Hypothesis Testing with 𝒕 using the p-Value method and TI-84 This is a copy of the other presentation, which uses the Classical, or Critical Value Method

Example 1 Acknowledgment: This problem was taken from “Elementary Statistics”, 10th edition, by Johnson & Kuby, p. 446ff. © 2007 Thomson Corporation. A commerical aircraft manufacturer buys rivets to use in assembling airliners. It requires that the mean shearing strength of the rivets must be at least 925 lbs. (μ≥925).

Example 1, continued We collect and test a sample of 𝑛=50 rivets and obtain 𝑥 =921.18 lb and 𝒔=𝟏𝟖 lb. Assume that we know from previous studies that the standard deviation of the shearing strength of rivets is 𝜎=18 lb. REQUIRED: Determine whether we have enough evidence, at the 𝛼=.05 level of significance, to conclude that we should reject this batch of rivets?

Example 1 initial direction We will use the 𝒕 Distribution and the 𝒕 values to do this test. Justification for that: We don’t know the population’s standard deviation And we have a “large” sample size: 𝑛≥30. If we had a “small” sample size, this method would also require that we know the rivet strength population is normally distributed.

Step 1. State the hypotheses The null hypothesis says that the rivets are ok: 𝐻 0 :𝜇≥925 The alternative hypothesis is a claim that the rivets are too weak: 𝐻 1 :𝜇<925

Step 2. Compute the Test Statistic Step 3. Compute its p-Value TI-84 Inputs Explanation of inputs 𝜇 0 from the null hypothesis 𝑥 ,𝑠,𝑛 from sample The Alternative Hypothesis Highlight Calculate and press ENTER.

Step 2: Compute the Test Value Step 3: Compute its p-Value TI-84 Inputs STAT, TESTS, 2:Ttest “Data” if data in L List Otherwise “Stats” 𝜇 0 from null hypothesis 𝜎 known already 𝑥 and 𝑛 and 𝒔 from our sample The claim of the Alternative Hypothesis

Step 2: Compute the Test Value Step 3: Compute its p-Value TI-84 Outputs From the “Calculate” It reminds us of the Alternative Hypothesis. 𝒕 = the test value, not needed, but interesting p = the p value of that test value. It repeats the sample data.

Step 4. Make a Decision Rules for a One-Tailed Test In this particular problem If the p value is less than the level of significance That is, if 𝑝<𝛼, then the decision is “Reject the Null Hypothesis” Otherwise, the decision is “Fail to reject the Null Hypothesis” Our p value is 0.0699 It is NOT < the 𝛼=.05 We “Fail to Reject” the null hypothesis.

A remark about our decision The rivets we tested in our sample had a lower shearing strength than advertised. But not SIGNIFICANTLY lower at the 𝛼=.05 level of significance. So it might just be the natural ups and downs of sampling. Lower, but not significantly so. We don’t have enough evidence to say “this is a bad lot”.

Step 5. Plain English conclusion The conclusion has to be suitable for a general audience. They don’t want to hear any Statistics lingo. Say something that a journalism school major could read in a news report. Here’s what we can say: “There is NOT enough evidence to conclude that these rivets are SIGNIFICANTLY weaker than the required strength.”

Example 2 Acknowledgment: This problem was adapted from “Elementary Statistics”, 10th edition, by Johnson & Kuby, p. 446ff. © 2007 Thomson Corporation. Suppose the statewide paramedic exam has an average score of 79.68 with a standard deviation unreported of 9.06. If 40 Darton State students took the exam and their mean score was 83.15 and standard deviation was 9.06, can we claim at the 𝛼=.01 level of significance that our students score higher than the rest of the state?

Example 2 remarks We scored higher, that’s for sure. 83.15 vs. 79.68 statewide. But we have to be careful before issuing a press release or using these results as a recruiting tool We want the Central Limit Theorem to tell us that these results are too good to be mere coincidence.

Example 2 initial direction We will use the Standard Normal Distribution and the z values to do this test. Justification for that: We know the population’s standard deviation And we have a “large” sample size: 𝑛≥30. If we had a “small” sample size, this method would also require that we know the rivet strength population is normally distributed.

Step 1. State the hypotheses The null hypothesis says that our EMT students are not significantly better when compared to the rest of the state: 𝐻 0 :𝜇≤79.68 The alternative hypothesis is a claim that our students performed extraordinarily well: 𝐻 1 :𝜇>79.68

Step 2. Compute the Test Value. Step 3. Compute its p Value

Step 4. Make a Decision Rules for a One-Tailed Test In this particular problem If the p value is less than the level of significance That is, if 𝑝<𝛼, then the decision is “Reject the Null Hypothesis” Otherwise, the decision is “Fail to reject the Null Hypothesis” Our p value is 0.010043305 It IS NOT < the 𝛼=.01 We “Fail to Reject the null hypothesis.”

Remarks about our decision We came very close to the 𝛼=.01 standard But not quite So we cannot reject the null hypothesis. This turned out differently from the z version of this problem, where we knew the population’s standard deviation.

Step 5. Plain English conclusion The conclusion has to be suitable for a general audience. They don’t want to hear any Statistics lingo. Say something that a journalism school major could read in a news report. Here’s what we can say: “Darton State College EMT students scored significantly higher than the statewide average in a recent examination.”

Example 3 Acknowledgment: This problem was adapted from “Elementary Statistics”, 10th edition, by Johnson & Kuby, p. 459. © 2007 Thomson Corporation. Someone claims that the average age of the three million horse racing fans in 55 years. We want to see if this claim is true here at the local track, at the 0.05 level of significance. We sample 35 patrons and finds the average age is 52.7 years . If we assume 𝜎=8 and 𝑠=8 years, what can we conclude?

Example 3 initial direction We will use the Standard Normal 𝑡 Distribution and the z t values to do this test. Justification for that: We don’t know the population’s standard deviation And we have a “large” sample size: 𝑛≥30. If we had a “small” sample size, this method would also require that we know the rivet strength population is normally distributed.

Step 1. State the hypotheses The null hypothesis says that the mean age is 55 years old: 𝐻 0 :𝜇=55 The alternative hypothesis is a claim that the mean age is not 55 years old, it’s different: 𝐻 1 :𝜇≠55

Step 2. Determine the Test Value. Step 3. Determine its p value.

Step 4. Make a Decision Rules for a Two-Tailed Test In this particular problem If 𝑝<𝛼, then the decision is “Reject the Null Hypoth.” Otherwise, the decision is “Fail to reject the Null Hypothesis” SAME AS ONE-TAILED TEST – the TI-84 quietly doubled the p value so we could just compare to alpha !!!! Our p value is 0.098 It is NOT < the 𝛼=.05 We “Fail to Reject the null hypothesis.”

Remarks about our decision The racing fans at our track were certainly younger than the supposed average age of 55. But it wasn’t strong enough evidence. So we let the null hypothesis stand. We did NOT “prove” the null hypothesis. We merely collected evidence that mildly disagreed with the null hypothesis.

Step 5. Plain English conclusion The conclusion has to be suitable for a general audience. They don’t want to hear any Statistics lingo. Say something that a journalism school major could read in a news report. Here’s what we can say: “We can’t disagree that the average age of a horse racing fan really is 55 years old, despite a little bit of evidence to the contrary.”