Question

With a simple random sample of n= 42, sample mean of 21.3 and sample standard deviation of 4.2 What is the 99% confidence interval for the population mean?

Answer 1

To find the 99% confidence interval for the population mean with a sample mean of 21.3 and sample standard deviation of 4.2 for n=42, we use the t-distribution to obtain a confidence interval of (19.55, 23.05).

Step-by-step explanation:

To calculate the 99% confidence interval for the population mean when the sample size is n=42, the sample mean is 21.3, and the sample standard deviation is 4.2, we can use the formula for the confidence interval which is:

Confidence Interval = Sample Mean ± (Critical Value × (Sample Standard Deviation / √n))

Since the sample size is less than 30 and we do not know the population standard deviation, we use the t-distribution to find the critical value. Using a t-table or software, we find the critical value for a 99% confidence interval with 41 degrees of freedom is approximately 2.704.

The error margin (E) is then calculated as follows:

Error Margin (E) = 2.704 × (4.2 / √42)

Now we calculate E:

Error Margin (E) = 2.704 × (4.2 / 6.4807)

Error Margin (E) = 2.704 × 0.6475 ≈ 1.75

The 99% confidence interval is then:

(21.3 ± 1.75), which gives us (19.55, 23.05).

We can say with 99% confidence that the true population mean lies between 19.55 and 23.05.