Answer to Question #120240 in Statistics and Probability for Duque

Question

a study claims that 98.6F is the average if the normal room temperature. The sample of body temperature for 25 houses are recorded in the figure. By using alpha = 0.05, can the claim be rejected
97.8 97.2 97.4 97.6 97.8
97.9 98.0 98.0 98.0 98.1
98.2 98.3 98.3 98.4 98.4
98.4 98.5 98.6 98.6 98.7
98.8 98.8 98.9 98.9 99.0

Answer 1

Let $X=$ body temperature.

The sample size is $n=25.$

$$\bar{x}={1\over n}\displaystyle\sum_{i=1}^nx_i=98.264$$

$$s^2 =\sqrt{{1\over n-1}\displaystyle\sum_{i=1}^n(x_i-\bar{x})^2}=0.2324$$

$$s=\sqrt{s^2}=0.4821$$

The following null and alternative hypotheses need to be tested:

$H_0:\mu=98.6$

$H_1:\mu\not=98.6$

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.

Based on the information provided, the significance level is $\alpha=0.05, df=25-1=24,$ and the critical value for a two-tailed test is $t_c=2.064.$

The rejection region for this two-tailed test is $R=\{t:|t|>2.064\}.$  

The t-statistic is computed as follows:

$$t={\bar{x}-\mu\over s/\sqrt{n}}={98.264-98.6\over 0.4821/\sqrt{25}}=-3.4848$$

Since it is observed that $|t|=3.4848>2.064=t_c,$ it is then concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that the population $\mu$ is different than 98.6, at the 0.05 significance level.

Using the P-value approach: The p-value for $df=24, t=-3.4848$ is $p=2(0.00095646)\approx0.001913,$ and since $p=0.001913<0.05,$ it is then concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that the population $\mu$ is different than 98.6, at the 0.05 significance level.

The claim can be rejected at the 0.05 significance level.