Question

An article in the ASCE Journal of Energy Engineering describes a study of the thermal inertial properties of autoclaved aerated concrete used as a building material. Five samples of the material were tested in a structure, and the five temperatures (◦C) reported were: 23.01, 22.22, 22.04, 22.62, and 22.59
Test the hypothesis of H0 : µ = 22.5 versus H1 : µ 6= 22.5.

Show and label the 5 steps.

Answer 1

Answer:

Null hypothesis:$\mu = 22.5$  

Alternative hypothesis:$\mu \neq 22.5$  

$t=\frac{22.496-22.5}{\frac{0.378}{\sqrt{5}}}=-0.0236$    

$p_v =2*P(t_{(4)}$  

If we compare the p value and the significance level given assumed for example $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the population mean is NOT significant different from 22.5 at 5% of significance.  

Step-by-step explanation:

Previous concepts  and data given

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

First we can calculate the average and the sample standard deviation with the following formulas:

$\bar X = \frac{\sum_{i=1}^n x_i}{n}$

$s= \sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}$

$\bar X=22.496$ represent the sample mean  

s=0.378 represent the sample standard deviation

n=5 represent the sample selected

$\alpha$ significance level  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is equal to 22.5, the system of hypothesis would be:  

Null hypothesis:$\mu = 22.5$  

Alternative hypothesis:$\mu \neq 22.5$  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

$t=\frac{22.496-22.5}{\frac{0.378}{\sqrt{5}}}=-0.0236$    

P-value

The first step is calculate the degrees of freedom, on this case:  

$df=n-1=5-1=4$  

Since is a two side test the p value would be:  

$p_v =2*P(t_{(4)}$  

Conclusion  

If we compare the p value and the significance level given assumed for example $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the population mean is NOT significant different from 22.5 at 5% of significance.