Answer to Question #169485 in Statistics and Probability for Sunera

Question

A random sample of 25 observations was drawn from a normal population whose standard deviation is 50. The sample mean was 200.

1. Estimate the population mean with 95% confidence.

2. Repeat part (a) changing the population standard deviation to 25.

3. Repeat part (a) changing the population standard deviation to 10.

4. Describe what happens to the confidence interval estimate when the standard deviation

is decreased.

Answer 1

We can estimate confidence interval for normal random variable using formula:

$Pr(X_{mean} - z\frac{\sigma}{\sqrt{n}}, X_{mean} + z\frac{\sigma}{\sqrt{n}}) = \gamma$ , where $X_{mean} = 200, \ \sigma = 50, \ z = \Phi^{-1}(1 - \frac{\alpha}{2}), \Phi(x)$ is CDF of normal random variable with mean = 0 and standard deviation = 1, $\alpha = 1 - \gamma$

$(X_{mean} - z\frac{\sigma}{\sqrt{n}}, X_{mean} + z\frac{\sigma}{\sqrt{n}})$ is confidence interval of sample mean

1. For $\gamma = 0.95$ : $z = \Phi^{-1}(1 - \frac{1 - 0.95}{2}) = 1.96$

Confidence interval for sample mean is:

$(200 - 1.96 \cdot \frac{50}{\sqrt{25}}, 200 + 1.96 \cdot \frac{50}{\sqrt{25}}) = (180.4, 219.6)$

2. For $\gamma = 0.95, \ \sigma=25$ : $z = \Phi^{-1}(1 - \frac{1 - 0.95}{2}) = 1.96$

Confidence interval for sample mean is:

$(200 - 1.96 \cdot \frac{25}{\sqrt{25}}, 200 + 1.96 \cdot \frac{25}{\sqrt{25}}) = (190.2, 209.8)$

3. For $\gamma = 0.95, \ \sigma=10$ : $z = \Phi^{-1}(1 - \frac{1 - 0.95}{2}) = 1.96$

Confidence interval for sample mean is:

$(200 - 1.96 \cdot \frac{10}{\sqrt{25}}, 200 + 1.96 \cdot \frac{10}{\sqrt{25}}) = (196.08, 203.92)$

4. So, when when the standard deviation is decreased, confidence interval size is getting smaller with linear rate.