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 lower boundary
To find the 95% 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 / √sample size)
First, let's find the critical value for a 95% confidence level. Since we assume a normal distribution, we can use the Z-table. The Z-value for a 95% confidence level is approximately 1.96.
Using the given information, we have:
Sample mean (x̄) = 109 kg
Sample standard deviation (σ) = 11 kg
Sample size (n) = 10
Plugging these values into the formula, we get:
Confidence Interval = 109 ± (1.96) * (11 / √10)
Calculating the standard error:
Standard Error = (sample standard deviation / √sample size)
Standard Error = 11 / √10 ≈ 3.484
Confidence Interval = 109 ± (1.96) * 3.484
Lower Boundary = 109 - (1.96) * 3.484
Lower Boundary ≈ 109 - 6.833
Lower Boundary ≈ 102.167
Therefore, the lower boundary of the 95% confidence interval for the true mean weight of all foot surgery patients is approximately 102.167 kg.
To find the confidence interval for the true mean weight of all foot surgery patients, we can use the following formula:
Confidence interval = sample mean ± (critical value * standard deviation / square root of sample size)
Since the sample size is 10 and we are looking for a 95% confidence interval, the critical value can be found using a t-distribution table or calculator. For a 95% confidence level with 9 degrees of freedom (n-1), the two-tailed critical value is approximately 2.262.
Plugging in the values into the formula:
Confidence interval = 109 ± (2.262 * 11 / √10)
Calculating the standard error (standard deviation divided by the square root of the sample size):
Standard error = 11 / √10 ≈ 3.480
Confidence interval = 109 ± (2.262 * 3.480)
Calculating the range:
2.262 * 3.480 ≈ 7.876
Lower boundary = 109 - 7.876 ≈ 101.124
Therefore, the lower boundary of the 95% confidence interval for the true mean weight of all foot surgery patients is approximately 101.124 kg.
To calculate the confidence interval for the true mean weight of all foot surgery patients, we can use the formula:
Confidence Interval = mean ± (critical value * standard deviation) / √(sample size)
Given that you are looking for a 95% confidence interval and assuming a normal distribution, we need to find the critical value associated with that confidence level. For a 95% confidence level, the critical value is commonly denoted as 1.96.
Let's plug in the values into the formula:
Confidence Interval = 109 ± (1.96 * 11) / √(10)
Calculating the values inside the square root:
√(10) = 3.162
Plugging the values into the formula:
Confidence Interval = 109 ± (1.96 * 11) / 3.162
Calculating the value inside the parentheses:
(1.96 * 11) = 21.56
Dividing by 3.162:
21.56 / 3.162 = 6.832
Now, let's calculate the confidence interval:
Confidence Interval = 109 ± 6.832
To find the lower boundary of the confidence interval, subtract the calculated value from the mean:
Lower Boundary = 109 - 6.832
Calculating the value:
Lower Boundary = 102.168
Therefore, the lower boundary of the 95% confidence interval for the true mean weight of all foot surgery patients is approximately 102.168 kg.