Hypothesis Testing with t Using the Normal Distribution t Distribution in a Hypothesis Test

Hypothesis Testing with t

Using the Normal Distributiont Distribution

in a Hypothesis Test

When and Why t Distribution & tinstead of Normal Distribution & z

Normal Distribution and z• When is known (the

population standard deviation)

• And you have sample size (or if , you know that the population is normally distributed)

t Distribution and t• When you don’t know and

therefore must use the sample standard deviation, instead.

• Again, should have sample size (or a normally-distributed population.)

• t builds in the extra uncertainty that comes with not knowning

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. ().

Example 1, continued

• We collect and test a sample of rivets and obtain lb.

• We don’t know the population’s but our sample has standard deviation lb.

• REQUIRED: Determine whether we have enough evidence, at the 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: .– 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:

• The alternative hypothesis is a claim that the rivets are too weak:

Step 2. Determine the Critical Value

• Left tail, Right tail, Two tail? Look at the direction of the null hypothesis. We have a LEFT-TAIL hypothesis test.

• Refer back to the . In this problem,


• Draw a picture of theStandard Normal Curve t Distribution for this situation: Left Tail, .


• The shaded part is called the CRITICAL REGION.

• Its area is • The rest of the area is

. Draw it!• What value is at the

boundary? Details next slide.• THE CRITICAL VALUE is


• Finding critical value:• Degrees of Freedom

• Or use TI-84 invT(.05,49) (unavailable on TI-83/Plus)

d.f. One tail, α α = 0.0545 1.67950 1.676

About that t table…

• Degrees of Freedom

• Use symmetry: left tailvalue is same as right, but with a negative sign.

• The table doesn’t have a row for df = 49.– Take the more conservative df = 45, t = -1.679– Or “interpolate”

d.f. One tail, α α = 0.0545 1.67950 1.676

Step 3. Compute the Test Statistic

Formula In this example,

The Test Statistic is

Step 4. Make a Decision

• If your Test Value is inside the CriticalRegion, then REJECT the null hypothesis.

• If your Test Value is outside the CriticalRegion, then “FAIL TO REJECT” the H0.

• Here, we “FAIL TO REJECT”.

A remark about z vs. t

• When we did this as a z problem, the critical z value was -1.651.

• When we did this as a t problem, the critical t value was -1.677.

• Other than that, the procedure was exactly the same.

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 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 unknown of 9.06. If 40 Darton State students took the exam and their mean score was 83.15 and their scores’ , can we claim at the 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.


• We will use the Distribution and the t values to do this test.


deviation– And we have a “large” sample size: .– If we had a “small” sample size, this method



• The null hypothesis says that our EMT students are not significantly better when compared to the rest of the state:

• The alternative hypothesis is a claim that our students performed extraordinarily well:


• Left tail, Right tail, Two tail? Look at the direction of the null hypothesis. We have a RIGHT-TAIL hypothesis test.

• Because our Alternative Hypothesis is making a claim of a mean that’s bigger than the mean stated in the Null Hypothesis.

• Refer back to the . In this problem, 1


• Draw a picture of thet Distribution for this situation: Right Tail, .

• The shaded part is called the CRITICAL REGION.

• Its area is • The rest of the area is. Draw it!


• What value is at the boundary? – Lookup in printed tables.

• Or use TI-84 invT(.99,39)• THE CRITICAL VALUE is

d.f. One tail, α 0.01

38 2.43440 2.429







• Here, we just barely “Fail to Reject H0”• When we did this with z, we did reject.






“Darton State College EMT students scored

higher than the statewide average in a recent

examination, but not at a statistically significant

level.”

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 with a sample standard deviation of years. What can we conclude?


• We will use the Distribution and the values to do this test.


deviation– We have a “large” sample size: .– If we had a “small” sample size, this method



• The null hypothesis says that the mean age is 55 years old:

• The alternative hypothesis is a claim that the mean age is not 55 years old, it’s different:


• Left tail, Right tail, Two tail? Look at the direction of the null hypothesis. We have a TWO-TAILED hypothesis test.

• Because our Alternative Hypothesis is making a claim of a mean that’s DIFFERENT THAN the mean stated in the Null Hypothesis.

• Higher, Lower, doesn’t matter, we’re just testing for “different”.

• Refer back to the . In this problem,

Step 2. Determine the Critical Values

• TI-84 invT(.025,34)

• Or the printed table:

• They’re opposites of each other: -2.03, +2.03

d.f. Two tails, α α=0.05

34 2.03236 2.028







• Here, we “FAIL TO REJECT THE NULL HYPOTH.”

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.






“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.”

Documents

Hypothesis Testing with t Using the Normal Distribution t Distribution in a Hypothesis Test