Question

In a study of the thermal inertial properties of aerated concrete used as building material, average interior temperatures were investigated. These temperatures are known to have an approximately normal distribution. The claim is the average interior temperature is 22.5 degrees. A consumer advocate believes this temperature is in fact less than 22.5 degreess. To this end, he collects a sample of 15 measurements and finds the average temperature of the sample to be 20.9 degrees with standard deviation of 3.5 degrees. Test this hypothesis at level of significance alpha = 0.04.

Answer 1

The values provided in the above question are as below

Sample mean = = 20.9

Sample standard deviation = = 3.5

Sample size = = 15

Population mean = = 22.5

The null and alternative hypothesis are

H₀ : There is not sufficient evidence to warrent rejection the claim that the average interior temperature is 22.5 degrees.

H₁ : There is sufficient evidence to warrent rejection the claim that the average interior temperature is 22.5 degrees.

Using parameter the null and alternative hypothesis are

We calculate the value of test statistic using following formula

(Round the answer to two decimal places)

The test statistic t = -1.77

Now, we calculate the critical value with level of significance alpha = 0.04 using Excel function

=TINV(probability, deg-freedom)

Note that, The Excel function provides the critical value of t based on two tails only, so if we have find the one tailed critical value then we multiplied the alpha by 2 and then find the critical value of t and also remember that if critical value of t is left tailed then we write it is with negative sign)

Here, probability = 2*alpha = 2*0.04 = 0.08 and deg-freedom = n - 1 = 15 - 1 = 14

probability = 0.08, deg-freedom = 14

Using above values the Excel function is as below

=TINV(0.08, 14) then press Enter

=1.887496

But the given test is left tailed then we write the above critical value with negative sign

=-1.887496 -1.89 (Round the answer to two decimal places)

Critical value of t = -1.89

By comparing the test statistic t with critical value of t we take decision of reject or fail to reject the null hypothesis H₀. using following way

The given test is left tailed test

We reject null hypothesis H₀, if test statistic t   critical value of t

Otherwise we fail to reject the null hypothesis H₀.

Here, the test statistic t = -1.77   Critical value of t = -1.89

We fail to reject the null hypothesis H₀.

That is, there is not sufficient evidence to warrent rejection the claim that the average interior temperature is 22.5 degrees.