Lesson 9: Hypothesis Test for One Mean (Sigma Known)

Design

We assume that at the time of the accident, the stability requirements were based on the Coast Guard criteria of a mean of 140 pounds per person. So, the Ethan Allen was supposed to be able to safely transport passengers and crew with a mean weight of 140 pounds. We want to investigate if 140 pounds is a reasonable value for the mean weight of tour boat passengers. This leads to the research question: “Is the mean weight of tour boat passengers greater than 140 pounds?”

We can rewrite the research question in a declarative sentence to obtain a hypothesis, or a testable statement about a population.

The first hypothesis we will write is that the Coast Guard criteria is appropriate: “The true mean weight of tour boat passengers is 140 pounds.” We call this the null hypothesis and label it as \(H_0\). The ‘’’null hypothesis’’’ is a statement of the “status quo”, or the value typically considered to be appropriate. The null hypothesis is expressed with a statement involving equality (\(=\)).

\(H_0:~~\mu = 140\) pounds

In contrast to the null hypothesis, we write the alternative hypothesis, denoted by \(H_a\). This is typically the statement that a researcher suspects is the actual truth and is often called the research hypothesis. In our case, we suspect that “The mean weight of tour boat passengers is greater than 140 pounds.”

\(H_a:~~\mu > 140\) pounds

In every hypothesis test we will perform this semester, the null hypothesis will be a statement involving equality. The alternative hypothesis can include greater than (\(>\)), less than (\(<\)), or not equal (\(\ne\)).

When we test hypotheses, we assume the null hypothesis is true. Because of this requirement, whenever we need to use \(\mu\) in a calculation, we can use the value specified in the null hypothesis. When we conduct a hypothesis test, we gather evidence against the null hypothesis, which we have assumed to be true. If we get enough evidence against the null hypothesis, we reject it and conclude it was false. If we do not have sufficient evidence against the null hypothesis, we do not reject it, i.e., we “fail to reject it.”

In summary, we {{Objective|state the null and alternative hypotheses|hyptwofirst|#Lesson_Outcomes}} for the true mean weight \(\mu\) of tour boat passengers to be

\(H_0:~~\mu = 140\) pounds

\(H_a:~~\mu > 140\) pounds

We also establish the [[Significance_Level|significance level]], \(\alpha\), to be \(\alpha = 0.05\).

Collect

How do we gather evidence against a null hypothesis? We collect data.

The marine accident report gives the weight (in pounds) of each of the passengers and the crew member. These values are reproduced below.

Ideally, we would have hoped for a simple random sample of all tour boat passengers. However, we will assume that this convenience sample is representative of the population of all tour boat passengers.

Describe

To help us understand the the data, we first summarize the data by {{Objective|creating an appropriate graphical summary which will support the hypothesis test|hypthreefirst|#Lesson_Outcomes}} that we will use to Make Inference in Step 4 of the Statistical Process.

Next, we {{Objective|compute appropriate numerical summaries that support the hypothesis test|hypthreesecond|#Lesson_Outcomes}} that we will use to Make Inference in the next step. The sample size is \(n=48\), and the sample mean is \(\bar x=177.6\) pounds. According to the CDC, the standard deviation of the weights of individuals in the United States is $ =26.7$ pounds.

Considering the data as a random sample of all possible tour boat passengers, it appears that the true mean weight of tour boat passengers might be greater than 140 pounds. However, we need to check this with a formal test of our hypotheses.

Make Inference

It is not sufficient to gain an intuitive sense for the data. We will use a statistical test to see if there is sufficient evidence to reject the null hypothesis that the true mean weight of tour boat passengers is 140 pounds.

An appropriate hypothesis test for these data is the one sample \(\sigma\) known hypothesis test. This is because the three {{Objective|requirements for the one mean \(\sigma\) known hypothesis test|hypfourfirst|#Lesson_Outcomes}} can be assumed to be met as (1) the data can be assumed to be a representative sample from the population of interest, (2) the sample size \(n=48\) is larger than 30 so the sampling distribution of \(\bar{x}\) can be assumed to be normal, and (3) \(\sigma=26.7\) pounds is known.

Assuming the null hypothesis is true (\(\mu=140\)), what is the probability that we would observe a sample mean (\(\bar{x}\)) as extreme or more extreme than the value of \(177.6\) we observed? This probability is called the \(P\)-value.

To find the \(P\)-value for the one sample \(\sigma\) known hypothesis test, we first calculate the number of standard deviations that the sample mean (\(\bar x = 177.6\)) is away from the assumed value of true mean (\(\mu=140\) pounds). This is our \(z\)-score. Then we use the applet to determine the probability of observing a value of \(z\) that is as large or larger than the value we observed. (This is the same as the probability of observing a value of $ x =177.6$ or more pounds, given that the true mean really is $ = 140$ pounds.)

We {{Objective|calculate the test-statistic (\(z\)) of the hypothesis test|hypfivefirst|#Lesson_Outcomes}} by computing

\[ \displaystyle{ z= \frac{ \bar x − \mu}{\sigma/\sqrt{n}} = \frac{177.6−140}{26.7/\sqrt{48}}=9.757 } \]

We {{Objective|calculate the p-value of the hypothesis test|hypfivesecond|#Lesson_Outcomes}} using the [http://byuimath.com/apps/normprob.html Normal Probability Applet] to determine the probability of observing a value of \(z\) that is as large or larger than \(z = 9.757\). Note that the alternative hypothesis (\(H_a: \mu > 140\)) specifies that we shade to the right of \(z\). While the following plot does this, we cannot actually see the shaded area on the right because the area is so small it is essentially zero!

{{NormalProbabilityApp|z=9.757|left=0|mid=0|right=1|width=500|height=325|fheight=350|static=1}}

The normal probability applet gives the area as “8.6086e-23.” Remember, this is scientific notation for the number 0.000000000000000000000086086! It is very unlikely that the mean of $ x =177.6$ pounds was observed just by chance.

We {{Objective|assess the statistical significance by comparing the p-value to the \(\alpha\)-level|hypsixfirst|#Lesson_Outcomes}}. Since the p-value (8.6086e-23) is certainly less than \(\alpha=0.05\), the evidence against \(H_0\) is strong enough to reject \(H_0\) and conclude \(H_a\). In other words, we {{Objective | draw the conclusion from this hypothesis test|hypsevenfirst|#Lesson_Outcomes}} that if every seat on the Ethan Allen was occupied, the boat would be carrying a greater load than it was certified to handle.

Take Action

As a result of this accident, the United States government took several actions. The Coast Guard stability regulations were changed, and the assumed average weight per person was increased to 185 pounds. As a result, the safety of public vessels has been improved.

Note

While this change in regulation was most likely the correct decision, it is important to keep in mind that whenever the null hypothesis is rejected there is always a small possibility that a Type I Error has been committed. A Type I Error is defined as rejecting the null hypothesis (\(H_0: \mu = 140\) in this case) when it was actually true. The {{Objective|interpretation of a Type I Error|eightfirst|#Lesson_Outcomes}} for this study would be that the assumed weight per person was incorrectly increased to 185 pounds and that in truth it should have been kept at 140 pounds (the null hypothesis). However, since we controlled the probability of a Type I Error at \(\alpha=0.05\), we are very confident that we have made a correct decision and not a Type I Error.