Question

The following scenario applies to questions 3-5: The weights of all of the Utah County Fair pigs have an unknown mean and known standard deviation of g = 18. A simple random sample of 100 pigs found to have a sample mean weight of x = 195. Question 3 3. Calculate a 95% confidence interval for the mean weight of all Utah County Fair pigs. (195, 200) (193, 204) (191, 199) (177, 213) Question 4 4. Suppose a sample of 200 was taken instead of 100. How will the margin of error change? the margin of error will increase in size the margin of error will decrease in size the margin of error will not change in size Question 5 5. If the researcher wanted to have 90% confidence in the results with a margin of error of 6.8, how many pigs must be sampled? 38 19 10 5

Answer 1

5

Explanation:

To calculate a 95% confidence interval for the mean weight of all Utah County Fair pigs, we use the formula:

Confidence Interval = Sample Mean ± Margin of Error

Given:

Sample Mean (x) = 195

Standard Deviation (σ) = 18

Sample Size (n) = 100

The margin of error can be calculated using the formula:

Margin of Error = (Z * σ) / √n

For a 95% confidence level, the Z-value for a two-tailed test is approximately 1.96.

Margin of Error = (1.96 * 18) / √100

= 3.528

Therefore, the confidence interval is:

(195 - 3.528, 195 + 3.528)

(191.472, 198.528)

The correct answer is (191, 199).

Question 4: If the sample size is increased from 100 to 200, the margin of error will decrease in size. The margin of error is inversely proportional to the square root of the sample size. As the sample size increases, the margin of error becomes smaller, resulting in a more precise estimate.

Question 5: To find out how many pigs must be sampled to have 90% confidence in the results with a margin of error of 6.8, we can use the formula:

Sample Size (n) = (Z^2 * σ^2) / E^2

Given:

Confidence Level (1 - α) = 90% (or 0.9)

Margin of Error (E) = 6.8

Standard Deviation (σ) = 18

For a 90% confidence level, the Z-value for a two-tailed test is approximately 1.645.

Sample Size (n) = (1.645^2 * 18^2) / 6.8^2

= 3.379

Therefore, the minimum number of pigs that must be sampled is approximately 4 (rounded up to the nearest whole number).

The correct answer is 5.