Question

Confidence intervals: A random sample of 6 bowlers at a state bowling tournament produced the following scores:

Steve: 185,

Hans: 165,

Brianna: 85,

Samantha: 135,

Paul: 80,

Jennifer: 225.

Assume these 6 bowlers are a representative sample of all bowlers in the tournament.

To answer this question, you will need to calculate:

a. The point estimate for the average score of all bowlers in the tournament, and

b. The point estimate for the standard deviation of scores for all bowlers in the tournament.

(Write down those numbers because you will be asked to data-enter them in the question that follows this question.)

Coach Swanson believes that the scores of all bowlers at the tournament are normally distributed.

What is the 90% confidence interval for the average score of a bowler at this tournament. Use Table 2.

(Round intermediate calculations to 4 decimal places, "sample mean" and "sample standard deviation" to 2 decimal places. Round final answers to 2 decimal places. Provide the low number first, to the high number second.)

  

Confidence interval

to

In the previous question, you calculated:

a. The point estimate for the average score of all of the bowlers in the tournament, and

b. The point estimate for the standard deviation of all scores for bowlers in the tournament.

What did you calculate for the (a) average score, and for the (b) standard deviation? (Provide those statistics now.)

Answer 1

(1)

(a)

From the given data, the following statistics are calculated:

n = Sample Size = 6

= Sample Mean =875/6 = 145.8333

s = Sample SD = 57.1329

The point estimate of the average score = 145.8333

(b)

The point estimate of the standard deviation score = 57.1329

(c)

SE = s/

= 57.1329/ = 23.3244

= 0.10

ndf = n - 1 = 6 - 1 =5

From Table, critical values of t = 2.0150

Confidence Interval:

145.8333 (2.0150 X 23.3244)

= 145.8333 46.9987

= (98.8346 , 192.8320)

So,

Answer is:

Confidence Interval

98.8346 to 192.8320

(a)

The point estimate of the average score = 145.8333

(b)

The point estimate of the standard deviation score = 57.1329