# A sample of 16 items provides a sample standard deviation of 9.5. Test the following hypotheses...

## Question:

A sample of 16 items provides a sample standard deviation of 9.5. Test the following hypotheses using {eq}\alpha = .05 {/eq}. What is your conclusion? Use both the p-value approach and the critical value approach.

{eq}H0: \sigma^2 \leq 50 {/eq}

{eq}Ha: \sigma^2 > 50 {/eq}

Test statistic =

p-value low from table =

p-value high from table =

Do you accept or reject the null hypothesis?

## p-value:

p-value is the probability that calculates the possibility of extreme events that could have been caused by the random sample errors assuming null hypothesis as true. It doesn?t not take into consideration the effect of alternative hypothesis. It just shows the strength of the assumed hypothesis.

## Answer and Explanation: 1

**Given information:**

{eq}\begin{align*} n = 16\\ s = 9.5\\ \alpha = 0.05 \end{align*} {/eq}

Null hypothesis

{eq}H0:{\sigma ^2} \le 50 {/eq}

Alternative hypothesis

{eq}H1:{\sigma ^2} > 50{/eq}

1)

Test statistics will be calculated as:

{eq}\begin{align*} TS = \dfrac{{\left( {n - 1} \right){s^2}}}{{{\sigma ^2}}}\\ = \dfrac{{15 \times {{9.5}^2}}}{{50}}\\ = 27.075 \end{align*} {/eq}

2)

p-value will be calculated as:

{eq}\begin{align*} p = P\left( {{\chi _{0.05,\left( {n - 1} \right)}} > TS} \right)\\ = P\left( {{\chi _{0.05,15}} > 27.075} \right)\\ = 0.03 \end{align*} {/eq}

p-value high from table is 0.03

p-value low from table is 0.97.

3)

p-value which is 0.03 is less than the alpha value which is 0.05.

Therefore, we reject the null hypothesis which implies that population variance is less than 50.

#### Learn more about this topic:

from

Chapter 6 / Lesson 9Definition of p-value and its comparison with alpha level. Learn how to find p-value using the p-value formula and a few commands to find p-value in excel.