The masses of 6 containers of Chow dog food are:
9.8
10.4
9.8
10
10.2
9.6
Find a 95% confidence interval for the mean mass of all such dog food. (Assume an approximate normal distribution). Give the lower limit.
To find the 95% confidence interval for the mean mass of all the containers of Chow dog food, we can use the t-distribution since the sample size is small (n = 6) and the population standard deviation is unknown.
First, let's calculate the sample mean:
Mean = (9.8 + 10.4 + 9.8 + 10 + 10.2 + 9.6) / 6 = 59.8 / 6 = 9.9667 (rounded to four decimal places)
Next, let's calculate the sample standard deviation (s):
To do this, we need to find the squared deviations from the mean for each data point, sum them up, divide by (n-1), and take the square root.
Squared Deviations from mean:
(9.8 - 9.9667)^2 = 0.002222
(10.4 - 9.9667)^2 = 0.018113
(9.8 - 9.9667)^2 = 0.002222
(10 - 9.9667)^2 = 0.001067
(10.2 - 9.9667)^2 = 0.005356
(9.6 - 9.9667)^2 = 0.133689
Sum of squared deviations:
0.002222 + 0.018113 + 0.002222 + 0.001067 + 0.005356 + 0.133689 = 0.162669
Sample variance (s^2) = Sum of squared deviations / (n-1) = 0.162669 / 5 = 0.032534 (rounded to six decimal places)
Sample standard deviation (s) = √(0.032534) = 0.180244 (rounded to six decimal places)
To calculate the t-critical value, we need to determine the degrees of freedom which is (n-1) = 6-1 = 5.
Using a t-table or calculator with a significance level (α) of 0.05 and degrees of freedom (df) of 5, the t-critical value is approximately 2.571.
Standard Error (SE) = s / √n = 0.180244 / √6 = 0.073523 (rounded to six decimal places)
Margin of Error (ME) = t-critical value * SE = 2.571 * 0.073523 = 0.188950 (rounded to six decimal places)
Finally, we can calculate the 95% confidence interval for the mean mass of all such dog food:
Lower Limit = Sample mean - Margin of Error = 9.9667 - 0.188950 = 9.777750 (rounded to six decimal places)
Therefore, the lower limit of the 95% confidence interval for the mean mass of all such dog food is approximately 9.777750.
To find the 95% confidence interval for the mean mass of all the dog food containers, we can use the formula:
Confidence Interval = Sample Mean +/- Margin of Error
First, let's calculate the sample mean:
Sample Mean = (Sum of all masses) / (Number of containers)
= (9.8 + 10.4 + 9.8 + 10 + 10.2 + 9.6) / 6
= 59.8 / 6
= 9.9667 (rounded to 4 decimal places)
Next, we need to calculate the standard deviation of the sample. For an approximate normal distribution, we can use the formula:
Sample Standard Deviation = Square root of [(Sum of (Xi - X̄)^2) / (n - 1)]
Where:
- Xi is the mass of each container
- X̄ is the sample mean
- n is the number of containers
Calculating the sum of (Xi - X̄)^2:
(9.8 - 9.9667)^2 + (10.4 - 9.9667)^2 + (9.8 - 9.9667)^2 + (10 - 9.9667)^2 + (10.2 - 9.9667)^2 + (9.6 - 9.9667)^2
= 0.00150667 + 0.00117778 + 0.00150667 + 0.00001333 + 0.00053222 + 0.00106667
= 0.00580333
Calculating the sample standard deviation:
Sample Standard Deviation = √(0.00580333 / (6 - 1))
= √(0.00580333 / 5)
= √(0.00116067)
= 0.03408 (rounded to 5 decimal places)
Now, we need to determine the margin of error. The formula for the margin of error is:
Margin of Error = Critical Value * (Sample Standard Deviation / √n)
Since we want a 95% confidence interval, and assuming an approximate normal distribution, the critical value is obtained by looking up the z-score in the standard normal distribution table. For a 95% confidence interval, the critical value is 1.96.
Calculating the margin of error:
Margin of Error = 1.96 * (0.03408 / √6)
= 1.96 * (0.03408 / √6)
= 0.02677 (rounded to 5 decimal places)
Finally, we can calculate the confidence interval:
Confidence Interval = Sample Mean +/- Margin of Error
= 9.9667 - 0.02677
= 9.9399 (rounded to 4 decimal places)
Therefore, the lower limit of the 95% confidence interval for the mean mass of all the dog food containers is approximately 9.9399.