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6b.1 - Steps in Conducting a Hypothesis Exam for \(\mu\)

Six Steps for Conducting a One-Sample Mean Hypothesis Exam

Steps 1-3 Section

Allow's apply the full general steps for hypothesis testing to the specific instance of testing a ane-sample mean.

Pace 1: Set upwardly the hypotheses and check atmospheric condition.

One Hateful t-test Hypotheses

Left-Tailed

\( H_0\colon \mu=\mu_0 \)
\( H_a\colon \mu<\mu_0\)

Right-Tailed

\( H_0\colon \mu=\mu_0 \)
\( H_a\colon \mu>\mu_0 \)

Two-Tailed

\( H_0: \mu=\mu_0 \)
\( H_a: \mu\ne \mu_0 \)

Weather: The data comes from an approximately normal distribution or the sample size is at least xxx.


Step two: Determine on the significance level, \(\alpha \).
Typically, five%. If \(\alpha\) is non specified, apply 5%.

Step 3: Calculate the test statistic.

Ane Mean t-test: \( t^*=\dfrac{\bar{x}-\mu_0}{\frac{due south}{\sqrt{due north}}} \)

The get-go few steps (Stride 1 - Step 3) are exactly the same as the rejection region or the p-value arroyo. The next part volition discuss steps 4 - 6 for both approaches.

Rejection Region Approach

Steps iv-vi Section

Step 4: Observe the advisable critical values for the tests. Write down clearly the rejection region for the problem.
𝛼 t 𝛼
Normal curve with a left tailed test shaded.

Decline \(H_0\) if \(t^* \le t_\alpha\)

𝛼 ane-𝛼 t 1-𝛼
Normal curve with a right tailed exam shaded.

Turn down \(H_0\) if \(t^* \ge t_{i-\blastoff}\)

𝛼/2 1-𝛼 t 1-𝛼/two 𝛼/2 t 𝛼/2
Normal curve with a 2-tailed test shaded

Decline \(H_0\) if \(|t^*| \ge |t_{\alpha/2}|\)


Step 5: Make a decision about the nada hypothesis.
Cheque to meet if the value of the test statistic falls in the rejection region. If it does, then turn down \(H_0 \) (and conclude \(H_a \)). If it does not fall in the rejection region, practice non reject \(H_0 \).

Step half-dozen: Land an overall decision.

P-Value Approach

Steps iv-6 Section

Stride iv: Compute the advisable p-value based on our alternative hypothesis.
  • If \(H_a \) is right-tailed, then the p-value is the probability the sample information produces a value equal to or greater than the observed examination statistic.
  • If \(H_a \) is left-tailed, then the p-value is the probability the sample information produces a value equal to or less than the observed test statistic.
  • If \(H_a \) is 2-tailed, then the p-value is two times the probability the sample data produces a value equal to or greater than the absolute value of the observed examination statistic.

Left-Tailed

\(P(t \le t^*)\)

Right-Tailed

\(P(t\ge t^*)\)

Two-Tailed

\(2\) x \(P(t \ge |t^*|)\)
Step v: Make a determination about the null hypothesis.
If the p-value is less than the significance level, \(\alpha\), then refuse \(H_0\) (and conclude \(H_a \)). If it is greater than the significance level, then do not refuse \(H_0 \).
Stride vi: Country an overall conclusion.

Example half-dozen-7 Length of Lumber Section

Continuing with our lumber example, the mean length of the lumber is supposed to be 8.5 feet. A architect wants to bank check whether the shipment of lumber she receives has a mean length different from viii.five feet. If the architect observes that the sample mean of 61 pieces of lumber is 8.three anxiety with a sample standard deviation of 1.2 feet, what will she conclude? Bear this test at a 1% level of significance.

Conduct the examination using the Rejection Region approach and the p-value approach.

Answer

Step i: Fix the hypotheses and check conditions.

Fix the hypotheses (since the research hypothesis is to check whether the mean is different from eight.5, we set information technology upwardly as a two-tailed exam):

\( H_0\colon \mu=8.5 \) vs. \(H_a\colon \mu\ne 8.5 \)

Can we use the t-exam? The answer is aye since the sample size of 61 is sufficiently big (greater than 30).

Step 2: Decide on the significance level, \(\blastoff \).
According to the question, \(\blastoff = 0.01 \).
Stride 3: Summate the test statistic.
\brainstorm{align}  t^*&=\dfrac{\bar{x}-\mu_0}{\frac{s}{\sqrt{northward}}}\\&=\dfrac{viii.iii-8.5}{\frac{1.2}{\sqrt{61}}}\\&=-1.3 \end{align}

Steps iv-6

Pace 4: Discover the appropriate disquisitional values for the tests. Write downwardly clearly the rejection region for the problem.
From the table and with degrees of freedom of 61-1=60, the critical value is \(t_{\alpha/ii}=t_{0.005}=2.660 \). The rejection region for the ii-tailed exam is given by:

\( t^*\le -2.660 \) or \(t^*\ge two.660 \)

*Recall how to apply to Minitab or t-table to find the t percentiles (Lesson 5.iv)
Step v: Brand a decision about the null hypothesis.
The observed t-value, or test statistic, is -one.3. Since \(t^* \) does not fall inside the rejection region, we fail to reject \(H_0 \).
Step six: Land an overall decision.
With a examination statistic of -1.3 and critical value of ± 2.660 at a one% level of significance, we do not have enough statistical testify to reject the null hypothesis. We conclude that there is not enough statistical evidence that indicates that the mean length of lumber differs from eight.5 feet.
Stride iv: Compute the appropriate p-value based on our culling hypothesis.
\brainstorm{align} \text{p-value}&=2P(T>|t^*|)\\&=2P\left(T>\left|\frac{\bar{x}-\mu_0}{\frac{s}{\sqrt{n}}}\right|\right)\\&=2P\left(T>\left|\frac{8.iii-8.5}{\frac{ane.ii}{\sqrt{61}}}\correct|\correct)\\&=2P(T>|-1.three|)\\&=2P(T>i.3) \terminate{align}.
From the t-table going across the row for 60 degrees of liberty, we exercise non find a value equal to 1.iii. Without software to find a more exact probability, the best we can practice from the t-tabular array is detect a range. We do see that the value falls betwixt 1.296 and 1.671. These 2 t-values represent to correct-tail probabilities of 0.ane and 0.05, respectively. Since 1.3 is between these 2 t-values, then it stands to reason that the probability to the right of ane.3 would fall between 0.05 and 0.i. Therefore, the p-value would be = 2×(0.05 and 0.i) or from 0.1 to 0.2.
Step 5: Brand a decision most the zip hypothesis.
With this range of possible p-values exceeding our 1% level of significance for the test, nosotros fail to reject the null hypothesis.
Step half dozen: State an overall determination.
With a test statistic of - 1.3 and p-value between 0.i to 0.2, we neglect to refuse the null hypothesis at a 1% level of significance since the p-value would exceed our significance level. We conclude that there is not enough statistical evidence that indicates that the mean length of lumber differs from 8.five feet.

Try it!

Emergency Room Wait Fourth dimension Section

Waiting room

The ambassador at your local infirmary states that on weekends the average wait time for emergency room visits is ten minutes. Based on discussions you have had with friends who have complained on how long they waited to be seen in the ER over a weekend, you dispute the administrator's claim. You decide to test your hypothesis. Over the course of a few weekends, you record the expect time for twoscore randomly selected patients. The boilerplate look time for these forty patients is 11 minutes with a standard deviation of 3 minutes.

Practise yous have enough testify to back up your hypothesis that the boilerplate ER wait time exceeds x minutes? You lot opt to bear the test at a 5% level of significance.

Stride 1: Set up the hypotheses and bank check atmospheric condition.

At this point we want to check whether nosotros tin can apply the key limit theorem. The sample size is greater than 30, so we should be okay.

This is a correct-tailed test.

\( H_0\colon \mu=x \) vs \(H_a\colon \mu>10 \)

Step 2: Make up one's mind on the significance level, \(\alpha \).
The problem states that \(\alpha=0.05 \).
Stride 3: Calculate the exam statistic.
\begin{align} t^*&=\dfrac{\bar{x}-\mu_0}{\frac{southward}{\sqrt{northward}}}\\&=\dfrac{eleven-x}{\frac{3}{\sqrt{40}}}\\&=2.11 \end{align}

Steps 4-half dozen

Footstep 4: Find the advisable critical values for the tests. Write down clearly the rejection region for the trouble.
The degrees of liberty for this exam are \(north-1=40-1=39 \). The alternative is right-tailed. Therefore, nosotros want to find the value, \(t_{0.05} \), such that \(P(T\ge t_{0.05})=0.05 \).

Using the table from the text, it shows 35 and 40 degrees of freedom. We would use 35 degrees of freedom. With \(\alpha=0.05 \) , nosotros see a value of ane.69. The critical value is 1.69 and the rejection region is any \(t^* \) such that \(t^*\ge ane.69 \) .

Note! If we used software (discussed in the next section), nosotros will find the critical value to be 1.685.

Step five: Make a decision about the nil hypothesis.
Our exam statistic, two.xi, is greater than our disquisitional value of 1.69 and therefore is in the rejection region. We would reject the null hypothesis.
Stride 6. Country the decision in words.
There is plenty evidence, at a significance level of 5%, to decline the goose egg hypothesis and conclude that the mean waiting fourth dimension is greater than ten minutes.
Step 4: Compute the appropriate p-value based on our alternative hypothesis.
Again, using the table with 35 degrees of liberty, our test statistic is 2.11 and is betwixt 2.030 and two.438. This corresponds to a p-value between 0.01 and 0.025.

Notation! If nosotros use software, the p-value is 0.0207.


Step 5: Brand a decision nearly the aught hypothesis.
Since our p-value is between 0.01 and 0.025, nosotros know it is less than our significance level, 5%. Therefore, we turn down the zippo hypothesis.

Step 6: State an overall conclusion.
In that location is enough evidence, at a significance level of 5%, to refuse the null hypothesis and conclude that the mean waiting fourth dimension is greater than 10 minutes.

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Source: https://online.stat.psu.edu/stat500/lesson/6b/6b.1

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