To construct a confidence interval for the population mean length of time, we will use the following formula:
Confidence interval = sample mean ± (critical value) * (standard deviation / sqrt(sample size))
First, let's calculate the sample mean:
Sample mean = (2.3 + 2.6 + 2.8 + 2.6 + 2.1 + 2.5 + 2.1 + 2.7 + 2) / 9 = 2.4222 (rounded to 4 decimal places)
Next, we need to calculate the standard deviation of the sample:
Step 1: Find the sample variance:
(x - mean)^2
(2.3 - 2.4222)^2 = 0.0144
(2.6 - 2.4222)^2 = 0.0316
(2.8 - 2.4222)^2 = 0.1416
(2.6 - 2.4222)^2 = 0.0316
(2.1 - 2.4222)^2 = 0.1037
(2.5 - 2.4222)^2 = 0.0061
(2.1 - 2.4222)^2 = 0.1037
(2.7 - 2.4222)^2 = 0.0816
(2 - 2.4222)^2 = 0.1772
Step 2: Sum up the variances:
0.0144 + 0.0316 + 0.1416 + 0.0316 + 0.1037 + 0.0061 + 0.1037 + 0.0816 + 0.1772 = 0.7315
Sample variance = sum of variances / (sample size - 1) = 0.7315 / (9 - 1) = 0.0914 (rounded to 4 decimal places)
Standard deviation = square root of sample variance = sqrt(0.0914) = 0.3023
Now, we need to find the critical value. We will use a t-distribution since our sample size (n = 9) is small and the population standard deviation is unknown. With a 90% confidence level and a sample size of 9, the degrees of freedom is 8 (n-1).
Using a t-table or calculator, the critical value for a 90% confidence level and 8 degrees of freedom is approximately 2.306.
Plugging in the values into the formula:
Confidence interval = 2.4222 ± (2.306) * (0.3023 / sqrt(9))
Confidence interval = 2.4222 ± (2.306) * (0.3023 / 3)
Confidence interval = 2.4222 ± (2.306) * 0.1008
Confidence interval = 2.4222 ± 0.2323
Therefore, the 90% confidence interval for the population mean length of time is (2.1899, 2.6545), which is an open-interval.
Suppose you are a researcher in a hospital. You are experimenting with a new sedative. You collect data from a random sample of 9 patients. The period of effectiveness of the sedative for each patient (in hours) is as follows:
Hours
2.3
2.6
2.8
2.6
2.1
2.5
2.1
2.7
2
Construct a 90% confidence interval for the population mean length of time. Enter your answer as an open-interval
1 answer