Solved on Jan 08, 2024

Determine the 90% confidence interval for the true mean weight of a population of dogs, given the sample mean of 69 ounces and population standard deviation of 5.1 ounces.

STEP 1

Assumptions
1. The sample size ( $n$ ) is 40.
2. The sample mean ( $\bar{x}$ ) is 69 ounces.
3. The population standard deviation ( $\sigma$ ) is 5.1 ounces.
4. The confidence level ( $CL$ ) is 90%.
5. We are assuming a normally distributed population or that the sample size is large enough for the Central Limit Theorem to apply.

STEP 2

To construct a confidence interval for the population mean, we will use the formula for a confidence interval when the population standard deviation is known:
$CI = \bar{x} \pm z \left( \frac{\sigma}{\sqrt{n}} \right)$
where $CI$ is the confidence interval, $\bar{x}$ is the sample mean, $z$ is the z-score corresponding to the desired confidence level, $\sigma$ is the population standard deviation, and $n$ is the sample size.

STEP 3

First, we need to find the z-score that corresponds to a 90% confidence level. Since the confidence interval is centered around the mean, we want the z-score that puts 5% of the distribution in each tail (since 100% - 90% = 10%, and 10%/2 = 5%).

STEP 4

We can look up the z-score for a 90% confidence interval in a standard normal distribution table, or use a calculator or software that provides this functionality.

STEP 5

The z-score that corresponds to the upper 5% of the standard normal distribution is approximately 1.645. This is because 50% + 40% (the confidence level) = 90%, so we look for the z-score that corresponds to the remaining 10% / 2 = 5% in the upper tail.

STEP 6

Now we can plug in the values into the confidence interval formula:
$CI = 69 \pm 1.645 \left( \frac{5.1}{\sqrt{40}} \right)$

STEP 7

Calculate the standard error ( $SE$ ) by dividing the population standard deviation by the square root of the sample size:
$SE = \frac{\sigma}{\sqrt{n}} = \frac{5.1}{\sqrt{40}}$

STEP 8

Calculate the standard error:
$SE = \frac{5.1}{\sqrt{40}} \approx \frac{5.1}{6.32} \approx 0.807$

STEP 9

Now, calculate the margin of error ( $ME$ ) by multiplying the z-score by the standard error:
$ME = z \times SE = 1.645 \times 0.807$

STEP 10

Calculate the margin of error:
$ME = 1.645 \times 0.807 \approx 1.327$

STEP 11

Finally, construct the confidence interval by subtracting and adding the margin of error from the sample mean:
$CI = 69 \pm 1.327$

STEP 12

Calculate the lower bound of the confidence interval:
$Lower\, bound = 69 - 1.327 \approx 67.673$

STEP 13

Calculate the upper bound of the confidence interval:
$Upper\, bound = 69 + 1.327 \approx 70.327$

STEP 14

Round the bounds to two decimal places:
$Lower\, bound \approx 67.67\, ounces$ $Upper\, bound \approx 70.33\, ounces$
The $90\%$ confidence interval for the true population mean dog weight is approximately $(67.67, 70.33)$ ounces.

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Determine the 90% confidence interval for the true mean weight of a population of dogs, given the sample mean of 69 ounces and population standard deviation of 5.1 ounces.

STEP 1

STEP 2

STEP 3

First, we need to find the z-score that corresponds to a 90% confidence level. Since the confidence interval is centered around the mean, we want the z-score that puts 5% of the distribution in each tail (since 100% - 90% = 10%, and 10%/2 = 5%).

STEP 4

We can look up the z-score for a 90% confidence interval in a standard normal distribution table, or use a calculator or software that provides this functionality.

STEP 5

The z-score that corresponds to the upper 5% of the standard normal distribution is approximately 1.645. This is because 50% + 40% (the confidence level) = 90%, so we look for the z-score that corresponds to the remaining 10% / 2 = 5% in the upper tail.

STEP 6

Now we can plug in the values into the confidence interval formula:
$CI = 69 \pm 1.645 \left( \frac{5.1}{\sqrt{40}} \right)$

STEP 7

Calculate the standard error ( $SE$ ) by dividing the population standard deviation by the square root of the sample size:
$SE = \frac{\sigma}{\sqrt{n}} = \frac{5.1}{\sqrt{40}}$

STEP 8

Calculate the standard error:
$SE = \frac{5.1}{\sqrt{40}} \approx \frac{5.1}{6.32} \approx 0.807$

STEP 9

Now, calculate the margin of error ( $ME$ ) by multiplying the z-score by the standard error:
$ME = z \times SE = 1.645 \times 0.807$

STEP 10

Calculate the margin of error:
$ME = 1.645 \times 0.807 \approx 1.327$

STEP 11

Finally, construct the confidence interval by subtracting and adding the margin of error from the sample mean:
$CI = 69 \pm 1.327$

STEP 12

Calculate the lower bound of the confidence interval:
$Lower\, bound = 69 - 1.327 \approx 67.673$

STEP 13

Calculate the upper bound of the confidence interval:
$Upper\, bound = 69 + 1.327 \approx 70.327$

STEP 14

Round the bounds to two decimal places:
$Lower\, bound \approx 67.67\, ounces$ $Upper\, bound \approx 70.33\, ounces$
The $90\%$ confidence interval for the true population mean dog weight is approximately $(67.67, 70.33)$ ounces.

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Determine the 90% confidence interval for the true mean weight of a population of dogs, given the sample mean of 69 ounces and population standard deviation of 5.1 ounces.

STEP 1

STEP 2

STEP 3

First, we need to find the z-score that corresponds to a 90% confidence level. Since the confidence interval is centered around the mean, we want the z-score that puts 5% of the distribution in each tail (since 100% - 90% = 10%, and 10%/2 = 5%).

STEP 4

We can look up the z-score for a 90% confidence interval in a standard normal distribution table, or use a calculator or software that provides this functionality.

STEP 5

The z-score that corresponds to the upper 5% of the standard normal distribution is approximately 1.645. This is because 50% + 40% (the confidence level) = 90%, so we look for the z-score that corresponds to the remaining 10% / 2 = 5% in the upper tail.

STEP 6

Now we can plug in the values into the confidence interval formula:CI=69±1.645(5.140)CI = 69 \pm 1.645 \left( \frac{5.1}{\sqrt{40}} \right)CI=69±1.645(40​5.1​)

STEP 7

Calculate the standard error (SESESE) by dividing the population standard deviation by the square root of the sample size:SE=σn=5.140SE = \frac{\sigma}{\sqrt{n}} = \frac{5.1}{\sqrt{40}}SE=n​σ​=40​5.1​

STEP 8

Calculate the standard error:SE=5.140≈5.16.32≈0.807SE = \frac{5.1}{\sqrt{40}} \approx \frac{5.1}{6.32} \approx 0.807SE=40​5.1​≈6.325.1​≈0.807

STEP 9

Now, calculate the margin of error (MEMEME) by multiplying the z-score by the standard error:ME=z×SE=1.645×0.807ME = z \times SE = 1.645 \times 0.807ME=z×SE=1.645×0.807

STEP 10

Calculate the margin of error:ME=1.645×0.807≈1.327ME = 1.645 \times 0.807 \approx 1.327ME=1.645×0.807≈1.327

STEP 11

Finally, construct the confidence interval by subtracting and adding the margin of error from the sample mean:CI=69±1.327CI = 69 \pm 1.327CI=69±1.327

STEP 12

Calculate the lower bound of the confidence interval:Lower bound=69−1.327≈67.673Lower\, bound = 69 - 1.327 \approx 67.673Lowerbound=69−1.327≈67.673

STEP 13

Calculate the upper bound of the confidence interval:Upper bound=69+1.327≈70.327Upper\, bound = 69 + 1.327 \approx 70.327Upperbound=69+1.327≈70.327

STEP 14

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Now we can plug in the values into the confidence interval formula:
$CI = 69 \pm 1.645 \left( \frac{5.1}{\sqrt{40}} \right)$

Calculate the standard error ( $SE$ ) by dividing the population standard deviation by the square root of the sample size:
$SE = \frac{\sigma}{\sqrt{n}} = \frac{5.1}{\sqrt{40}}$

Calculate the standard error:
$SE = \frac{5.1}{\sqrt{40}} \approx \frac{5.1}{6.32} \approx 0.807$

Now, calculate the margin of error ( $ME$ ) by multiplying the z-score by the standard error:
$ME = z \times SE = 1.645 \times 0.807$

Calculate the margin of error:
$ME = 1.645 \times 0.807 \approx 1.327$

Finally, construct the confidence interval by subtracting and adding the margin of error from the sample mean:
$CI = 69 \pm 1.327$

Calculate the lower bound of the confidence interval:
$Lower\, bound = 69 - 1.327 \approx 67.673$

Calculate the upper bound of the confidence interval:
$Upper\, bound = 69 + 1.327 \approx 70.327$