A sample of 134 golfers showed that their average score on a particular golf course was 87.43 with a standard deviation of 4.53.
Answer each of the following (show all work
and state the final answer to at least two decimal places.):
(A) Find the 99% confidence interval of the mean score for all 134 golfers.
(B) Find the 99% confidence interval of the mean score for all golfers if this is a sample of 105 golfers instead of a sample of 134.
(C) Which confidence interval is smaller and why?
To find the confidence intervals, we will use the formula:
Confidence Interval = Sample Mean ± (Critical Value * Standard Error)
where the critical value is determined by the desired confidence level and the standard error is the standard deviation divided by the square root of the sample size.
(A) Find the 99% confidence interval of the mean score for all 134 golfers:
Sample size (n) = 134
Sample mean (x̄) = 87.43
Standard deviation (σ) = 4.53
The critical value for a 99% confidence level can be found using the Z-table or standard normal distribution table. For a 99% confidence level, the critical value is approximately 2.576.
Standard error (SE) = σ / √n = 4.53 / √134 ≈ 0.391
Confidence Interval = 87.43 ± (2.576 * 0.391)
Confidence Interval ≈ 87.43 ± 1.006
Confidence Interval ≈ (86.424, 88.436)
The 99% confidence interval for the mean score of all 134 golfers is approximately (86.424, 88.436).
(B) Find the 99% confidence interval of the mean score for all golfers if this is a sample of 105 golfers instead of a sample of 134:
Sample size (n) = 105
Sample mean (x̄) is not given, so we will use the previously calculated x̄ = 87.43
Standard deviation (σ) = 4.53
Standard error (SE) = σ / √n = 4.53 / √105 ≈ 0.442
Confidence Interval = 87.43 ± (2.576 * 0.442)
Confidence Interval ≈ 87.43 ± 1.139
Confidence Interval ≈ (86.291, 88.569)
The 99% confidence interval for the mean score of all 105 golfers is approximately (86.291, 88.569).
(C) To determine which confidence interval is smaller, we compare the standard errors. In this case, the confidence interval for the smaller sample size of 105 golfers has a larger standard error (0.442) compared to the confidence interval for the larger sample size of 134 golfers (0.391).
A smaller confidence interval indicates greater precision in estimating the population mean, and this is achieved with a smaller standard error. Therefore, the confidence interval for the sample of 134 golfers is smaller because it has a smaller standard error, indicating a more precise estimate of the population mean.
To answer the questions, we can use the formula for calculating confidence intervals for a population mean:
Confidence interval = sample mean ± (critical value * standard deviation / √sample size)
Now let's solve each question step by step:
(A) Find the 99% confidence interval of the mean score for all 134 golfers.
1. Calculate the critical value using the z-table for a 99% confidence level. Since this is a two-tailed test, we divide the significance level by 2, resulting in an alpha value of 0.01/2 = 0.005. Looking up this value in the z-table, we find that the critical value corresponding to a 99% confidence level is approximately 2.58.
2. Plug the values into the formula:
Confidence interval = 87.43 ± (2.58 * 4.53 / √134)
3. Calculate the confidence interval:
Confidence interval = 87.43 ± (2.58 * 4.53 / √134)
Confidence interval ≈ 87.43 ± (2.58 * 4.53 / 11.56)
Confidence interval ≈ 87.43 ± (11.7026 / 11.56)
Confidence interval ≈ 87.43 ± 1.01
Therefore, the 99% confidence interval for the mean score for all 134 golfers is approximately 86.42 to 88.44.
(B) Find the 99% confidence interval of the mean score for all golfers if this is a sample of 105 instead of 134.
We can use the same formula as in question A, but with a different sample size of 105.
Confidence interval = sample mean ± (critical value * standard deviation / √sample size)
1. Calculate the critical value using the z-table for a 99% confidence level. We already know the critical value is 2.58.
2. Plug the values into the formula:
Confidence interval = 87.43 ± (2.58 * 4.53 / √105)
3. Calculate the confidence interval:
Confidence interval = 87.43 ± (2.58 * 4.53 / √105)
Confidence interval ≈ 87.43 ± (2.58 * 4.53 / 10.247)
Confidence interval ≈ 87.43 ± (11.6834 / 10.247)
Confidence interval ≈ 87.43 ± 1.14
Therefore, the 99% confidence interval for the mean score for all 105 golfers is approximately 86.29 to 88.57.
(C) Which confidence interval is smaller and why?
The confidence interval from question B (sample size of 105) is smaller than the confidence interval from question A (sample size of 134). This difference is because the sample size directly affects the width of the confidence interval: the larger the sample size, the narrower the interval. With a smaller sample size, there is more uncertainty, resulting in a wider confidence interval.