Question

The lengths of text messages are normally distributed with a population standard deviation of 6 characters and an unknown population mean. If a random sample of 21 text messages is taken and results in a sample mean of 30 characters, find a 80% confidence interval for the population mean. Round your answers to two decimal places. z0.10 z0.05 z0.04 z0.025 z0.01 z0.005 1.282 1.645 1.751 1.960 2.326 2.576

Answer 1

Answer:

The 80% confidence interval for the population mean is between 28.32 characters and 31.68 characters.

Step-by-step explanation:

We have the standard deviation for the population, so we can use the normal distribution. If we had the standard deviation for the sample, we would have to use the t-distribution.

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.8}{2} = 0.1$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$.

So it is z with a pvalue of $1-0.5 = 0.9$, so $z = 1.282$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.282*\frac{6}{\sqrt{21}} = 1.68$

The lower end of the interval is the sample mean subtracted by M. So it is 30 - 1.68 = 28.32 characters.

The upper end of the interval is the sample mean added to M. So it is 30 + 1.68 = 31.68 characters.

The 80% confidence interval for the population mean is between 28.32 characters and 31.68 characters.