Confidence intervals: A random sample of 6 bowlers at a state bowling tournament

Question

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

Steve: 175,

Hans: 125,

Brianna: 180,

Samantha: 220,

Paul: 240,

Jennifer: 245.

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 99% 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.)

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.)

Explanation / Answer

a)

Point estimate for the average score = 197.5

b)

point estimate of std. dev. = 46.1248

c)

CI for 99%

n = 6

mean = 197.5

t-value of 99% CI = 4.0321

std. dev. = 46.1248

SE = std.dev./sqrt(n)

= 46.1248/sqrt(6)

= 18.8304

ME = t*SE

= 4.0321*18.8304

= 75.9268

Lower Limit = Mean - ME = 197.5 - 75.9268 = 121.573

Upper Limit = Mean + ME = 197.5 + 75.9268 = 273.42680

99% CI (121.5732 , 273.4268 )

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