Question

A random sample of 130 units have average 36 and standard deviation 0.7. a) Find a 90% confidence interval. b) How large a sample do we need if we want to be 90% sure that sample mean is within 0.05 of the true mean?

Answer 1

To find a 90% confidence interval, use the Z-distribution and calculate the margin of error. For this given sample, the 90% confidence interval is (35.91, 36.09). To have a 90% confidence that the sample mean is within 0.05 of the true mean, a sample size of at least 1081 units is needed.

Step-by-step explanation:

To find a 90% confidence interval, we first determine the critical value for a 90% confidence level. Since the sample size is large (130 units), we can use the Z-distribution. The critical value for a 90% confidence level is approximately 1.645.

Next, we calculate the margin of error (ME) using the formula: ME = Z * (σ/√n), where Z is the critical value, σ is the standard deviation, and n is the sample size. In this case, Z = 1.645, σ = 0.7, and n = 130. Plugging in these values, we get: ME = 1.645 * (0.7/√130) ≈ 0.09.

Finally, we construct the confidence interval by adding and subtracting the margin of error from the sample mean. The 90% confidence interval is therefore: (36 - 0.09, 36 + 0.09), which simplifies to (35.91, 36.09).

b) To determine the required sample size to have a 90% confidence that the sample mean is within 0.05 of the true mean, we can rearrange the margin of error formula: n = (Z * σ / ME)^2. Plugging in the values Z = 1.645, σ = 0.7, and ME = 0.05, we get: n = (1.645 * 0.7 / 0.05)^2 ≈ 1081. To be 90% confident, we would need a sample size of at least 1081 units.