Question

Suppose you have a sample of size 42 with a mean of 30 and a population standard deviation of 7.4. What is the maximal margin of error associated with a 95% confidence interval for the true population mean? As in the reading, in your calculations, use z = 2. Give your answer as a decimal, to two places

Answer 1

Answer:

The maximal margin of error associated with a 95% confidence interval for the true population mean is 2.238.

Step-by-step explanation:

We have given,

The sample size n=42

The sample mean $\bar{x}=30$

The population standard deviation $\sigma=7.4$

Let $\alpha$ be the level of significance = 0.05

Using the z-distribution table,

The critical value at 5% level of significance and two tailed z-distribution is

$\pm z_{\frac{0.05}{2}}=\pm 1.96$

The value of margin of error is

$ME=z_{\alpha/2}(\frac{\sigma}{\sqrt{n}})$

$ME=1.96(\frac{7.4}{\sqrt{42}})$

$ME=1.96(1.1418)$

$ME=2.238$

The maximal margin of error associated with a 95% confidence interval for the true population mean is 2.238.