traditional method one mean, sigma known. the problem in 2004, the average monthly social security...

Traditional Method

One mean, sigma known

The Problem

In 2004, the average monthly Social Security benefit for retired workers was $954.90 with a standard deviation of $98.00. In a sample of 50 retired workers in 2005, their average Social Security benefit was $970.30. If the standard deviation has not changed, is there sufficient evidence at α=.05 to support a claim that the benefits have increased? Use the traditional method.

Source: Bluman, Elemenary Statistics, eighth edition, citing the New York Times Almanac.

Option to work on your own and just check answerIf you want to work through this

problem on your own and just check your answer, click the person to the left when you’re ready to check your work.

Otherwise, click away from the person or press the space bar,and we’ll work through this problem together.

Set-upThis is a test about 1 mean, the mean in 2005.

This person is confused; she thought it was about 2 means, one for 2004 and one for 2005! Click on her if you share her confusion. Otherwise, move the mouse away from her and click (or just hit the space bar) to keep going.

Set-up, slide 2

Here’s what we know:

Populationμ= ?

This is what the hypotheses will be about!

Set-up, slide 3

Here’s what we know:

Populationμ= ?σ=98

Since we are told to assume the standard deviation has not changed, we can use the population standard deviation from 2004 as the population standard deviation for 2005.

Set-up, slide 4

Population μ= ?σ=98

Sample

n= 50

Step 1:State the hypotheses and identify the claim.

We are asked to evaluate the claim that benefits have increased. That is:

𝑏𝑒𝑛𝑒𝑓𝑖𝑡𝑠 𝑖𝑛2005>𝑏𝑒𝑛𝑒𝑓𝑖𝑡𝑠 𝑖𝑛2004

That’s μ! That’s 954.90

Step 1, slide 2So the claim is….

𝜇>954.90

Do you see an equals sign?

Nope. It must be the Alternate Hypothesis.

Step 1 slide 3Step 1

• The Null Hypothesis has to have an equals sign, since the Null always claims there is no difference between things.

• The Null Hypothesis will compare the same quantities as the Alternate Hypothesis (in this case, μ and 954.90).

Step 1, slide 4

Step (*)

Draw the picture and label the area in the critical region.

STOP!

Step (*) slide 2

Do we know we have a normal Distribution?

Step (*) slide 3Yes!! We have a normal distribution because our sample size (50) is big enough---it is at least 30.

Step (*) slide 4Step (*) Since we have a normal distribution,

draw a normal curve.

Top level: Area

Middle level: standard units

We always use z-values when we know the population standard deviation, σ.

(z)

Step (*)slide 5Step (*): Since we have a normal distribution,


Top level: Area

Middle Level: Standard Units (z)

The center is always 0 in standard units. Label this whenever you draw the picture.

0

Step (*) slide 6:Step (*): Since we have a normal distribution,


0

Top level: Area


Bottom level: Actual Units ($)

In this case, the actual units are dollars, since our hypothesis is about the average monthly benefit, which is measured in dollars.

Step (*) slide 7Step (*): Since we have a normal distribution,


0

Top level: Area


Bottom level: Actual Units ($) 954.90

The number from the Null Hypothesis always goes in the center of the bottom level; that’s because we’re drawing the picture as if the Null is true.

Step (*) slide 8Then remember:

The raditional MethodT

is op-downT

Step (*) slide 9

Step (*):(continued)

Once you’ve drawn the picture, start at the Top level and label the area in the critical region.

Standard Units (z) 0

Actual Units ($) 954.90

Top level: Area

.05

Step (*) slide 10Step (*): Once you’ve drawn the picture,

start at the Top level and mark off the area in the critical region.


Actual Units ($)

.05

Top level: Area

This is a right-tailed test since includes a greater than (>) α=.05 = area in right tail

Step 2:



.05

Middle Level

Put critical value here!

Move down to the middle level.Label the critical value, which is the boundary between the critical and non-critical regions.

Picking Table E or Table FWe can find the critical value using either Table E or Table F. Click on the table you want to use.

TABLE E

Table E gives us the z-values associated with certain areas under the standard normal curve

Table F

The bottom row of table F gives us the z-values associated with the area in the tail/s.

Step 2 slide 2Our picture looks like this: (we know the area to the right of the critical value, and want to know the critical value.)

.05

0?

To use Table E, we want to have our picture match this one, where we know the area to the left of the critical value.

We can subtract the area in the right tail from the total area (1) to get the area to the left!

1-.05=.95

Step 2 slide 3Now we can look up .9500 in the area part of Table E.

Area

Let’s zoom in!

Step 2 slide 6

The two areas closest to .9500 are .9495 and .9505. Since they are equally close to .9500, pick the bigger one, .9505.

Step 2 slide 7

The z-value associated with the area .9505 is 1.65.

Finishing up step 2:



.05


Adding the critical value to the picture



.05


1.65

Step 3: Move down to the bottom level.Mark off the observed value (.



.05

1.65

Bottom level

Step 3 slide 2

0

954.90

1.65

.05

Our observed value is 970.30. That’s bigger than 954.90,

so it must go here.

Why not here? It’s also to the right of 954.90, but it’s in the critical region.

Step 3 slide 3How can we figure out whether the observed value will be to the left or right of the critical value?

Step 3 slide 4We can’t compare the observed value and the critical value as long as they are measured in different units. So we’ll convert the observed value to standard units.

Step 3 slide 5The result is called the test value, and we can easily see whether it is bigger or smaller than the critical value!

Step 3 slide 6𝑧=

𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑𝑣𝑎𝑙𝑢𝑒−𝑒𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑣𝑎𝑙𝑢𝑒𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑒𝑟𝑟𝑜𝑟

¿ 𝑋−𝜇

( 𝜎√𝑛 )¿970.30−954.90

( 98√50 )1

Step 3 slide 7

Standard units (z) 0

Actual units ($) 954.90

1.65

.05

1.111

1.11<1.65, so it goes somewhereIn the region between 0 and 1.65

970.30

Line up the observed value with the test value; note that it is not in the critical region.

Step 4: Decide whether or not to reject the Null.

𝐻0

I throw myself on the mercy of this court.

Step 4 slide 2



1.65

.05

1.111

970.30

Since the observed value is not in the critical region, we don’t reject the Null.

Step 5: Answer the question.• Talk about the claim.• Since the claim is the Alternate

Hypothesis, use the language of support.

• We didn’t reject the Null, so we don’t support the claim.

Step 5 slide 2There is not enough evidence to support the claim that benefits increased from 2004 to 2005.

Let’s see a quick summary! Good idea!

SummaryEach click will take you to the next step; step (*) is broken into two clicks.

Step 1: 𝐻0 :𝜇=954.90𝐻1 :𝜇>954.90(𝑐𝑙𝑎𝑖𝑚)

Step (*)



.05

Step 21.65

Step 31.111

970.30

Step 4: Don’t reject Null.

Step 5: There’s not enough evidence to support the claim.

Celebration: we did it!

And there was much rejoicing.

End of regular slide show; subsequent slides explain how this is about one mean, not two

Press the escape key to exit the slide show. If you keep clicking through, you’ll see the slides explaining why this was a test about one mean, rather than 2.

Explanation of why this is about 1 mean, slide 1

It’s true that there are two means in this problem, but there’s a very important difference between them:

Explanation about 1 mean, slide 2It’s true that there are two means in this problem, but there’s a very important difference between them:

We know the mean for 2004, so there’s no need to make and test a hypothesis about it.

Explanation about 1mean, slide 3It’s true that there are two means in this problem, but there’s a very important difference between them:

We know the mean for 2004, so there’s no need to make and test a hypothesis about it.

We don’t know the mean for 2005, so we have to form a hypothesis about it and test that hypothesis.

Explanation about 1 mean, slide 4 We say the hypothesis test is about one mean when there is just one mean that we don’t know, even if we are comparing it to a known mean.

When we say a test is about two means, that will indicate that there are two means and we don’t know either of them.

Click anywhere on this slide to return to the hypothesis test. Don’t just hit the space bar or you’ll go to the wrong slide!

Using Table F to find the critical value:

We need to look at the top part of Table F to determine which column will contain our z-value.

Since this is a one-tailed test, look for α = .05 in this row.

Using Table F to find the critical value, slide 2

Be sure to go all the way to the bottom row of Table F; this is the only row that gives us z-values!

Z = 1.645

Finishing up step 2:



.05


Adding the critical value to the picture, Table F version



.05


1.645

Step 3: Table F VersionStep 3: Move down to the bottom level.

Mark off the observed value (.



.05

1.645

Bottom level

Step 3 slide 2, table F version

0

954.90

1.645

.05

Our observed value is 970.30. That’s bigger than 954.90,

so it must go here.

Why not here? It’s also to the right of 954.90, but it’s in the critical region.

Step 3 slide 3, Table F versionHow can we figure out whether the observed value will be to the left or right of the critical value?

Step 3 slide 4, Table F versionWe can’t compare the observed value and the critical value as long as they are measured in different units. So we’ll convert the observed value to standard units.

Step 3 slide 5, Table F versionThe result is called the test value, and we can easily see whether it is bigger or smaller than the critical value!

Step 3 slide 6, Table F version𝑧=

𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑𝑣𝑎𝑙𝑢𝑒−𝑒𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑣𝑎𝑙𝑢𝑒𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑒𝑟𝑟𝑜𝑟

¿ 𝑋−𝜇

( 𝜎√𝑛 )¿970.30−954.90

( 98√50 )1

Step 3 slide 7, Table F version



1.645

.05

1.111

1.11<1.645, so it goes somewhereIn the region between 0 and 1.645

970.30

Line up the observed value with the test value; note that it is not in the critical region.

Step 4: Decide whether or not to reject the Null, Table F version

Step 4: Decide whether or not to reject the Null.

𝐻0

I throw myself on the mercy of this court.

Step 4 slide 2, Table F version



1.645

.05

1.111

970.30

Since the observed value is not in the critical region, we don’t reject the Null.

Step 5: Answer the question. (Table F version)

Step 5: Answer the question.

• Talk about the claim.• Since the claim is the Alternate

Hypothesis, use the language of support.

• We didn’t reject the Null, so we don’t support the claim.

Step 5 slide 2, Table F versionThere is not enough evidence to support the claim that benefits increased from 2004 to 2005.

Let’s see a quick summary! Good idea!

Summary, Table F versionEach click will take you to the next step; step (*) is broken into two clicks.

Step 1: 𝐻0 :𝜇=954.90𝐻1 :𝜇>954.90(𝑐𝑙𝑎𝑖𝑚)

Step (*)



.05

Step 21.645

Step 31.111

970.30

Step 4: Don’t reject Null.

Step 5: There’s not enough evidence to support the claim.

Celebration: we did it! (Table F version)

And there was much rejoicing.

traditional method one mean, sigma known. the problem in 2004, the average monthly social security...

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