Suppose that a committee is studying whether or not there is waste of time in our judicial system. It is interested in the mean amount of time individuals waste at the courthouse waiting to be called for jury duty. The committee randomly surveyed 81 people who recently served as jurors. The sample mean wait time was 7 hours with a sample standard deviation of 3.3 hours.

a. Which distribution should you use for this problem? Explain your choice.

1. The Student's t-distribution for 80 degrees of freedom should be used because we do not know the population standard deviation.
2. The Student's t-distribution for 81 degrees of freedom should be used because the sample standard deviation is known.
3. The standard normal distribution should be used because the sample standard deviation is known.
4. The standard normal distribution should be used because the population standard deviation is known.

b. Construct a 95% confidence interval for the population mean time wasted.
c. Explain in a complete sentence what the confidence interval means.

1. We are 95% confident that a wait time at the courthouse lies within this interval.
2. There is a 95% chance that a wait time at the courthouse lies within this interval.
3. We are 95% confident that the mean wait time at the courthouse of the sample of 81 individuals waiting at the courthouse lies within this interval.
4. We are 95% confident that the true population mean wait time at the courthouse lies within this interval.

a) Option 1 is correct.

The Student's t-distribution for 80 degrees of freedom should be used because we do not know the population standard deviation.

b) Confidence interval for the population is given as (6.27, 7.73)

c) Option 4 is correct.

We are 95% confident that the true population mean wait time at the courthouse lies within this interval.

Step-by-step explanation:

Sample Mean = 7 hours

Sample standard deviation = 3.3 hours.

Sample size = 81

To construct the confidence interval for the population mean time wasted,

Since there's no information on the population standard deviation, the t-distribution will be used.

The t-distribution is used when there is no information on the population mean or standard deviation. We then assume that the sample standard deviation is a good estimate of the population standard deviation. And that the sample mean too, is a good estimate of the population mean.

The degree of freedom for t-distribution is given as (n-1).

df = 81 - 1 = 80

Hence, the correct answer for (a) is Option 1.

The Student's t-distribution for 80 degrees of freedom should be used because we do not know the population standard deviation.

b) To construct the 95% confidence interval for the population mean time wasted

Confidence interval = (Mean) ± (Margin of error)

Mean = 7 hours

Margin of Error = (Critical value) × (Standard error)

Critical value = t-score for a significance level of (0.05/2) and a degree of freedom of 80.

Note that the significance level.is obtained by subtracting the confidence interval from 100% and dividing it by 2 to accommodate significance at the start and end of the distribution.

So, t (0.025, 80) = 1.990 (obtained from literature)

Standard error = (σ/√n) = (3.3/√81) = 0.3667

Margin of Error = (Critical value) × (Standard error)

= 1.990 × 0.3667 = 0.7297

Confidence interval = (Mean) ± (Margin of error)

= 7 ± 0.7297 = (6.2703, 7.7297) ≈ (6.27, 7.73)

c) The interpretation for confidence interval is that the range given by the confidence interval is understood to contain the true mean required (in this case, the population mean), with a certain confidence level.

So, it is clear that the right answer is "we are 95% confident that the true population mean wait time at the courthouse lies within this interval".

Hope this Helps

coin flip

non-answer:

choosing

sorry if this isnt what you're looking for i dont really understand the question

it will cost dollar $540

the living room is 3 which means if we divide 325/3 then we get $108

therefore if 1part of the living room cost $108 and 3 costed $325

then total 5part of the room will cost 5*108 which gives $540