Math / Data & Statistics

QuestionConfidence interval for the population mean: Use of the $t$ distribution
A corporation that maintains a large fleet of company cars for the use of its sales staff is interested in the mean distance driven monthly per salesperson. The Español following list gives the monthly distances in miles driven by a random sample of 15 salespeople. $2346,2026,2561,2437,2244,2180,2108,2271,2382,1953,2356,2309,2391,1958,2605$ Send data to calculator Send data to Excel
Based on this sample, find a $99 \%$ confidence interval for the mean number of miles driven monthly by members of the sales staff, assuming that monthly driving distances are normally distributed. Give the lower limit and upper limit of the $99 \%$ confidence interval.
Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list of formulas.)
Lower limit: $\square$ Upper limit: $\square$

Studdy Solution

STEP 1

What is this asking? We need to figure out a range where we're 99% sure the true average monthly miles driven by all salespeople falls, based on the miles driven by a small group of 15 salespeople. Watch out! Don't mix up the sample mean (average of the 15 salespeople) with the population mean (average of all salespeople).
We're trying to estimate the population mean!
Also, remember to use the t-distribution since our sample size is small and we don't know the true population standard deviation.

STEP 2

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

STEP 3

Let's add up all the miles driven by our 15 salespeople: $2346 + 2026 + 2561 + 2437 + 2244 + 2180 + 2108 + 2271 + 2382 + 1953 + 2356 + 2309 + 2391 + 1958 + 2605 = 33527$ .

STEP 4

Now, divide this sum by the number of salespeople (15) to get the sample mean, often denoted as $\bar{x}$ : $\bar{x} = \frac{33527}{15} = 2235.133$ So, our sample salespeople drove an average of 2235.133 miles per month.

STEP 5

For each salesperson's mileage, we need to find how far it is from our sample mean of 2235.133, square that difference, and then add all those squared differences together.
This gives us: $\sum_{i=1}^{15} (x_i - \bar{x})^2 = 1231823.733$ .

STEP 6

Next, we divide that big sum by one less than the number of salespeople (15 - 1 = 14).
This is called the sample variance: $s^2 = \frac{1231823.733}{14} = 87987.410$

STEP 7

Finally, we take the square root of the sample variance to get the sample standard deviation, denoted as $s$ : $s = \sqrt{87987.410} = 296.627$ This tells us how spread out the mileage is among our sample salespeople.

STEP 8

Since we want a 99% confidence interval, this means we have 0.5% of the area in each tail of the t-distribution.
So, we're looking for the t-score that corresponds to an area of 0.995 (1 - 0.005).

STEP 9

We also need something called degrees of freedom, which is just the number of salespeople minus 1 (15 - 1 = 14).

STEP 10

Using a t-table or calculator with these values, we find our t-score is approximately 2.977.

STEP 11

The margin of error tells us how much "wiggle room" we need around our sample mean.
We calculate it by multiplying the t-score (2.977) by the sample standard deviation (296.627) and then dividing by the square root of the number of salespeople: $\text{Margin of Error} = 2.977 \cdot \frac{296.627}{\sqrt{15}} = 228.218$

STEP 12

To get the lower limit of our confidence interval, we subtract the margin of error (228.218) from the sample mean (2235.133): $\text{Lower Limit} = 2235.133 - 228.218 = 2006.9$

STEP 13

To get the upper limit, we add the margin of error to the sample mean: $\text{Upper Limit} = 2235.133 + 228.218 = 2463.3$

STEP 14

We are 99% confident that the true average monthly mileage for all salespeople falls between 2006.9 miles and 2463.3 miles.

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Studdy Solution

STEP 1

STEP 2

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

STEP 3

Let's add up all the miles driven by our 15 salespeople: $2346 + 2026 + 2561 + 2437 + 2244 + 2180 + 2108 + 2271 + 2382 + 1953 + 2356 + 2309 + 2391 + 1958 + 2605 = 33527$ .

STEP 4

Now, divide this sum by the number of salespeople (15) to get the sample mean, often denoted as $\bar{x}$ : $\bar{x} = \frac{33527}{15} = 2235.133$ So, our sample salespeople drove an average of 2235.133 miles per month.

STEP 5

For each salesperson's mileage, we need to find how far it is from our sample mean of 2235.133, square that difference, and then add all those squared differences together.
This gives us: $\sum_{i=1}^{15} (x_i - \bar{x})^2 = 1231823.733$ .

STEP 6

Next, we divide that big sum by one less than the number of salespeople (15 - 1 = 14).
This is called the sample variance: $s^2 = \frac{1231823.733}{14} = 87987.410$

STEP 7

Finally, we take the square root of the sample variance to get the sample standard deviation, denoted as $s$ : $s = \sqrt{87987.410} = 296.627$ This tells us how spread out the mileage is among our sample salespeople.

STEP 8

Since we want a 99% confidence interval, this means we have 0.5% of the area in each tail of the t-distribution.
So, we're looking for the t-score that corresponds to an area of 0.995 (1 - 0.005).

STEP 9

We also need something called degrees of freedom, which is just the number of salespeople minus 1 (15 - 1 = 14).

STEP 10

Using a t-table or calculator with these values, we find our t-score is approximately 2.977.

STEP 11

STEP 12

To get the lower limit of our confidence interval, we subtract the margin of error (228.218) from the sample mean (2235.133): $\text{Lower Limit} = 2235.133 - 228.218 = 2006.9$

STEP 13

To get the upper limit, we add the margin of error to the sample mean: $\text{Upper Limit} = 2235.133 + 228.218 = 2463.3$

STEP 14

We are 99% confident that the true average monthly mileage for all salespeople falls between 2006.9 miles and 2463.3 miles.

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Studdy Solution

STEP 1

STEP 2

1. Calculate the sample mean2. Calculate the sample standard deviation3. Find the t-score4. Calculate the margin of error5. Calculate the confidence interval

STEP 3

STEP 4

Now, **divide** this sum by the **number of salespeople (15)** to get the **sample mean**, often denoted as xˉ\bar{x}xˉ: xˉ=3352715=2235.133\bar{x} = \frac{33527}{15} = 2235.133xˉ=1533527​=2235.133 So, our sample salespeople drove an average of **2235.133 miles** per month.

STEP 5

STEP 6

Next, we **divide** that big sum by **one less than the number of salespeople (15 - 1 = 14)**.This is called the **sample variance**: s2=1231823.73314=87987.410s^2 = \frac{1231823.733}{14} = 87987.410s2=141231823.733​=87987.410

STEP 7

Finally, we take the **square root** of the **sample variance** to get the **sample standard deviation**, denoted as sss: s=87987.410=296.627s = \sqrt{87987.410} = 296.627s=87987.410​=296.627 This tells us how spread out the mileage is among our sample salespeople.

STEP 8

Since we want a **99% confidence interval**, this means we have **0.5%** of the area in each tail of the *t*-distribution.So, we're looking for the *t*-score that corresponds to an area of **0.995** (1 - 0.005).

STEP 9

We also need something called **degrees of freedom**, which is just the **number of salespeople minus 1 (15 - 1 = 14)**.

STEP 10

Using a *t*-table or calculator with these values, we find our **t-score** is approximately **2.977**.

STEP 11

STEP 12

To get the **lower limit** of our confidence interval, we **subtract** the **margin of error (228.218)** from the **sample mean (2235.133)**: Lower Limit=2235.133−228.218=2006.9\text{Lower Limit} = 2235.133 - 228.218 = 2006.9 Lower Limit=2235.133−228.218=2006.9

STEP 13

To get the **upper limit**, we **add** the **margin of error** to the **sample mean**: Upper Limit=2235.133+228.218=2463.3\text{Upper Limit} = 2235.133 + 228.218 = 2463.3 Upper Limit=2235.133+228.218=2463.3

STEP 14

We are 99% confident that the true average monthly mileage for all salespeople falls between **2006.9 miles** and **2463.3 miles**.

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

Now, divide this sum by the number of salespeople (15) to get the sample mean, often denoted as $\bar{x}$ : $\bar{x} = \frac{33527}{15} = 2235.133$ So, our sample salespeople drove an average of 2235.133 miles per month.

Next, we divide that big sum by one less than the number of salespeople (15 - 1 = 14).
This is called the sample variance: $s^2 = \frac{1231823.733}{14} = 87987.410$

Finally, we take the square root of the sample variance to get the sample standard deviation, denoted as $s$ : $s = \sqrt{87987.410} = 296.627$ This tells us how spread out the mileage is among our sample salespeople.

Since we want a 99% confidence interval, this means we have 0.5% of the area in each tail of the t-distribution.
So, we're looking for the t-score that corresponds to an area of 0.995 (1 - 0.005).

We also need something called degrees of freedom, which is just the number of salespeople minus 1 (15 - 1 = 14).

Using a t-table or calculator with these values, we find our t-score is approximately 2.977.

To get the lower limit of our confidence interval, we subtract the margin of error (228.218) from the sample mean (2235.133): $\text{Lower Limit} = 2235.133 - 228.218 = 2006.9$

To get the upper limit, we add the margin of error to the sample mean: $\text{Upper Limit} = 2235.133 + 228.218 = 2463.3$

We are 99% confident that the true average monthly mileage for all salespeople falls between 2006.9 miles and 2463.3 miles.