A group of 10 foot surgery patients had a mean weight of 109 kg. The sample standard deviation was 11 kg. Find a confidence interval for a sample for the true mean weight of all foot surgery patients. Find a 95% Confidence Interval and assume normal distribution. Give the upper boundary.
To find the 95% confidence interval for the true mean weight of all foot surgery patients, we can use the formula:
Confidence Interval = mean ± (critical value) * (standard deviation / √n)
Given:
Mean weight of the sample (x̄) = 109 kg
Sample standard deviation (s) = 11 kg
Sample size (n) = 10
The critical value for a 95% confidence interval with a normal distribution can be found using a t-distribution table or a calculator. For a sample size of 10 and a 95% confidence level, the critical value is approximately 2.262.
Confidence Interval = 109 ± (2.262) * (11 / √10)
Confidence Interval = 109 ± 8.66
Therefore, the 95% confidence interval for the true mean weight of all foot surgery patients is (100.34, 117.66).
The upper boundary of this confidence interval is 117.66 kg.
To find the confidence interval for the true mean weight of all foot surgery patients, we can use the formula:
Confidence Interval = sample mean ± (critical value) * (sample standard deviation / √n),
where
- sample mean: 109 kg (given)
- sample standard deviation: 11 kg (given)
- n: number of patients in the sample = 10 (given)
- critical value: depends on the desired confidence level (95% in this case)
First, let's find the critical value using a standard normal distribution table. Since we want a 95% confidence level, the confidence level will be split equally into two tails, meaning each tail will have 2.5% probability. Looking at the table, the critical value is approximately 1.96.
Next, we can plug the values into the formula to find the confidence interval:
Confidence Interval = 109 ± 1.96 * (11 / √10)
Now, let's do the calculations:
Confidence Interval = 109 ± 1.96 * (11 / √10)
= 109 ± 1.96 * 3.479
Multiplying 1.96 and 3.479 gives us approximately 6.8142.
Therefore, the confidence interval for the true mean weight of all foot surgery patients is approximately:
95% Confidence Interval: (109 - 6.8142, 109 + 6.8142)
= (102.19, 115.81)
So, the upper boundary of the confidence interval is 115.81 kg.
To find a confidence interval for the true mean weight of all foot surgery patients, we can use the formula:
Confidence Interval = mean ± (critical value) * (standard deviation / sqrt(sample size))
In this case, we are looking for a 95% confidence interval, which means we want to find the critical value associated with a 95% confidence level. Since we are assuming a normal distribution, we can find this critical value using a z-table or a statistical software.
For a 95% confidence level, the critical value is approximately 1.96.
Given:
Mean weight (x̄) = 109 kg
Sample standard deviation (s) = 11 kg
Sample size (n) = 10
Now we can calculate the confidence interval upper boundary using the formula:
Confidence Interval = mean ± (critical value) * (standard deviation / sqrt(sample size))
Confidence Interval = 109 ± (1.96) * (11 / sqrt(10))
To find the upper boundary, we add the result of the calculation to the mean weight:
Upper Boundary = 109 + (1.96) * (11 / sqrt(10))
Now, we can calculate the upper boundary:
Upper Boundary ≈ 109 + (1.96) * (11 / sqrt(10)) ≈ 109 + (1.96) * (3.48) ≈ 109 + 6.81 ≈ 115.81 kg
Therefore, the upper boundary of the 95% confidence interval for the true mean weight of all foot surgery patients is approximately 115.81 kg.