Question

The state test scores for 12randomly selected high school seniors are shown on the right.

Assume the population is normally distributed.

2.5              3.5                2.8

2.0              0.7                4.0

2.3              1.2                3.7

0.2              2.1                3.2

A-Find the sample mean

B-Find sample standard deviation

Round to one decimal place as​ needed.

C-To find a 99% confidence interval for the population​ mean, first determine whether the​ t-distribution or the normal distribution should be used in constructing the interval. Refer to the flowchart below.

​1) Determine the sample size. Recall that n equals 12. Continue down the left side of the flowchart.

2) The assumption that the population is normally distributed is given. Continue down the left side of the flowchart.

he population standard​ deviation, sigma , is not known. Continue down the left side of the flowchart.

​Therefore, the​ t-distribution should be used.

Construct a 99% confidence interval for the population mean

In order to determine the critical value t c, we need to know the degrees of freedom d.f. and the confidence level c. The degrees of freedom are given by the following formula.

Determine the degrees of freedom for this problem.

Now determine the confidence level c. Recall that we are constructing a 99% confidence interval.

Use technology or a​ t-distribution table to determine the critical value with 11degrees of freedom and a confidence level of 0.99. If using a​ t-distribution table look in the row associated with 11

degrees of freedom and the column labeled 0.99 for the confidence level.

Substitute the known values into the margin of error formula to determine the margin of error E.

​Finally, construct a 99% confidence interval with

x overbar equals 2.3 and E equals1.07

.

Answer 1

(A)

Sample Mean () is found as follows:

(B)
Sample Standard Deviation (s) is found as follows:

(C)

(1) Sample Size = 12

(2) Degrees of Freedom = d.f. = n - 1 = 12 - 1 = 11

= 0.01

Using Technology, critical values of t = 3.1058

SE = s/

= 1.1935/

= 0.3445

Margin of Error = 3.1058 X 0.3445 = 1.07

Confidence Interval:

2.35 1.07

= ( 1.28 ,3.42)

Confidence Interval:

1.28 < < 3.42