One Sample Test

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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In this post, we will learn about “Testing a claim about a Mean” using a Large sample. Before going to the main topic, we need to understand some related basics.

Hypothesis Testing

When a hypothesis test involves a claim about a population parameter (in our case mean/average), we draw a representative sample from the target population and compute the sample mean to test the claim about population. If the sample drawn is large enough ($n\ge 30$), then the Central Limit Theorem (CLT) applies, and the distribution of the sample mean is assumed to be approximately normal, that is we have $\mu_{\overline{x}} = \mu$ and $\sigma_{\overline{x}} = \frac{\sigma}{\sqrt{n}} \approx \frac{s}{\sqrt{c}}$.

Testing a Claim about a Mean

It is worth noting that $s$ and $n$ are known from the sample data, and we have a good estimate of $\sigma_{\overline{x}}$ but the population mean $\mu$ is not known to us. The $\mu$ is the parameter that we are testing a claim about a mean. To have a value for $\mu$, we will always assume that the null hypothesis is true in any hypothesis test.

It is also worth noting that the null hypothesis must be of one of the following types:

• $H_0:\mu = \mu_o$
• $H_0:\mu \ge \mu_0$
• $H_0:\mu \le \mu_0$

where $\mu_0$ is a constant, and we will always assume that the purpose of our test is that $\mu=mu_0$.

Standardized Test Statistic

To determine whether to reject or not reject the null hypothesis, we have two methods namely (i) a standardized value and (ii) a p-value. In both cases, it will be more convenient to convert the sample mean $\overline{x}$ to a Z-score called the standardized test statistic/score.

Since, we assumed that $\mu=\mu_0$, and we have $\mu_{\overline{x}} =\mu_0$, then the standardized statistic is:

$$Z = \frac{\overline{x} – \mu _{\overline{x}}} {\sigma_{\overline{x}} } = \frac{\overline{x} – \mu _{\overline{x}}} {\frac{s}{\sqrt{n}} }$$

As long as $\mu=\mu_0$ is assumed, the distribution standardized test statistics $Z$ is Standard Normal Distribution.

Example: Testing a Claim about an Average/ Mean

Suppose the average body temperature of a healthy person is less than the commonly accepted temperature of $98.6^{o}F$. Assume that a sample of 60 healthy persons is drawn. The average temperature of these 60 persons is $\overline{x}=98.2^oF$ and the sample standard deviation is $s=1.1^oF$.

The hypothesis of the above statement/claim would be

$H_0:\mu\ge 98.6$
$H_1:\mu < 98.6$

Note that from the alternative hypothesis, we have a left-tailed test with $\mu_0=98.6$.

Based on our sample data, the standardized test statistic is

\begin{align*}
Z &= \frac{\overline{x} – \mu _{\overline{x} } } {\frac{s}{\sqrt{n} } }\\
&=\frac{98.2 – 98.6}{\frac{1.1}{\sqrt{60}}} \approx -2.82
\end{align*}

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