Using the z-distribution, it is found that since the p-value of the test is less than 0.05, we reject the null hypothesis and find that there is enough evidence to conclude that the true proportion of all life insurance claims made to this company that are settled within 30 days is less than 85%.

<h3>What are the hypothesis tested?</h3>

At the null hypothesis, it is tested if there is not enough evidence that the proportion is below 85%, that is:

$H_0: p \geq 0.85$

At the alternative hypothesis, it is tested if there is enough evidence that the proportion is below 85%, that is:

$H_0: p < 0.85$

<h3>What is the test statistic?</h3>

The test statistic is given by:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

In which:

$\overline{p}$ is the sample proportion.

p is the proportion tested at the null hypothesis.

n is the sample size.

For this problem, the parameters are:

$p = 0.85, n = 250, \overline{p} = \frac{203}{250} = 0.812$

Hence the test statistic is:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

$z = \frac{0.812 - 0.85}{\sqrt{\frac{0.85(0.15)}{250}}}$

z = -1.68

<h3>What is the p-value and the conclusion</h3>

Using a z-distribution calculator, with a left-tailed test, as we are testing if the proportion is less than a value, and z = -1.68, the p-value is of 0.0465.

Since the p-value of the test is less than 0.05, we reject the null hypothesis and find that there is enough evidence to conclude that the true proportion of all life insurance claims made to this company that are settled within 30 days is less than 85%.

More can be learned about the z-distribution at brainly.com/question/16313918

#SPJ1

4 0

1 year ago