Help me simplify this

A random sample of 40 binomial trials resulted in 14 successes. test the claim that the population proportion of successes does

mylen [45]

Using the z-distribution, since the p-value of the test is of 0.057 > 0.05, there is not enough evidence that the population proportion of successes does not equal 0.50.

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

At the null hypotheses, we test if the proportion of successes equals 0.5, hence:

$H_0: p = 0.5$

At the alternative hypotheses, we test if it does not equal, hence:

$H_1: p \neq 0.5$

<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 given by:

$n = 40, \overline{p} = \frac{14}{40} = 0.35, p = 0.5$

Hence the test statistic is given by:

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

z = (0.35 - 0.5)/(0.5/sqrt(40))

z = -1.9.

<h3>What is the decision?</h3>

Using a z-distribution calculator, considering a two-tailed test, as we are testing if the proportion is different of a value, with z = -1.9, we get that the p-value of the test is of 0.057.

Since the p-value of the test is of 0.057 > 0.05, there is not enough evidence that the population proportion of successes does not equal 0.50.

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

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1 year ago

Two shooters, Rodney and Philip, practice at a shooting range. They fire rounds each at separate targets. The targets are mark

Sedaia [141]

Complete Question

Answer:

a

$SE = 0.66}$