Question

The number of jelly beans placed in jars by a class of kindergarteners is normally distributed with an unknown population
mean and standard deviation. A random sample of 20 jars is taken and results in a sample mean of 43 jelly beans and
sample standard deviation of 6 jelly beans.

Answer 1

Using the t-distribution, the 80% confidence interval for the mean number of jelly beans is (41.22, 44.78).

What is a t-distribution confidence interval?

The confidence interval is:

$\overline{x} \pm t\frac{s}{\sqrt{n}}$

In which:

• $\overline{x}$ is the sample mean.

• t is the critical value.

• n is the sample size.

• s is the standard deviation for the sample.

The critical value, using a t-distribution calculator, for a two-tailed 80% confidence interval, with 20 - 1 = 19 df, is t = 1.3277.

The other parameters are given as follows.

$\overline{x} = 43, s = 6, n = 20$.

Hence, the bounds of the interval are given by:

$\overline{x} - t\frac{s}{\sqrt{n}} = 43 - 1.3277\frac{6}{\sqrt{20}} = 41.22$

$\overline{x} - t\frac{s}{\sqrt{n}} = 43 + 1.3277\frac{6}{\sqrt{20}} = 44.78$

The 80% confidence interval for the mean number of jelly beans is (41.22, 44.78).

More can be learned about the t-distribution at https://brainly.com/question/16162795