weight of the boxes in our sample.

$\bar{x} = \frac{13.01 + 14.96 + 13.10 + 13.11 + 13.09 + 13.01 + 13.14 + 14.96 + 13.04 + 13.03 + 13.10 + 13.11}{12}$

$\bar{x} = \frac{156.66}{12} \approx 13.055$ So, our sample mean is approximately $13.055$ ounces!

Now, let's see how spread out our weights are.
We'll use the sample standard deviation, usually shown as $s$ , to measure this spread.

First, we find the variance, $s^2$ , by calculating the average of the squared differences between each weight and the sample mean: $s^2 = \frac{\sum_{i=1}^{12}(x_i - \bar{x})^2}{12-1}$

Plugging in our values, we get: $s^2 = \frac{(13.01-13.055)^2 + (14.96-13.055)^2 + \dots + (13.11-13.055)^2}{11} \approx 0.602$

The sample standard deviation, $s$ , is the square root of the variance: $s = \sqrt{s^2} = \sqrt{0.602} \approx 0.776$ So, our sample standard deviation is about $0.776$ ounces.

We have $12 - 1 = 11$ degrees of freedom.
Looking up the t-value for a 90% confidence interval with 11 degrees of freedom in a t-table (or using a calculator), we find $t_{\alpha/2} \approx 1.796$ .

The margin of error, $E$ , tells us how much our sample mean might differ from the true population mean.
It's calculated as: $E = t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}$ where $s$ is the sample standard deviation and $n$ is the sample size.

Plugging in our values, we get: $E = 1.796 \cdot \frac{0.776}{\sqrt{12}} \approx 0.403$ Our margin of error is approximately $0.403$ ounces.

Finally, we can build our confidence interval!
It's centered around our sample mean and extends out by the margin of error on either side.

The formula is: $\bar{x} - E < \mu < \bar{x} + E$

Plugging in our values: $13.055 - 0.403 < \mu < 13.055 + 0.403$ $12.652 < \mu < 13.458$

We are 90% confident that the true mean weight of the cereal boxes lies between $\bf{12.652}$ ounces and $\bf{13.458}$ ounces.