Answer to Question #240681 in Statistics and Probability for Marie

Question

An article in the ASCE Journal of Energy Engineering (1999, Vol. 125, pp.59-75) describes a study of the thermal inertia properties of autoclaved aerated concrete used as a building material. Five samples of the material were tested in a structure, and the average interior temperatures (°C) reported were as follows: 23.01, 22.22, 22.04, 22.62, and 22.59. Test that the average interior temperature is equal to 22.5°C using alpha (a) = 0.05.

Answer 1

Calculate the mean and standard deviation of the available data.

$\bar{x} = \frac{\sum x_i}{n} \\ = \frac{23.01 + 22.22+...+22.59}{5} \\ = 22.496 \\ s = \sqrt{ \frac{\sum (x_i - \bar{x})^2}{n-1}} \\ = \sqrt{\frac{(23.01 -22.496)^2 +(22.22-22.496)^2 +...+(22.59-22.496)^2}{4}} \\ = 0.3783 \\ H_0: \mu = 22.5 \\ H_1: \mu ≠ 22.5$

When n<30 and $\sigma$ is unknown a one sample t-test is used.

Test-statistic:

$t = \frac{\bar{x} - \mu}{s/ \sqrt{n}} \\ = \frac{22.496 -22.5}{0.3783 / \sqrt{5}} \\ = -0.02$

The absolute calculated t value is,

|t| = 0.02

Calculate the degrees of freedom.

$df = n - 1 \\ = 5-1 \\ = 4 \\ α=0.05$

From the standard t table values, observe that the critical value of t for the two tails test at the 5% level of significance and 4 degrees of freedom is 2.776

Compare the absolute calculated t value with the critical value.

Here, the absolute calculated t value is less than the critical value of t so fails to reject the Null hypothesis.

Hence, conclude that the true mean of average interior temperatures is 22.50