E. canis infection is a tick-borne disease of dogs that is sometimes contracted by humans. Among infected humans,the distribution of white blood cell counts has an unknown mean $\mu$ and a standard deviation $\sigma$. In the general population, the mean white blood cell count is 7250/mm3. It is believed that person infected with E. canis must on average have lower white blood cell counts.

(a) What are the null and alternative hypothesis for a one-sided test?

(b) For a sample of 15 infected persons, the mean white blood cell count is $\overline{x}=4767/mm3$ and standard deviation is s = 3204/mm3. Conduct the test at the $\alpha=0.05$ level?

(c) What do you conclude?

#### Solution

(a) The hypothesis testing problem is $H_0:\mu=7250$ against $H_a:\mu< 7250$.

(b) Following are the steps for testing above hypothesis problem.

Given that the sample size $n = 15$, sample mean $\overline{x}= 4767/mm3$ and sample standard deviation $s = 3204 / mm3$.

#### Step 1 Hypothesis Testing Problem

The hypothesis testing problem is

$H_0 : \mu = 7250$ against $H_1 : \mu < 7250$ ($\text{left-tailed}$)

#### Step 2 Test Statistic

The test statistic is

` $$ \begin{aligned} t& =\frac{\overline{x} -\mu}{s/\sqrt{n}} \end{aligned} $$ `

which follows $t$ distribution with $n-1$ degrees of freedom.

#### Step 3 Significance Level

The significance level is $\alpha = 0.05$.

#### Step 4 Critical Value(s)

As the alternative hypothesis is $\text{left-tailed}$, the critical value of $t$ $\text{is}$ $-1.761$.

The rejection region (i.e. critical region) is $\text{t < -1.761}$.

#### Step 5 Computation

The test statistic under the null hypothesis is

` $$ \begin{aligned} t&=\frac{ \overline{x} -\mu_0}{s/\sqrt{n}}\\ &= \frac{4767-7250}{3204/ \sqrt{15 }}\\ &= -3.001 \end{aligned} $$ `

#### Step 6 Decision

*Traditional approach:*

The test statistic is $t =-3.001$ which falls $\text{inside}$ the critical region, we $\text{reject}$ the null hypothesis.

**OR**

#### Step 6 Decision

*$p$-value Approach*

This is a $\text{left-tailed}$ test, so the p-value is the area to the left of the test statistic ($t=-3.001$). Thus the p-value = $0.0048$.

The p-value is $0.0048$ which is $\text{less than}$ the significance level of $\alpha = 0.05$, we $\text{reject}$ the null hypothesis.

(c) There is sufficient evidence to conclude that $\mu < 7250$ / mm3. We conclude that person infected with E. canis must on average have lower white blood cell counts.

#### Further Reading

- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators