A random sample of 19 residents of Escambia County in Florida has a mean annual income of \$ 56,800 and a standard deviation of \$ 7600. In Miami-Dade County of Florida, a random sample of 21 residents has a mean annual income of \$ 53,900 and a standard deviation of \$ 7425. Test the claim at $\alpha = 0.01$ that the mean annual incomes in Escambia and Miami-Dade counties are not the same. Assume the population variances are not equal.

#### Solution

Given that the sample size $n_1 = 19$, $n_2 = 21$, sample mean $\overline{x}_1= 56800$,

$\overline{x}_2= 53900$, sample standard deviation $s_1 = 7600$ and $s_2 = 7425$.

**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 for testing above hypothesis testing problem is

` $$ \begin{aligned} t=\frac{(\overline{x}_1 -\overline{x}_1)-(\mu_1 - \mu_2)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}} \end{aligned} $$ `

(Assuming that the population variances are not equal).

The test statistic $t$ follows Students' $t$ distribution with $\nu$ degrees of freedom, where

` $$ \begin{aligned} \nu = \frac{\bigg(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\bigg)^2}{\frac{s_1^4}{n_1^2(n_1-1)}+\frac{s_2^4}{n_2^2(n_2-1)}}=37 \end{aligned} $$ `

rounded to nearest integer.

**Step 3 Level of significance**

The significance level is $\alpha = 0.01$.

**Step 4 Determine the critical value**

As the alternative hypothesis is $\textit{two-tailed}$, the critical value of $t$ using $\alpha = 0.01$ and degrees of freedom $=37$ $\text{are}$ $\text{-2.715 and 2.715}$.

The rejection region (i.e. critical region) is $\text{t < -2.715 or t > 2.715}$.

**Step 5 Computation**

The test statistic for testing above hypothesis testing problem under the null hypothesis is

` $$ \begin{aligned} t&=\frac{(\overline{x}_1 -\overline{x}_1)-0}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ &= \frac{(56800-53900)}{\sqrt{\frac{7600^2}{19}+\frac{7425^2}{21}}}\\ &= 1.2184 \end{aligned} $$ `

**Step 6 Decision**

Traditional approach:

The rejection region (i.e. critical region) is $\text{t < -2.715 or t > 2.715}$. The test statistic is $t =1.2184$ which falls $\text{outside}$ the critical region, we $\textit{fail to reject}$ the null hypothesis.

OR

$p$-value approach:

The test is $\textit{two-tailed}$ test, so p-value is the area to the $\textit{extreme}$ of the test statistic ($t=1.2184$). That is p-value = $2*P(t\geq 1.2184 ) = 0.2308$.

The p-value is $0.2308$ which is $\textit{greater than}$ the significance level of $\alpha = 0.01$, we $\textit{fail to reject}$ the null hypothesis.

There is no sufficient evidence to conclude that the mean annual incomes in Escambia and Miami-Dade counties are not the same.

#### Further Reading

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