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2021-10-16

Suppose we are testing ${H}_{o}:p=0.20$ vs ${H}_{a}:p$ does not equal 0.20 and $TS=2.34$ . What is the p-value?

Alix Ortiz

Skilled2021-10-17Added 109 answers

Step 1

From the information, the test statistics is 2.34. The null and alternative hypotheses are given as:

Null hypothesis:

${H}_{0}:p=0.20$

${H}_{A}:p\ne 0.20$

Using the test statistic the p-value is to be calculated.

The null hypothesis states that there is evidence that the population proportion is equal to 0.20. And alternative hypothesis states that there is no evidence that the population proportion is equal to 0.20. The test is two-tailed test and the test statistic is obtained by using the standard normal distribution. The main objective of the problem is to obtain two sided p-value.

Step 2

The p-value is obtained as shown below:

From the given information, the test statistic is,$z=2.34$ and the test is two tailed test.

The p-value is,

$p-value=2P\left(Z>2.34\right)$

$=2[1-P(Z\le 2.34)]$

From ’Standard Normal Table’ the area to the left of$Z=2.34$ is 0.09904. Hence,

$p-value=2[1-P(Z\le 2.34)]$

$=2[1-0.9904]$

$=2\times 0.0096$

$=0.0192$

The p-value is 0.0192

The p-value is obtained by multiplication of the probability of$Z>2.34$ with 2. First calculate the probability of $Z\le 2.34$ . It is calculated by taking the value that is intersected with the row containing the value 2.30 and the column containing the value 0.04, from the standard normal table.

From the information, the test statistics is 2.34. The null and alternative hypotheses are given as:

Null hypothesis:

Using the test statistic the p-value is to be calculated.

The null hypothesis states that there is evidence that the population proportion is equal to 0.20. And alternative hypothesis states that there is no evidence that the population proportion is equal to 0.20. The test is two-tailed test and the test statistic is obtained by using the standard normal distribution. The main objective of the problem is to obtain two sided p-value.

Step 2

The p-value is obtained as shown below:

From the given information, the test statistic is,

The p-value is,

From ’Standard Normal Table’ the area to the left of

The p-value is 0.0192

The p-value is obtained by multiplication of the probability of

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