A student suspected the average cost of a date was no longer \$30.00 To test her hypothesis, she randomly selected 16 men and asked them how much they had spent on dates the previous weekend. She found the average cost was \$31.17. The standard deviation of the samples was \$5.51. At $\alpha=0.05$, is there enough evidence to support her claim?
Solution
Given that the sample size $n = 16$, sample mean $\overline{x}= 31.17$ and sample standard deviation $s = 5.51$.
Hypothesis Testing Problem
The hypothesis testing problem is
$H_0 : \mu = 30$ against $H_1 : \mu \neq 30$ ($\text{two-tailed}$)
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.
Significance Level
The significance level is $\alpha = 0.05$.
Critical Value
As the alternative hypothesis is $\text{two-tailed}$, the critical value of $t$ for $15$ degrees of freedom $\text{are}$ $-2.131 and 2.131$.
The rejection region (i.e. critical region) is $\text{t < -2.131 or t > 2.131}$.
Computation
The test statistic under the null hypothesis is
$$ \begin{aligned} t&=\frac{ \overline{x} -\mu_0}{s/\sqrt{n}}\\ &= \frac{31.17-30}{5.51/ \sqrt{16 }}\\ &= 0.849 \end{aligned} $$
Decision
Traditional approach:
The test statistic is $t =0.849$ which falls $\text{outside}$ the critical region, we $\text{fail to reject}$ the null hypothesis.
$p$-value Approach
This is a $\text{two-tailed}$ test, so the p-value is the
area to the left of the test statistic ($t=0.849$) is p-value = $0.409$.
The p-value is $0.409$ which is $\text{greater than}$ the significance level of $\alpha = 0.05$, we $\text{fail to reject}$ the null hypothesis.
There no enough evidence to support her claim.
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
