A researcher claims that average body temperature is different for the two genders. A sample of 65 men and 65 women yields average body temperatures of 98.105 and 98.394 oF, respectively. Based on historical data, the standard deviation of body temperature is known to be 0.699 and 0.743 degrees Fahrenheit, respectively. At a 5% level of significance, can you conclude that average body temperature differs with gender? Would your answer change for a 1% level of significance?
Solution
Given that the sample size $n_1 = 65$, $n_2 = 65$, sample mean $\overline{x}_1= 98.105$,
$\overline{x}_2= 98.394$, standard deviation $\sigma_1 = 0.699$ and $\sigma_2 = 0.743$.
Step 1 State the hypothesis testing problem
The hypothesis testing problem is
$H_0 : \mu_1 = \mu_2$ against $H_1 : \mu_1 \neq \mu_2$ ($\textit{two-tailed}$)
Step 2 Define test statistic
The test statistic is
$$ \begin{aligned} Z=\frac{(\overline{x}_1 -\overline{x}_1)-(\mu_1 -\mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}. \end{aligned} $$
The test statistic $Z$ follows standard normal distribution $N(0,1)$.
Step 3 Specify the level of significance
The significance level is $\alpha = 0.05$.
Step 4 Determine the critical value
As the alternative hypothesis is $\textit{two-tailed}$, the critical value of $Z$ $\text{are}$ $\text{-1.96 and 1.96}$.
The rejection region (i.e. critical region) is $\text{Z < -1.96 or Z > 1.96}$.
Step 5 Computation
The test statistic for testing above hypothesis under the null hypothesis is
$$ \begin{aligned} Z_{obs}&=\frac{(\overline{x}_1 -\overline{x}_1)-0}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}\\ &= \frac{98.105-98.394}{\sqrt{\frac{0.699^2}{65}+\frac{0.743^2}{65}}}\\ &= -2.284 \end{aligned} $$
Step 6 Decision
Traditional approach:
The rejection region (i.e. critical region) is $\text{Z < -1.96 or Z > 1.96}$. The test statistic is $Z_{obs} =-2.284$ which falls $inside$ the critical region, we $\textit{reject}$ the null hypothesis.
OR
$p$-value approach:
The test is $\text{two-tailed}$ test, so the p-value is the area to the $\text{extreme}$ of the test statistic ($Z_{obs}=-2.284$) is p-value = $0.0224$.
The p-value is $0.0224$ which is $\textit{less than}$ the significance level of $\alpha = 0.05$, we $\textit{reject}$ the null hypothesis.
We conclude that at 5% level of sginificance the average body temperature differs with gender.
While for 1% level of significance as the p-value $0.0224$ is not less than 0.01, we conclude that the average body temperature do not differ significantly.
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
