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Section 11.4 Hypothesis Test for one mean

In this section, you will consider the following options for null hypothesis and corresponding alternate hypothesis with respect to the unknown value \(\mu\text{:}\)

\begin{equation*} H_0 : \mu = \mu_0 \\ H_a: \mu \neq \mu_0 \end{equation*}

or

\begin{equation*} H_0 : \mu \le \mu_0 \\ H_a: \mu \gt \mu_0 \end{equation*}

or

\begin{equation*} H_0 : \mu \ge \mu_0 \\ H_a:\mu \lt \mu_0 \end{equation*}

Again, we choose only one of these three options and as before the first is called a "two-tailed" test and the last two are called "one-tailed" tests. Note that some people will write all of these null hypothesis options using equality but the alternate hypothesis determines the number of tails.

Once again, if the test sample size is sufficiently large and the standard deviation \(\sigma\) of the underlying distribution is known, one can use the normal distribution to compute probabilities. If the test sample size is relatively small or if \(\sigma\) is not known (and therefore is approximated by the sample standard deviation s), then the t-distribution can be utilized to compute probabilities. We will only consider using the t-distribution to compute p-values.

The Standard Error \(\sigma_e\) is given by

\begin{equation*} \sigma_e = \frac{s}{\sqrt{n}} \end{equation*}

where s is the standard deviation of the sample. For this presentation, we will assume that the actual population is relatively large relative to the sample size. In cases where this is not true, an adjustment (not presented here) will need to be made when computing \(\sigma_e\text{.}\)

To determine the p-value for a given sample with sample mean \(\overline{x}\text{,}\)

\begin{equation*} t = \frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}} \end{equation*}

is a t-variable with n-1 degrees of freedom. Therefore, probabilities on t can be computed using the t-distribution.

Consider a Two-tailed Hypothesis test for \(\mu\) using

\begin{equation*} H_0 : \mu = 200 \\ H_a: \mu \neq 200 \end{equation*}

using a sample of size n = 49 and with a resulting mean of \(\overline{x} = 206\text{,}\) a sample standard deviation of \(s = 15\text{,}\) and a significance level of \(\alpha = 0.01\text{.}\)

The standard error for this test is

\begin{equation*} \sigma_e = \frac{15}{\sqrt{49}} = \frac{15}{7} \end{equation*}

and so using the t-distribution with degrees of freedom n-1 = 48 yields a t-statistic of

\begin{equation*} t = \frac{\overline{x} - 200}{\sigma_e} = \frac{206-200}{\frac{15}{7}} = \frac{14}{5} = 2.80. \end{equation*}

To compute the p-value,

\begin{equation*} P(t \gt 2.80) + P(t \lt 2.80) \approx 0.0037 + 0.0037 = 0.0074. \end{equation*}

Since this p-value is less than our significance level \(\alpha = 0.01\) then you can reject the null hypothesis and accept the alternate.

Checkpoint 11.4.1.
Checkpoint 11.4.2. WebWorK - Another Two Tailed Test.

Consider One-tailed Hypothesis test for \(\mu\) using an interesting application:

Suppose that a manufacturer bottling a delicious beverage and the label indicates that the bottle contains 16 fluid ounces. Since providing the customer too little product might cause a significant negative reaction relative to the modest additional cost of providing a little too much, consider the following hypothesis test:

\begin{equation*} H_0 : \mu = 16 \\ H_a: \mu \gt 200. \end{equation*}

To test this hypothesis at significance level \(\alpha = 0.05\text{,}\) you randomly pull out 20 bottles from the production line and accurately measure the amount of produce in each bottle. If the resulting average of these measurements is \(\overline{x} = 16.05\) ounces with a standard deviation of \(s = 0.08\) ounces, determine if you can safely make it known that more product is actually delivered in general to each consumer.

The standard error for this test is

\begin{equation*} \sigma_e = \frac{0.08}{\sqrt{20}} \approx 0.01789 \end{equation*}

and using the t-distribution with n-1 = 19 degrees of freedom yields a t-statistic of

\begin{equation*} t = \frac{\overline{x} - 16}{\sigma_e} = \frac{16.05-16}{0.01789} \approx 2.795. \end{equation*}

To compute the p-value,

\begin{equation*} P(t \gt 2.795) \approx 0.0058. \end{equation*}

Since this p-value is less than our significance level \(\alpha = 0.05\) (by a lot) then you can reject the null hypothesis and accept the alternate. It is safe therefore to say that customers can expect at least 16 ounces! However, note that some folks will still be stiffed since the standard deviation of 0.08 certainly means that some bottles have less than 16.05-0.08 ounces of beverage.

Checkpoint 11.4.3. WebWorK - one tailed test.