75 packages are randomly selected from packages received by a parcel service. The sample has a mean weight of 10.2 pounds and a standard deviation of 2.9 pounds. What is the 95% confidence interval for the true mean weight of all packages by the parcel service?
Use same process as in you first post, but for ±.025.
To find the 95% confidence interval for the true mean weight of all packages, we can use the formula:
Confidence Interval = Sample Mean ± (Critical Value × Standard Error)
First, let's calculate the standard error, which is the standard deviation of the sample divided by the square root of the sample size:
Standard Error = Standard Deviation / √Sample Size
In this case, the sample mean weight is 10.2 pounds, the standard deviation is 2.9 pounds, and the sample size is 75. Therefore:
Standard Error = 2.9 / √75
Next, we need to find the critical value for a 95% confidence interval. Since we don't know the population standard deviation, we can use the t-distribution and the corresponding degrees of freedom. The degrees of freedom for a sample size of 75 is 74. We can use a t-distribution table or a statistical calculator to find the critical value.
For a 95% confidence level with 74 degrees of freedom, the critical value is approximately 1.990.
Now, let's substitute the values into the confidence interval formula:
Confidence Interval = 10.2 ± (1.990 × (2.9 / √75))
Calculating the value inside the parentheses:
1.990 × (2.9 / √75) ≈ 1.990 × 0.334 ≈ 0.665
Therefore, the confidence interval for the true mean weight of all packages by the parcel service is:
Confidence Interval ≈ 10.2 ± 0.665
To find the specific range of the confidence interval, we add and subtract the value from the sample mean:
Lower Limit = 10.2 - 0.665 ≈ 9.535
Upper Limit = 10.2 + 0.665 ≈ 10.865
So, the 95% confidence interval for the true mean weight of all packages by the parcel service is approximately 9.535 to 10.865 pounds.