Question

John wishes to study the mean human body temperature. John organizes a simple random sample which allows him to measure the human body temperature of 45 people at school. His calculations show that his sample has a mean human body temperature of 98.40°F and a standard deviation of 0.62°F. Prior studies indicate that human body temperatures are normally distributed with a standard deviation of 0.50°F. Use the p-value method and a 2% significance level to test the claim that the mean human body temperature of the population is equal to 98.6°F as is commonly believed.

What population parameter is being tested?

Answer 1

$H_0: a=a_0=98.6, H_1: a\neq a_0=98.6\text{ (two-sided)}\\ a\text{ is population mean}\\ \alpha=0.02\\ N=45\\ \overline{x}=98.40\\ \sigma=0.50\\ \text{We will use the following random variable:}\\ U=\frac{(\overline{X}-a_0)\sqrt{n}}{\sigma}\\ u_{obs}=\frac{(98.40-98.6)\sqrt{45}}{0.50}\approx -2.683\\ \Phi(u_{cr})=\frac{1-\alpha}{2}=0.49\\ u_{cr}=2.33\\ \Phi(x)=\frac{1}{\sqrt{2\pi}}\int_0^x e^{-\frac{t^2}{2}}dt\text{ --- Laplace function}\\ p-value=2(0.5-\Phi(2.683))\approx 0.0073<\alpha=0.02\\ \text{So we reject } H_0.\\ \text{At a significance level 0.02 we reject the hypothesis that }\\ \text{the mean human body temperature of the population is equal}\\ \text{to 98.6°F and accept the hypothesis that the mean human body}\\ \text{temperature of the population is not equal to 98.6°F}.\\ \text{Here we test population mean}.$