Solved on Mar 26, 2024

A study found that a new sneaker increased the average jump height of 32 male athletes by $1.45 \pm 0.91$ inches. What is the $95\%$ confidence interval margin of error for the average jump increase?

STEP 1

1. The sample of 32 male athletes is a simple random sample from the population of interest.
2. The distribution of jump increases is approximately normal or the sample size is large enough to apply the Central Limit Theorem.
3. The standard deviation of 0.91 inches is the sample standard deviation, not the population standard deviation.
4. The margin of error for the $95\%$ confidence interval can be calculated using the formula $ME = z \cdot \frac{\sigma}{\sqrt{n}}$ , where $z$ is the z-score corresponding to the desired confidence level, $\sigma$ is the sample standard deviation, and $n$ is the sample size.
5. Since the population standard deviation is unknown and the sample size is less than 30, we should use the t-distribution instead of the z-distribution. However, since the sample size is 32, which is close to 30, we can approximate using the z-distribution for simplicity and because the t-distribution approaches the z-distribution as the sample size increases.

STEP 2

1. Determine the appropriate z-score for a $95\%$ confidence interval.
2. Calculate the margin of error using the z-score and the given standard deviation and sample size.

STEP 3

Find the z-score corresponding to a $95\%$ confidence interval. This is the value of $z$ for which the area under the standard normal distribution to the left of $z$ is $0.975$ (since $0.95/2 = 0.025$ and $1 - 0.025 = 0.975$ ).
$z_{0.975}$

STEP 4

Use a standard normal distribution table or a calculator to find the z-score that corresponds to an area of $0.975$ .
$z_{0.975} \approx 1.96$

STEP 5

Calculate the margin of error (ME) using the formula:
$ME = z \cdot \frac{\sigma}{\sqrt{n}}$
where $z$ is the z-score from Step 2, $\sigma$ is the sample standard deviation, and $n$ is the sample size.

STEP 6

Substitute the known values into the margin of error formula:
$ME = 1.96 \cdot \frac{0.91}{\sqrt{32}}$

STEP 7

Calculate the denominator $\sqrt{32}$ .
$\sqrt{32} \approx 5.6569$

STEP 8

Divide the sample standard deviation by the square root of the sample size.
$\frac{0.91}{5.6569} \approx 0.1609$

STEP 9

Multiply the result from Step 6 by the z-score to find the margin of error.
$ME = 1.96 \cdot 0.1609 \approx 0.3154$

STEP 10

Round the margin of error to four decimal places as requested.
$ME \approx 0.3154$
The margin of error of the $95\%$ confidence interval for the average jump increase is approximately $0.3154$ inches when rounded to four decimal places.

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A study found that a new sneaker increased the average jump height of 32 male athletes by $1.45 \pm 0.91$ inches. What is the $95\%$ confidence interval margin of error for the average jump increase?

STEP 1

STEP 2

1. Determine the appropriate z-score for a $95\%$ confidence interval.
2. Calculate the margin of error using the z-score and the given standard deviation and sample size.

STEP 3

Find the z-score corresponding to a $95\%$ confidence interval. This is the value of $z$ for which the area under the standard normal distribution to the left of $z$ is $0.975$ (since $0.95/2 = 0.025$ and $1 - 0.025 = 0.975$ ).
$z_{0.975}$

STEP 4

Use a standard normal distribution table or a calculator to find the z-score that corresponds to an area of $0.975$ .
$z_{0.975} \approx 1.96$

STEP 5

Calculate the margin of error (ME) using the formula:
$ME = z \cdot \frac{\sigma}{\sqrt{n}}$
where $z$ is the z-score from Step 2, $\sigma$ is the sample standard deviation, and $n$ is the sample size.

STEP 6

Substitute the known values into the margin of error formula:
$ME = 1.96 \cdot \frac{0.91}{\sqrt{32}}$

STEP 7

Calculate the denominator $\sqrt{32}$ .
$\sqrt{32} \approx 5.6569$

STEP 8

Divide the sample standard deviation by the square root of the sample size.
$\frac{0.91}{5.6569} \approx 0.1609$

STEP 9

Multiply the result from Step 6 by the z-score to find the margin of error.
$ME = 1.96 \cdot 0.1609 \approx 0.3154$

STEP 10

Round the margin of error to four decimal places as requested.
$ME \approx 0.3154$
The margin of error of the $95\%$ confidence interval for the average jump increase is approximately $0.3154$ inches when rounded to four decimal places.

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A study found that a new sneaker increased the average jump height of 32 male athletes by 1.45±0.911.45 \pm 0.911.45±0.91 inches. What is the 95%95\%95% confidence interval margin of error for the average jump increase?

STEP 1

STEP 2

1. Determine the appropriate z-score for a 95%95\%95% confidence interval.2. Calculate the margin of error using the z-score and the given standard deviation and sample size.

STEP 3

STEP 4

Use a standard normal distribution table or a calculator to find the z-score that corresponds to an area of 0.9750.9750.975.z0.975≈1.96 z_{0.975} \approx 1.96 z0.975​≈1.96

STEP 5

Calculate the margin of error (ME) using the formula:ME=z⋅σn ME = z \cdot \frac{\sigma}{\sqrt{n}} ME=z⋅n​σ​where zzz is the z-score from Step 2, σ\sigmaσ is the sample standard deviation, and nnn is the sample size.

STEP 6

Substitute the known values into the margin of error formula:ME=1.96⋅0.9132 ME = 1.96 \cdot \frac{0.91}{\sqrt{32}} ME=1.96⋅32​0.91​

STEP 7

Calculate the denominator 32\sqrt{32}32​.32≈5.6569 \sqrt{32} \approx 5.6569 32​≈5.6569

STEP 8

Divide the sample standard deviation by the square root of the sample size.0.915.6569≈0.1609 \frac{0.91}{5.6569} \approx 0.1609 5.65690.91​≈0.1609

STEP 9

Multiply the result from Step 6 by the z-score to find the margin of error.ME=1.96⋅0.1609≈0.3154 ME = 1.96 \cdot 0.1609 \approx 0.3154 ME=1.96⋅0.1609≈0.3154

STEP 10

Round the margin of error to four decimal places as requested.ME≈0.3154 ME \approx 0.3154 ME≈0.3154 The margin of error of the 95%95\%95% confidence interval for the average jump increase is approximately 0.31540.31540.3154 inches when rounded to four decimal places.

Start learning now

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

A study found that a new sneaker increased the average jump height of 32 male athletes by $1.45 \pm 0.91$ inches. What is the $95\%$ confidence interval margin of error for the average jump increase?

1. Determine the appropriate z-score for a $95\%$ confidence interval.
2. Calculate the margin of error using the z-score and the given standard deviation and sample size.

Use a standard normal distribution table or a calculator to find the z-score that corresponds to an area of $0.975$ .
$z_{0.975} \approx 1.96$

Calculate the margin of error (ME) using the formula:
$ME = z \cdot \frac{\sigma}{\sqrt{n}}$
where $z$ is the z-score from Step 2, $\sigma$ is the sample standard deviation, and $n$ is the sample size.

Substitute the known values into the margin of error formula:
$ME = 1.96 \cdot \frac{0.91}{\sqrt{32}}$

Calculate the denominator $\sqrt{32}$ .
$\sqrt{32} \approx 5.6569$

Divide the sample standard deviation by the square root of the sample size.
$\frac{0.91}{5.6569} \approx 0.1609$

Multiply the result from Step 6 by the z-score to find the margin of error.
$ME = 1.96 \cdot 0.1609 \approx 0.3154$

Round the margin of error to four decimal places as requested.
$ME \approx 0.3154$
The margin of error of the $95\%$ confidence interval for the average jump increase is approximately $0.3154$ inches when rounded to four decimal places.