# Week8 livelecture2010 follow_up

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27-Jul-2015Category

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### Transcript of Week8 livelecture2010 follow_up

1. B Heard

Lecture for the Final ExamStatistics For Decision Making

Follow-Up

With Selected Excel Explanations

Note these are not all

of the charts originally

presented

Not to be used, posted, etc. without my expressed permission.B Heard

2. Be able to understand the normal distribution and how it relates to the mean, standard deviation, variance, etc.

For example, I did an analysis and found the mean number of failures was 7 and the standard deviation was 1.5. Answer the two questions below.

How many standard deviations is 10 from the mean?10 7 = 3, 3/1.5 = 2 (your answer)

How many standard deviations is 6.25 from the mean? 6.25 7 =- .75,- .75/1.5 = -0.5 (your answer)

Final Exam Review

Use Calculator

3. Be able to use the Standard Normal Distribution Tables or

Excel to find probability values and z scores.

Examples:

Find the following probability involving the Standard Normal Distribution. What is P(z -.60)?

1 0.2743 =0.7257(using table or Excel 1 =NORMDIST(-0.6,0,1,TRUE))

Final Exam Review

See Following Charts

4. Explanation

Go to Template Site

http://highered.mcgraw-hill.com/sites/0070620164/student_view0/excel_templates.html

Save Normal Distribution File to your computer

Open the file and make sure you Unprotect as I described in previous presentations

Take out all the values in the green cells

5. Explanation

I put 0 in for the mean and 1 for the standard deviation.

I worked the first one on the top left and the second on the top middle. (Only input into green cells)

6. A researcher is performing a hypothesis teston a claim about a population proportion. Usingan alpha = .04 and n = 80, what is the rejection region if the alternate hypothesisis Ha: p > 0.70?

Alternate hypothesis test shows that this is a Right

Tailed test (since its p > 0.70) with a right tail area of .04

(since alpha = .04).Therefore we are going to reject Ho

if z > 1.75 (looked in standard normal table to find z

score for a probability of 0.96, z of 1.75 was the closest)

Final Exam Review

Covered in Week 7 Lecture

7. A researcher is performing a hypothesis teston a claim about a population proportion. Usingan alpha = .03 and n = 95, what two critical values determine the rejection regionif the null hypothesis is: Ho: p = 0.44?

Since Ho: p = 0.44, this is a two tailed test. Each tail has

an area = .03/2 = .015. The z-values that correspond to

this area in the tail is +/- 2.17. (You can see this by

finding the z score for either .015 or .985 realizing its two

tailed)

Final Exam Review

Covered in Week 7 Lecture

8. A manufacturer claims that the mean lifetime of its computer component is 1100 hours. A buyers researcher selects 49 of these components and finds the mean lifetime to be 1105 hours with a standard deviation of 30 hours. Test the manufacturer's claim. Use alpha = .02.

Answer on following chart

Final Exam Review

9. Ho: mu = 1100 hours (claim);

Ha: mu does not = 1100 hours; two tailed test, therefore, .01 is in left tail

and .01 is in right tail; thus critical values are 2.33;

test statistic is

z = (xbar - mu) [sigma sqrt(n)] = (1105 - 1100) [30 sqrt(49)] = 5 [30 7] = 5 4.29 = 1.17which is in the do not reject area because p value corresponding to z= +1.17 is 0.879

Fail toreject Ho. (Because 1.17 is in the bounds of the critical values

2.33)There is not enough evidence to reject the manufacturer's claim that

the mean lifetime is 1100 hours.

Final Exam Review

Covered in Week 7 Lecture

10. A Pizza Delivery Service claims thatit will get its pizzas delivered in less than 30 minutes. A random selection of 49 service times was collected, and their mean was calculated to be 28.6 minutes. The standard deviation was 4.7 minutes. Is there enough evidence to support the claim at alpha = .10. Perform an appropriate hypothesis test, showing each important step.(Note: 1st Step:WriteHo and Ha; 2nd Step: Determine Rejection Region; etc.)

Answer following chart

Final Exam Review

11. Ho: mu >= 30 min.

Ha: mu < 30 min. (claim). Therefore, it is a left-tailed test.

n=49; x-bar=28.6; s=4.7; alpha=0.10

Since alpha = 0.10, then the critical z value will be zc = -1.28

since n>30 then s can be used in place of sigma.

Standardized test statistic z = (x-bar - mu)/(s/sqrt(n)) z = (28.6-30)/(4.7/sqrt(49)) z = -2.085

since -2.085 < -1.28, we REJECT Ho.

That is, at alpha = 0.10, There is enough evidence to support the

Pizza Delivery Services claim.

(p-value method could have also been used)

Final Exam Review

Covered in Week 7 Lecture

12. Determine the minimum required sample size if you want to be 90% confident that the sample mean is within 5 units of the population mean given sigma = 8.4. Assume the population is normally distributed.

n = (Zc*sigma/E)^2 = [(1.645 * 8.4)/ 5]^2 = (2.7636)^2= 7.64= 8 (always round up sample sizes)

Final Exam Review

Covered in Week 7 Lecture

13. The failure times of a component are listed in hours. {100, 95, 120, 190, 200, 200,280}.Find the mean, median, mode, variance, and range.Do you think this sample might have come from a normal population? Why or why not?

mean = 169.3

median =190

mode = 200

variance = 4553.6

range = 185

Doubtful it came from a normal, compare mean, median,

mode, etc.

Final Exam Review

Use "average", "median", "mode",

"var" etc. functions

in Excel (Don't forget

to visually inspect the

mode for multiples)