Question

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 8 hours with a sample standard deviation of 4 hours.a.
i. x¯= 8
ii. sx = 4
iii. n = 81
iv. n – 1 = 80

b. Define the random variables X and X¯in words. X is amount of time individuals waste at the courthouse waiting to be called for jury duty. X- is mean amount of time that sample of 81 people who recently served at jurors.

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

d. Construct a 95% confidence interval for the population mean time wasted.
i. State the confidence interval.
ii. Sketch the graph.
iii. Calculate the error bound.

e. Explain in a complete sentence what the confidence interval means.

Answer 1

NOTE:

As Part (a) and Part (b) are already correctly solved in the question, I will be solving Part (c), Part (d) and Part (e).

Answer:

Part (c):

As we are not given the standard deviation of the population, we will use

t-distribution for this scenario.

The degrees of freedom will be n-1 = 80, hence our distribution parameter will be t$_{_{80}}$.

Part (d):

The critical value (t) for 95% confidence level is at 80 degree of freedom is obtained from the attached table.

t = 1.99

Now, let us first calculate the error bound

E = t * (sx)/(√n)

E = 1.99 * (4)/(√81)

E = 0.884                 ===========> Error Bound

Confidence interval is obtained as follows:

(X¯ - E), (X¯ + E)

(8 - 0.884), (8 + 0.884)

(7.116, 8.884)           ===========> Confidence Interval

Graph is attached as an image.

Part (e):

The confidence interval indicates that we can predict with 95% confidence that the population mean time wasted at the court, waiting to be called for jury duty, will lie between 7.1155 and 8.8845 hours.

Answer 2

The sample mean wait time was, which is 8 hours.

Standard deviation of the sample, which is of 4 hours. This is the sample mean, which is 8 hours.

The degrees of freedom will be n - 1 = 80, hence our distribution parameter will be t.

The value of error bound is E = 0.884 and the value on confidence interval is (7.116, 8.884)  .

The confidence interval indicates that we can predict with 95% confidence that the population mean time wasted at the court, waiting to be called for jury duty, will lie between 7.1155 and 8.8845 hours.

Given that,

The committee randomly surveyed 81 people who recently served as jurors.

The sample mean wait time was 8 hours with a sample standard deviation of 4 hours.

• The sample mean wait time was, which is 8 hours.

X is amount of time individuals waste at the courthouse waiting to be called for jury duty.

. X- is mean amount of time that sample of 81 people who recently served at jurors.

Standard deviation of the sample, which is of 4 hours. This is the sample mean, which is 8 hours.

• As we are not given the standard deviation of the population, we will use.

t-distribution for this scenario.

The degrees of freedom will be n - 1 = 80, hence our distribution parameter will be t.

• The critical value (t) for 95% confidence level is at 80 degree of freedom is obtained from the attached table.  t = 1.99

Now, let us first calculate the error bound,

$E = \frac{T.(sx)}{\sqrt{N} } \\\\E = \frac{1.99. (4)}{\sqrt{81} }$

E = 0.884

The value of error bound is E = 0.884

Confidence interval is obtained as follows:

(X¯ - E), (X¯ + E)

(8 - 0.884), (8 + 0.884)

(7.116, 8.884)

The value on confidence interval is (7.116, 8.884)  

The value of error bound is E = 0.884 and the value on confidence interval is (7.116, 8.884)  

• The confidence interval indicates that we can predict with 95% confidence that the population mean time wasted at the court, waiting to be called for jury duty, will lie between 7.1155 and 8.8845 hours.

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