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.
Table of Contents
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*}
