# Consider the following hypotheses: H_0: mu less than or equal to 62.2 H_A: mu greater than 62.2....

## Question:

Consider the following hypotheses:

{eq}H_{0} : \mu \leq 62.2 \\ H_{A} : \mu > 62.2 {/eq}

A sample of 44 observations yields a sample mean of 63.3. Assume that the sample is drawn from a normal population with a known population standard deviation of 5.1.

a. Calculate the p-value.

A. p-value ≥ 0.10

B. p-value < 0.01

C. 0.01 ≥ p-value < 0.025

D. 0.025 ≤ p-value < 0.05

E. 0.05 ≤ p-value < 0.10

b. What is the conclusion if α = 0.01?

A. Reject H0 since the p-value is greater than α.

B. Reject H0 since the p-value is smaller than α.

C. Do not reject H0 since the p-value is greater than α.

D. Do not reject H0 since the p-value is smaller than α.

## Hypothesis testing

Hypothesis testing is a test conducted to check whether the claim made in the null hypothesis is rejected or not by comparing the test statistics p-value with a certain level of significance. If the p-value of test statistics is less than the level of significance under the null hypothesis true, there is sufficient evidence against the null hypothesis and when the p-value is greater than level of significance there is insufficient evidence against the null hypothesis.

Given information:

{eq}\begin{align*} {H_0}&:\mu \le 62.2\\ {H_1}&:\mu > 62.2\\ n &= 44\\ \bar X &= 63.3\\ \sigma &= 5.1 \end{align*} {/eq}

The test statistics to conduct the hypothesis testing is written below:

{eq}Z = \dfrac{{\bar X - \mu }}{{\dfrac{\sigma }{{\sqrt n }}}} {/eq}

A

The p- value for the above test will be calculated as follows:

{eq}\begin{align*} {\rm{P - Value}} &= P\left( {Z > \dfrac{{\bar X - \mu }}{{\dfrac{\sigma }{{\sqrt n }}}}} \right)\\ &= P\left( {Z > \dfrac{{63.3 - 62.2}}{{\dfrac{{5.1}}{{\sqrt {44} }}}}} \right)\\ &= P\left( {Z > 1.43} \right)\\ &= 1 - P\left( {Z < 1.43} \right)\\ &= 1 - 0.9236 \Rightarrow .076359 \end{align*} {/eq}

The, p- value of the test is 0.076359.

Hence, Option E. 0.05 p-value < 0.10 is correct.

B

The P-value of the test is greater than the level of significance of 0.01 which indicates that there is no sufficient evidence against the null hypothesis.

Hence, option C Does not reject H0 since the p-value is greater than is correct as the p-value is greater than the level of significance which results in non- rejection of the null hypothesis.