Math / Data & Statistics

QuestionIn a class survey, students are asked how many hours they sleep per night. In the sample of 22 students, the mean was 5.77 hours with a standard deviation of 1.572 hours. a) Construct $90 \%, 95 \%$ , and $99 \%$ confidence interval for the mean number of hours slept per night in the population. b) Compare the results of all these confidence intervals, how does increasing the level of confidence affect the size of the margin of error?

Studdy Solution

STEP 1

What is this asking? We want to estimate how many hours students sleep, based on a small group of students, with different levels of certainty. Watch out! Don't mix up the standard deviation of the sample with the standard deviation of the population!
Also, remember that higher confidence means a wider interval, not a narrower one.

STEP 2

1. Calculate the margin of error for each confidence level.
2. Construct the confidence intervals.
3. Compare the results.

STEP 3

Alright, let's kick things off by finding our critical t-scores!
Since we don't know the population standard deviation, we're using the t-distribution.
Our sample size is $\text{n} = \textbf{22}$ , so the degrees of freedom are $\text{df} = \text{n} - 1 = \textbf{22} - 1 = \textbf{21}$ .
For a 90% confidence level, our alpha is $1 - 0.90 = \textbf{0.10}$ , so alpha/2 is $\textbf{0.05}$ .
Looking up the t-table for $\text{df} = \textbf{21}$ and $\alpha/2 = \textbf{0.05}$ , we find $t_{0.05, 21} = \textbf{1.721}$ .
Similarly, for 95% confidence ( $\alpha/2 = \textbf{0.025}$ ), $t_{0.025, 21} = \textbf{2.080}$ , and for 99% confidence ( $\alpha/2 = \textbf{0.005}$ ), $t_{0.005, 21} = \textbf{2.831}$ .
These t-scores tell us how many standard errors wide our confidence interval needs to be.

STEP 4

The standard error measures the variability of our sample mean.
It's calculated as the sample standard deviation divided by the square root of the sample size: $SE = \frac{s}{\sqrt{n}} = \frac{\textbf{1.572}}{\sqrt{\textbf{22}}} \approx \textbf{0.335}$

STEP 5

Now, let's multiply each t-score by the standard error to get the margin of error for each confidence level. For 90%: $ME_{90} = \textbf{1.721} \cdot \textbf{0.335} \approx \textbf{0.575}$ For 95%: $ME_{95} = \textbf{2.080} \cdot \textbf{0.335} \approx \textbf{0.697}$ For 99%: $ME_{99} = \textbf{2.831} \cdot \textbf{0.335} \approx \textbf{0.948}$

STEP 6

To construct the confidence intervals, we add and subtract the margin of error from the sample mean. 90% CI: $\textbf{5.77} \pm \textbf{0.575} = (\textbf{5.195}, \textbf{6.345})$ 95% CI: $\textbf{5.77} \pm \textbf{0.697} = (\textbf{5.073}, \textbf{6.467})$ 99% CI: $\textbf{5.77} \pm \textbf{0.948} = (\textbf{4.822}, \textbf{6.718})$

STEP 7

As the confidence level increases (from 90% to 99%), the margin of error also increases.
This makes sense, because a wider interval is more likely to capture the true population mean.
We're more confident, but our estimate is less precise!

STEP 8

a) The 90%, 95%, and 99% confidence intervals for the mean number of hours slept are (5.195, 6.345), (5.073, 6.467), and (4.822, 6.718) respectively. b) Increasing the confidence level increases the margin of error, resulting in a wider confidence interval.

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Studdy Solution

STEP 1

STEP 2

1. Calculate the margin of error for each confidence level.
2. Construct the confidence intervals.
3. Compare the results.

STEP 3

STEP 4

The standard error measures the variability of our sample mean.
It's calculated as the sample standard deviation divided by the square root of the sample size: $SE = \frac{s}{\sqrt{n}} = \frac{\textbf{1.572}}{\sqrt{\textbf{22}}} \approx \textbf{0.335}$

STEP 5

STEP 6

STEP 7

As the confidence level increases (from 90% to 99%), the margin of error also increases.
This makes sense, because a wider interval is more likely to capture the true population mean.
We're more confident, but our estimate is less precise!

STEP 8

a) The 90%, 95%, and 99% confidence intervals for the mean number of hours slept are (5.195, 6.345), (5.073, 6.467), and (4.822, 6.718) respectively. b) Increasing the confidence level increases the margin of error, resulting in a wider confidence interval.

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Studdy Solution

STEP 1

STEP 2

1. Calculate the margin of error for each confidence level.2. Construct the confidence intervals.3. Compare the results.

STEP 3

STEP 4

STEP 5

STEP 6

STEP 7

As the confidence level increases (from 90% to 99%), the margin of error also increases.This makes sense, because a wider interval is more likely to capture the true population mean.We're more confident, but our estimate is less precise!

STEP 8

a) The 90%, 95%, and 99% confidence intervals for the mean number of hours slept are (5.195, 6.345), (5.073, 6.467), and (4.822, 6.718) respectively. b) Increasing the confidence level increases the margin of error, resulting in a wider confidence interval.

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.

1. Calculate the margin of error for each confidence level.
2. Construct the confidence intervals.
3. Compare the results.

As the confidence level increases (from 90% to 99%), the margin of error also increases.
This makes sense, because a wider interval is more likely to capture the true population mean.
We're more confident, but our estimate is less precise!