Week 7 homework help 2015 13 15_19

WEEK 7 HOMEWORK HELPNUMBERS 13,15 & 19

B HeardNot to be copied, posted, shared without my permission. Students can download one copy for their personal use.

WEEK 7 HOMEWORK HELP (MORE)

.

They say “it is 60 mg” that means they are claiming the caffeine content is = 60,

soH0: µ = 60 (claim)Ha: µ ≠ 60


.

Remember you are looking at the Alternative Hypothesis to determine “tails”

Ha: µ ≠ 60 implies we have a two-tailed test


.

As noted in the problem, our alpha is 0.06Use Minitab to find the critical valuesGraph >> Probability Distribution Plots >> Click View Probability


.

Normal DistributionMean of 0Standard Deviation of 1Then Click Shaded Area Tab


.

Probability Radial ButtonBoth TailsAlpha is 0.06 (The area in the tails)Input for ProbabilityClick OK


.

Look at GraphSee values at bottomThey generally round toTwo decimals placesSo our z values would be-1.88 and 1.88

0.4

0.3

0.2

0.1

0.0

X

De

nsit

y

-1.881

0.03

1.881

0.03

0

Distribution PlotNormal, Mean=0, StDev=1


.

Find the standardized test statistic (to compare and see if it is in the critical region)

MinitabStat >> Basic Statistics >> 1 Sample z


.

See my inputs and choices based to the problem

1 – alpha(100- 6% =

94%)

Alternative was not equal to


.

See Session Window after clicking ok’s

Standardized Test Statistic is 1.90


.

See Session Window after clicking ok’s

0.4

0.3

0.2

0.1

0.0

X

De

nsit

y

-1.881

0.03

1.881

0.03

0


1.90 would be barely in the rejection region (beyond +1.88), but it

IS in the rejection region

What do we have?

Remember what we hadH0: µ = 60 (claim)Ha: µ ≠ 60Therefore we must REJECT the null that the caffeine content is 60 and say there is evidence that it is not 60, so we rejected the null (and the claim in this one) So we could say that there is enough evidence to reject the claim.


.

They say “more than 1 pound per person per day” that means they are claiming it’s more than

1, soH0: µ ≤ 1 Ha: µ > 1 (claim)


.

Sample size less than 30, so we will use t distribution


.

We will use Minitab


.

As noted in the problem, the sample size is 20, we willuse Minitab to find the critical t valuesGraph >> Probability Distribution Plots >> Click View Probability


.

t distributionDegrees of Freedom isSample Size Minus 120 -1 = 19Then Click Shaded Area Tab


.

We will reject anything to the right of 1.729

0.4

0.3

0.2

0.1

0.0

X

De

nsit

y

1.729

0.05

0

Distribution PlotT, df=19


.

We will use Minitab


.

Find the standardized test statistic (to compare and see if it is in the critical region)

MinitabStat >> Basic Statistics >> 1 Sample t


.

Make selections based on problem

1 – alpha(100- 5% =

95%)

Alternative was GREATER THAN


Find your standardized test statistic

Here it is, compare to my graph


.

It’s to the right of 1.729

0.4

0.3

0.2

0.1

0.0

X

De

nsit

y

1.729

0.05

0

Distribution PlotT, df=19

4.47 would be way to the right, so we are

going to reject the null


.

Remember what we hadH0: µ ≤ 1 Ha: µ > 1 (claim)By rejecting the null on this one, we are saying that there IS enough evidence to support the claim (because the claim is the alternative on this one)


.

They are claiming “more than 20%” or p>0.20, so we set up our null and alternative

H0: p ≤ 0.20Ha: µ > 0.20 (claim)This one screams proportion…. We do three types (1 Sample Z, 1 Sample t and 1 Proportion – this is a “1 Proportion”)


The z test for a proportion is easy, you are just multiplying n times p and n times q to make sure their products are large enough


We have our null and alternative

H0: p ≤ 0.20

Ha: µ > 0.20 (claim)


These are similar to 1 sample z when finding the critical values. We know alpha is equal to 0.03

So in Minitab, go to Graph, Probability Distribution Plots, Click View Probability, etc.

Normal

Mean of 0

SD of 1


Click Shaded Area Tab, we know alpha is equal to 0.03

Right Tail, becauseHa: µ > 0.20

Alpha


Graph

0.4

0.3

0.2

0.1

0.0

X

De

nsit

y

1.881

0.03

0


We will reject anything to the right

of this value(Like a z value they

generally round these to two decimal

places) So 1.88


Now get your standardized test statistic to compare.

Minitab

Stat >> Basic Statistics >> 1 Proportion


.

In a random sample of 1150 consumers, you find 22% have

stopped buying the product…

(0.22)(1150) = 253

Environmental Agency CLAIMED more than 20% or 0.20

After inputting everything and

checking Perform Hypothesis Test,

Click Options Button


After clicking the options button, change confidence to 97% (1-alpha) and Alternative to Greater Than AND YOU MUST CHECK BOX THAT SAYS USE TEST AND INTERVAL BASED ON NORMAL DISTRIBUTION

Greater Than, because

Ha: µ > 0.20

Alpha = 0.031 – 0.03 = .97 or

97%

You MUST Check this Box

Here is our result including our standardized test statistic

Here it is … 1.70, now compare to

graph


Graph

0.4

0.3

0.2

0.1

0.0

X

De

nsit

y

1.881

0.03

0


1.70 WOULD NOT BE IN THE REJECTION REGION. SO WE DO NOT REJECT THE

NULL HYPOTHESIS


.

REMEMBERH0: p ≤ 0.20Ha: µ > 0.20 (claim)By failing to reject the null, we are not supporting the claim… or saying there is not enough evidence to support the claim (because the claim is the alternative on this one)


Notice on our results that they also give us the p value…. You could use either, but they almost ALWAYS specify which they want you to use…

Here is the p value


This extra stuff was “extra”… Best to you…

Week 7 homework help 2015 13 15_19

Education