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

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

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_{(0.05)/(2)}=\pm 1.96`

The value of margin of error is

`ME=z_(\alpha/2)((\sigma)/(√(n)))`

`ME=1.96((7.4)/(√(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.