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) * (standard deviation / sqrt(sample size))
The critical value for a 95% confidence interval with a normal distribution can be found using a z-table. The corresponding critical value for a 95% confidence interval is 1.96.
Sample mean = 109 kg
Standard deviation = 11 kg
Sample size = 10
Critical value = 1.96
Confidence Interval= 109 +/- 1.96 * (11 / sqrt(10))
Confidence Interval= 109 +/- 1.96 * (11 / 3.16)
Confidence Interval= 109 +/- 1.96 * 3.484
Confidence Interval= 109 +/- 6.824
The lower boundary of the confidence interval is the sample mean minus the confidence interval width:
Lower Boundary = 109 - 6.824
Lower Boundary ≈ 102.176
Therefore, the lower boundary for the 95% confidence interval is approximately 102.176 kg.
To find the confidence interval for the true mean weight of all foot surgery patients, we can use the formula:
Confidence Interval = Mean ± (Critical Value * Standard Error)
Given:
Mean weight (x̄) = 109 kg
Sample standard deviation (s) = 11 kg
Sample size (n) = 10
Confidence level = 95% (which means α = 0.05)
First, we need to calculate the critical value for a 95% confidence level. Since we assume a normal distribution, we can use the Z-table to find the critical value.
1. Look up the Z-score for a 95% confidence level (α = 0.05) in the Z-table.
The critical Z-score for a 95% confidence level in a two-tailed test is approximately 1.96.
Next, we need to calculate the standard error (SE) of the sample mean.
Standard Error = s / √n
2. Calculate the standard error:
SE = 11 kg / √10 ≈ 3.48 kg
Finally, we can calculate the confidence interval:
Confidence Interval = 109 kg ± (1.96 * 3.48 kg)
3. Calculate the confidence interval:
Confidence Interval = 109 kg ± (1.96 * 3.48 kg)
≈ 109 kg ± 6.82 kg
To find the lower boundary of the confidence interval, subtract the value obtained in step 3 from the mean weight:
Lower Boundary = 109 kg - 6.82 kg
4. Calculate the lower boundary:
Lower Boundary ≈ 102.18 kg
The lower boundary of the 95% confidence interval for the true mean weight of all foot surgery patients is approximately 102.18 kg.
To find the confidence interval for the true mean weight of all foot surgery patients, we will use the formula:
Confidence Interval = mean ± (critical value) * (standard deviation / sqrt(sample size))
Given that we want a 95% confidence interval, we need to calculate the critical value. For a 95% confidence level and a large sample size (greater than 30), we can use the Z-table to find the critical value.
Step 1: Determine the critical value.
A 95% confidence level corresponds to a significance level (α) of 0.05. Since we have a large sample size (10 patients), we can use the Z-table to find the critical value.
The critical value for a 95% confidence level corresponds to a Z-value of 1.96. This means that 95% of the data falls within 1.96 standard deviations of the mean in a standard normal distribution.
Step 2: Calculate the confidence interval.
Using the formula mentioned above:
Confidence Interval = mean ± (critical value) * (standard deviation / sqrt(sample size))
Confidence Interval = 109 kg ± (1.96) * (11 kg / sqrt(10))
Calculating the square root of 10 gives us approximately 3.1623.
Confidence Interval = 109 kg ± (1.96) * (11 kg / 3.1623)
Calculating (1.96) * (11 kg / 3.1623) gives us approximately 6.8949.
Confidence Interval = 109 kg ± 6.8949 kg
Finally, we need to determine the lower boundary of the confidence interval by subtracting the result from the mean weight:
Lower boundary = 109 kg - 6.8949 kg
Lower boundary = 102.1051 kg
Therefore, the lower boundary of the 95% confidence interval for the true mean weight of all foot surgery patients is approximately 102.1051 kg.