Math / Data & Statistics

QuestionModule Knowlsdge Chsck Question 9 Sophi
Pilots who cannot maintain regular sleep hours due to their work schedule often suffer from insominia. A random sample of 26 commercial airline pilots was taken, and the pilots in the sample reported the time at which they went to sleep on their most recent working day. These times measured in hours after midnight. (Thus, if the pilot reported going to sleep at 11 p.m., the measurement was -1 .) The sample mean was 0.5 hours, and the standard deviation was 19 hours. Assume that the sample is drawn from a normally distributed population. Find a $90 \%$ confidence interval for the population standard deviation, that is, the standard deviation of the time (hours after midnight) at which pilots go to sleep on their work days. Then give its lower limit and upper limit. Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places. (If necessary, consult a list of formulas.) $\square$

Studdy Solution

STEP 1

What is this asking? We need to find the 90% confidence interval for the standard deviation of pilot sleep times, given a sample of 26 pilots. Watch out! We're dealing with the standard deviation, not the mean!
Also, make sure to use the right chi-squared distribution values.

STEP 2

1. Calculate Degrees of Freedom
2. Find Chi-Squared Values
3. Calculate the Confidence Interval
4. Determine Lower and Upper Limits

STEP 3

Alright, first things first!
We need our degrees of freedom, which is just our sample size minus 1.
We have a sample of 26 pilots, so our degrees of freedom is $26 - 1 = 25$ .
Easy peasy!

STEP 4

Now, for the fun part!
We're looking for a 90% confidence interval, which means we'll have 5% in each tail of our chi-squared distribution.
So, we need to find the chi-squared values for 0.05 and 0.95 with 25 degrees of freedom.

STEP 5

Looking up these values in a chi-squared table (or using a calculator), we find: * $\chi^2_{0.05, 25} = 37.652$ * $\chi^2_{0.95, 25} = 14.611$

STEP 6

The formula for the confidence interval of the standard deviation is: $\sqrt{\frac{(n-1) \cdot s^2}{\chi^2_{\alpha/2, n-1}}} < \sigma < \sqrt{\frac{(n-1) \cdot s^2}{\chi^2_{1-\alpha/2, n-1}}}$ Where: * $n$ is the sample size ( $26$ ) * $s$ is the sample standard deviation ( $1.9$ ) * $\chi^2_{\alpha/2, n-1}$ and $\chi^2_{1-\alpha/2, n-1}$ are our chi-squared values

STEP 7

Let's plug in our values: $\sqrt{\frac{(26-1) \cdot (1.9)^2}{37.652}} < \sigma < \sqrt{\frac{(26-1) \cdot (1.9)^2}{14.611}}$

STEP 8

Simplifying: $\sqrt{\frac{25 \cdot 3.61}{37.652}} < \sigma < \sqrt{\frac{25 \cdot 3.61}{14.611}}$ $\sqrt{\frac{90.25}{37.652}} < \sigma < \sqrt{\frac{90.25}{14.611}}$ $\sqrt{2.400} < \sigma < \sqrt{6.177}$

STEP 9

Taking the square roots, we get our confidence interval: $1.549 < \sigma < 2.485$

STEP 10

So, our lower limit is 1.55 and our upper limit is 2.49, rounded to two decimal places.

STEP 11

The 90% confidence interval for the population standard deviation is (1.55, 2.49).

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Studdy Solution

STEP 1

What is this asking? We need to find the 90% confidence interval for the standard deviation of pilot sleep times, given a sample of 26 pilots. Watch out! We're dealing with the standard deviation, not the mean!
Also, make sure to use the right chi-squared distribution values.

STEP 2

1. Calculate Degrees of Freedom
2. Find Chi-Squared Values
3. Calculate the Confidence Interval
4. Determine Lower and Upper Limits

STEP 3

Alright, first things first!
We need our degrees of freedom, which is just our sample size minus 1.
We have a sample of 26 pilots, so our degrees of freedom is $26 - 1 = 25$ .
Easy peasy!

STEP 4

Now, for the fun part!
We're looking for a 90% confidence interval, which means we'll have 5% in each tail of our chi-squared distribution.
So, we need to find the chi-squared values for 0.05 and 0.95 with 25 degrees of freedom.

STEP 5

Looking up these values in a chi-squared table (or using a calculator), we find: * $\chi^2_{0.05, 25} = 37.652$ * $\chi^2_{0.95, 25} = 14.611$

STEP 6

STEP 7

Let's plug in our values: $\sqrt{\frac{(26-1) \cdot (1.9)^2}{37.652}} < \sigma < \sqrt{\frac{(26-1) \cdot (1.9)^2}{14.611}}$

STEP 8

Simplifying: $\sqrt{\frac{25 \cdot 3.61}{37.652}} < \sigma < \sqrt{\frac{25 \cdot 3.61}{14.611}}$ $\sqrt{\frac{90.25}{37.652}} < \sigma < \sqrt{\frac{90.25}{14.611}}$ $\sqrt{2.400} < \sigma < \sqrt{6.177}$

STEP 9

Taking the square roots, we get our confidence interval: $1.549 < \sigma < 2.485$

STEP 10

So, our lower limit is 1.55 and our upper limit is 2.49, rounded to two decimal places.

STEP 11

The 90% confidence interval for the population standard deviation is (1.55, 2.49).

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Studdy Solution

STEP 1

What is this asking? We need to find the 90% confidence interval for the standard deviation of pilot sleep times, given a sample of 26 pilots. Watch out! We're dealing with the standard deviation, *not* the mean!Also, make sure to use the right chi-squared distribution values.

STEP 2

1. Calculate Degrees of Freedom2. Find Chi-Squared Values3. Calculate the Confidence Interval4. Determine Lower and Upper Limits

STEP 3

Alright, first things first!We need our **degrees of freedom**, which is just our **sample size** minus 1.We have a sample of **26** pilots, so our degrees of freedom is 26−1=2526 - 1 = 2526−1=25.Easy peasy!

STEP 4

Now, for the fun part!We're looking for a **90% confidence interval**, which means we'll have **5%** in each tail of our chi-squared distribution.So, we need to find the chi-squared values for **0.05** and **0.95** with **25** degrees of freedom.

STEP 5

Looking up these values in a chi-squared table (or using a calculator), we find: * χ0.05,252=37.652\chi^2_{0.05, 25} = 37.652χ0.05,252​=37.652 * χ0.95,252=14.611\chi^2_{0.95, 25} = 14.611χ0.95,252​=14.611

STEP 6

STEP 7

Let's plug in our values: (26−1)⋅(1.9)237.652<σ<(26−1)⋅(1.9)214.611 \sqrt{\frac{(26-1) \cdot (1.9)^2}{37.652}} < \sigma < \sqrt{\frac{(26-1) \cdot (1.9)^2}{14.611}} 37.652(26−1)⋅(1.9)2​​<σ<14.611(26−1)⋅(1.9)2​​

STEP 8

STEP 9

Taking the square roots, we get our **confidence interval**: 1.549<σ<2.485 1.549 < \sigma < 2.485 1.549<σ<2.485

STEP 10

So, our **lower limit** is **1.55** and our **upper limit** is **2.49**, rounded to two decimal places.

STEP 11

The 90% confidence interval for the population standard deviation is (1.55, 2.49).

Get Studdy

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.

What is this asking? We need to find the 90% confidence interval for the standard deviation of pilot sleep times, given a sample of 26 pilots. Watch out! We're dealing with the standard deviation, not the mean!
Also, make sure to use the right chi-squared distribution values.

1. Calculate Degrees of Freedom
2. Find Chi-Squared Values
3. Calculate the Confidence Interval
4. Determine Lower and Upper Limits

Alright, first things first!
We need our degrees of freedom, which is just our sample size minus 1.
We have a sample of 26 pilots, so our degrees of freedom is $26 - 1 = 25$ .
Easy peasy!

Now, for the fun part!
We're looking for a 90% confidence interval, which means we'll have 5% in each tail of our chi-squared distribution.
So, we need to find the chi-squared values for 0.05 and 0.95 with 25 degrees of freedom.

Looking up these values in a chi-squared table (or using a calculator), we find: * $\chi^2_{0.05, 25} = 37.652$ * $\chi^2_{0.95, 25} = 14.611$

Let's plug in our values: $\sqrt{\frac{(26-1) \cdot (1.9)^2}{37.652}} < \sigma < \sqrt{\frac{(26-1) \cdot (1.9)^2}{14.611}}$

Taking the square roots, we get our confidence interval: $1.549 < \sigma < 2.485$

So, our lower limit is 1.55 and our upper limit is 2.49, rounded to two decimal places.