Math / Data & Statistics

QuestionThe weights of four randomly and independently selected bags of potatoes labeled 20.0 pounds were found to be 20.8, $21.4,20.9$ , and 21.2 pounds. $\square$ Assume Normality. Answer parts (a) and (b) below. a. Find a $95 \%$ confidence interval for the mean weight of all bags of potatoes. $(20.64$ , 21.51) (Type integers or decimals rounded to the nearest hundredth as needed. Use ascending order.) b. Does the interval capture 20.0 pounds? Is there enough evidence to reject a mean weight of 20.0 pounds? A. The interval captures 20.0 pounds, su there is not enough evidence to reject a mean weight of 20.0 pounds. It is plausible the population mean weight is 20.0 pounds. B. The interval does not capture 20.0 pounds, so there not is enough evidence to reject a mean weight of 20.0 pounds. It is plausible the population mean weight is 20.0 pounds. C. The interval captures 20.0 pounds, so there is enough evidence to reject a mean weight of 20.0 pounds. It is not plausible the population mean weight is 20.0 pounds. D. The interval does not capture 20.0 pounds, so there is enough evidence to reject a mean weight of 20.0 pounds. It is not plausible the population mean weight is 20.0 pounds. E. There is insufficient information to make a decision regarding the rejection of 20.0 pounds. The sample size of 4 bags is less than the required 25.

Studdy Solution

STEP 1

What is this asking? We're checking if the average weight of potato bags is actually 20 pounds, given the weights of four bags, and we want to be 95% sure about our conclusion! Watch out! Don't mix up the sample mean with the population mean.
Also, remember that a confidence interval gives a range of plausible values, not just a single number.

STEP 2

1. Calculate the sample mean and standard deviation.
2. Find the margin of error.
3. Construct the confidence interval.
4. Check if 20.0 pounds is within the interval.

STEP 3

Alright, let's kick things off by finding the average weight of the four bags.
We add up all the weights and divide by the number of bags, which is 4.
So, $20.8 + 21.4 + 20.9 + 21.2/4 = 84.3/4 = 21.075$.
The sample mean, $\bar{x}$ , is 21.075 pounds.

STEP 4

Next, we need to see how spread out our weights are.
This is where the standard deviation comes in!
First, we find the difference between each weight and the sample mean (21.075), square each difference, add them all up, divide by $n-1 = 4-1 = 3$ (that's one less than the number of bags), and finally take the square root.
Here's the calculation: $\sqrt{\frac{(20.8-21.075)^2 + (21.4-21.075)^2 + (20.9-21.075)^2 + (21.2-21.075)^2}{3}} = \sqrt{\frac{0.075625 + 0.105625 + 0.030625 + 0.015625}{3}} = \sqrt{\frac{0.2275}{3}} \approx \sqrt{0.075833} \approx 0.275$ So, our sample standard deviation, $s$ , is approximately 0.275 pounds.

STEP 5

For a 95% confidence interval with 3 degrees of freedom (remember, it's $n-1 = 4-1 = 3$ ), our critical value ( $t^*$ ) from the t-distribution table is approximately 3.182.

STEP 6

The margin of error tells us how much our sample mean might differ from the true population mean.
We calculate it by multiplying the critical value by the standard error, which is the sample standard deviation divided by the square root of the sample size: $3.182 \cdot \frac{0.275}{\sqrt{4}} \approx 3.182 \cdot \frac{0.275}{2} \approx 3.182 \cdot 0.1375 \approx 0.436$ Our margin of error is approximately 0.436 pounds.

STEP 7

To get the confidence interval, we subtract the margin of error from the sample mean to get the lower bound

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Studdy Solution

STEP 1

STEP 2

1. Calculate the sample mean and standard deviation.
2. Find the margin of error.
3. Construct the confidence interval.
4. Check if 20.0 pounds is within the interval.

STEP 3

Alright, let's kick things off by finding the average weight of the four bags.
We add up all the weights and divide by the number of bags, which is 4.
So, \(20.8 + 21.4 + 20.9 + 21.2/4 = 84.3/4 = 21.075\).
The sample mean, $\bar{x}$ , is 21.075 pounds.

STEP 4

STEP 5

For a 95% confidence interval with 3 degrees of freedom (remember, it's $n-1 = 4-1 = 3$ ), our critical value ( $t^*$ ) from the t-distribution table is approximately 3.182.

STEP 6

STEP 7

To get the confidence interval, we subtract the margin of error from the sample mean to get the lower bound

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Studdy Solution

STEP 1

STEP 2

1. Calculate the sample mean and standard deviation.2. Find the margin of error.3. Construct the confidence interval.4. Check if 20.0 pounds is within the interval.

STEP 3

Alright, let's **kick things off** by finding the average weight of the four bags.We **add up** all the weights and **divide** by the number of bags, which is 4.So, \(20.8 + 21.4 + 20.9 + 21.2/4 = 84.3/4 = 21.075\).The **sample mean**, xˉ\bar{x}xˉ, is **21.075** pounds.

STEP 4

STEP 5

For a 95% confidence interval with 3 degrees of freedom (remember, it's n−1=4−1=3n-1 = 4-1 = 3n−1=4−1=3), our **critical value** (t∗t^*t∗) from the t-distribution table is approximately **3.182**.

STEP 6

STEP 7

To get the **confidence interval**, we subtract the **margin of error** from the **sample mean** to get the lower bound

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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. Calculate the sample mean and standard deviation.
2. Find the margin of error.
3. Construct the confidence interval.
4. Check if 20.0 pounds is within the interval.

Alright, let's kick things off by finding the average weight of the four bags.
We add up all the weights and divide by the number of bags, which is 4.
So, \(20.8 + 21.4 + 20.9 + 21.2/4 = 84.3/4 = 21.075\).
The sample mean, $\bar{x}$ , is 21.075 pounds.

For a 95% confidence interval with 3 degrees of freedom (remember, it's $n-1 = 4-1 = 3$ ), our critical value ( $t^*$ ) from the t-distribution table is approximately 3.182.

To get the confidence interval, we subtract the margin of error from the sample mean to get the lower bound