Introduction: One Sample Hypothesis Test
In this post, I will discuss One Sample Hypothesis Test (One Sample t-test). When testing a claim about the mean using sample data with a small number of observations (i.e., sample size), the appropriate t-distribution instead of the standard normal distribution should be used to determine the standardized test statistic, critical values, rejection region, and p-values.
Recall that if the sample of values drawn follows the normal distribution, the sample size (number of observations in the sample) is less than 30, and the population standard deviation is unknown, then the random variable
$$t=\frac{\overline{x} – \mu}{\frac{s}{\sqrt{n}}}$$
has the Student’s t-distribution with $n-1$ degrees of freedom.
The procedure of locating the rejection regions for a t-distribution hypotheses test is the same as for the normal distribution tests, however, the critical values will differ. To find the critical value(s) $t_0$ for a test, determine if the test is one-tailed or two-tailed and the significance level ($\alpha$). The critical values can be found in the t-distribution table by looking up the entry in the column giving the level of significance and the row showing the degrees of freedom.
Note that:
- For a right-tailed test, $t_0$ is the positive value in the table
- For a left-tailed test, $t_0$ is the negative of the value in the table
- For a two-tailed test, there are two critical values $t_0$ both the value and its opposite.
Assumptions of the One Sample Hypothesis Test (t-test)
- Independence: Observations in the sample should be independent of each other.
- Normality: The population from which the sample is drawn should be normally distributed. However, the t-test is relatively robust to violations of normality, especially for larger sample sizes.
- Random Sampling: The sample should be a random sample from the population.
One Sample Hypothesis Test for Mean
Example 1: SAT Math scores are normally distributed. A sample of SAT Math scores for 16 students has an average score of 522.8 with a sample standard deviation of 154.5. Suppose, one wishes to support the claim that the average SAT Math score exceeds 500 using a level of significance of 0.05.
Solution
Step 1: The null and alternative hypotheses test in this case are
$H_0: \mu \le 500$ vs $H_1: \mu > 500$
From the alternative hypothesis, the test is right-hand-tailed with $\mu_0=500$.
Step 2: Level of Significance is 5% = 0.05
Step 3: Critical Value
Using the t-distribution table with one Tail, 5% level of significance, and $n-1=16-1=15$ degrees of freedom, the critical value is $t_0=1.753$. The rejection region is thus $t\ge 1.7535$.
Step 4: Test Statistics
The standardized test statistic is
\begin{align*}
t&=\frac{\overline{X} – \mu_0 }{\frac{s}{\sqrt{n}}}\\
&= \frac{522.8 – 500}{\frac{154.5}{\sqrt{16}}}\\
&= \frac{22.8}{38.625} = 0.59
\end{align*}
Step 5: Interpretation of One Sample Mean Test
Since the standardized test statistic is not in the region of rejection, therefore, one should not reject $H_0$ and so the sample data is not sufficient to support the claim that the average exceeds 500 at the 0.05 level of significance.
Example 2: A biologist measures the weights of anesthetized female grizzly bears during winter. A sample of 14 bears is found to have an average weight of $\overline{X} = 376.6$lbs. with a sample standard deviation of $s=32.5$lbs. Is there sufficient evidence to support the claim that the average weight of all female bears in the area is less than 400 lbs? Use $\alpha=0.01$ level of significance.
Solution:
Step 1: The null and alternative hypotheses in this case
$H_0:\mu \ge 400$ vs $H_1:\mu < 400$
From the alternative hypothesis, the test is left-hand-tailed with $\mu_0=400$.
Step 2: Level of Significance is 1% = 0.01
Step 3: Critical Value
Using the Student’s t-distribution table with one tail, the level of significance $\alpha = 0.01$, and $n-1=14-1=13$ degrees of freedom, the critical value $t_0=-2.650$. The region of rejection is thus $t\le -2.650$.
Step 4: The Test Statistics is
\begin{align*}
t&=\frac{\overline{X} – \mu_0 }{\frac{s}{\sqrt{n}}}\\
&= \frac{376.6 – 400}{\frac{32.5}{\sqrt{14}}}\\
&= \frac{-23.4}{8.686} = -2.694
\end{align*}
Since the standardized test statistic is in the rejection region, one should reject the null hypothesis ($H_0$), which supports the claim that the average bear weight is less than 400 lbs.
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