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Use the Table provided in the hand out to find the required

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Use the Table provided in the hand out to find the required x2 - values. Illustrate your work graphically. 1. For a x2 - curve with 19 degrees of freedom, find the x2 - value having area. a. 0.025 to its right b. 0.95 to its right 2. For an x2 - curve with df = 10, determine: a. x20.05 b. x20.0975 3. Consider an x2 - curve with df = 8. Obtain the x2 - value having area a. 0.01 to its left. b. 0.95 to its left 4. Determine the two x2 - values that divide the area under the curve into the middle 0.95 area and two outside 0.025 areas for a x2 - curve with a. df = 5 b. df= 26 5. Determine the two x2 - values that divide the area under the curve into a middle 0.90 area and two outside 0.05 areas for a x2 - value with a. df= 11 b. df = 28 6. The table below provides a relative-frequency distribution for the number of years of school completed by Canadian residents, 25 years old and over. /15 marks Years ReI. Freq. 8 or less 0.183 9 -11 0.153 12 0.346 13 - 15 0.157 16 or more 0.161 A random sample of Ontario residents, 25 years old and over, gave the following statistics: Years Frequency 8 or less 83 13 - 15 36 16 or more 38 9 -11 48 12 95 Do the data provide sufficient evidence to conclude that the distribution of the number of years of school completed by Ontario residents is different from the national distribution? Show all steps. 7. A national survey was conducted to obtain information on the alcohol consumption patterns of Canadian adults by marital status. A random sample of 1772 residents, 18 years old and over, yielded the data displayed in the Table below, for instance, that of the 1772 adults sampled, 1173 are married, 590 abstained, and 411 are married and abstained. / 15 marks Abstain 1-60 Over 60 Total Single 67 213 74 354 Married 411(NNN) NNN-NNNNWidowed 85 51 7 143 Divorced 27 60 15 103 Total Your task is to determine whether there is an association between marital status and alcohol consumption. Determine: (Show all steps) Ho: Marital status and alcohol consumption are statistically independent Ha: Marital status and alcohol consumption are statistically dependent Test for the two assumptions (all expected frequencies are at least 1 and at most 20% of the expected frequencies are less than 5)and indicate in each step what is being performed.

I believe a little bit of your question is lost in translation. I am assuming that the first 5 questions relate to a chi squared distribution, not an x2 distribution. The lowercase (and uppercase) Greek symbols for the letter chi look very much like an English x (or X). (PS: That’s how the term “Xmas” came about. The Greek word for Christ would look, with English letters, like Xristos and would be pronounced “Khris-tos). The first letter, chi, gives the KR sound.)

If you are doing any reports with your homework or answers, chi, (and any Greek letter) can be typed in any MS Office product (Word, Excel, PowerPoint) by merely typing the corresponding English letter and then changing the font of just that letter to the Symbol font. The letter chi used in a chi-squared distribution is the lowercase chi. Type a lowercase “c” and change the font of just that letter to show the Greek lowercase “chi” symbol. To show chi-squared, type “c2”. Then change the font of the “c” to Symbol and change the font of the “2” to superscript.

Your instructions said “Use the Table provided in the hand out to find the required x2 - values. Illustrate your work graphically.”

Presumably you have a chi-squared table as a handout. I also assume your table has a picture of a chi-squared distribution above the table. Since I do not have access to your handout, I’d direct you to http://www.medcalc.org/manual/chi-square-table.php so we can both look at the same table. The values in this table should match the values in your table; although your table may have more or fewer rows and/or columns.

Whenever using a table like this it is imperative to always look at the picture first. In the medcalc table I referenced on the web, the shaded area is to the right of the chi-squared value of concern. Notice the “P” pointing to the shaded area. That tells us that for whatever value of chi-squared we are concerned with, the probability values shown within the table will be the total of all probabilities to the RIGHT of that value, or the area under the curve to the right of your chi-squared value in question.

Now, let’s answer your questions.

1 - For a chi-squared curve with 19 degrees of freedom, find the x2 - value having area.

a. 0.025 to its right

If we want a probability of 0.025 (2.5%) to the right of our chi-squared value, look on the top row of numbers and find 0.025. Now look down the left column of numbers (degrees of freedom) and find 19. Find the cell where df=19 intersects with P=0.025. The number in the cell where these rows and columns intersect contain the number 32.852. This is the corresponding chi-squared value.

In plain English, for a chi-squared value of 32.852 and 19 degrees of freedom, we find that only 2.5% of the total area under the curve (or only 2.5% of the total probability) will be beyond this chi-squared value of 32.852. (Note: the table supplied as a handout may have lesser or greater decimal place accuracy. Use your table and answer to the level of accuracy given in your table.)

You are also asked to illustrate your work graphically. If the chi-squared axis on your handout picture contains a horizontal scale with numbers, these would be chi-squared values. Find the point where chi-squared is 32.852 and draw a vertical line up from there to intersect the curve. Shade in the area to the right of your line. Show 0.025 as the probability to the right of your line.

b. 0.95 to its right

Our medcalc table doesn’t show data for probabilities of 0.95 (95%) to the right. Please allow me to direct you to http://people.richland.edu/james/lecture/m170/tbl-chi.html. This table from Richland does have 0.95. This table does NOT have a picture, but we can note from the text just below the table title that says, “The areas given across the top are the areas to the right of the critical value.” We want 95% of the area under the curve to be to the right of our chi-squared value, so we find the intersection of the column with probability of 0.95 row with df=19. At this intersection we find that chi-squared = 10.117.

Remember to also show your work graphically.

2 - For an x2 - curve with df = 10, determine:

a. x2, 0.05

b. x, 20.0975

Your text in these two questions has lost a little in translation (or copy and paste).

I am assuming for (a) we want the chi-squared value where the probability to the right is 0.05 and df=10. The Richland table from the web tells me the chi-squared value is 18.307

For (b), a probability to the right of 0.0975 (9.75%) is not on any of the tables I can find. (This is an unusual value for a probability.) If I assume a typo and assume the actual probability desired is 0.975 (97.5%), the chi-squared value with this probability to the right when df = 10 is 3.247

Don’t forget to show your work graphically.

3. Consider an x2 - curve with df = 8. Obtain the x2 - value having area

a. 0.01 to its left.

b. 0.95 to its left

For (a), if we want 0.01 (1%) to the left, we just subtract 0.01 from the 100% total area under the curve (1.00) and get 0.99 probability to the right. For this probability and df=8, chi-squared = 1.646

For (b) we are looking for 0.05 to the right (1.00 total - 0.95 to the left = 0.05 to the right). For 0.05 to the right and df=8. Chi-squared = 15.507.

Remember to show your work graphically.

4. Determine the two x2 - values that divide the area under the curve into the middle 0.95 area and two outside 0.025 areas for a x2 - curve with

a. df = 5

b. df= 26

The first line will be at the point where 97.5% (0.975) of the probability (the area under the curve) is to the right.

The second line will be at the point where only 2.5% (0.025) of the probability is to the right.

The area in between these two lines should represent the middle 95% of the probabilities. The area on each side of this middle area, the area in each tail, will be 0.025 or 2.5% each. (Note: this is commonly used for 95% confidence levels.)

(a, line 1) P=0.975, df=5, chi-squared = 0.831

(a, line 2) P=0.025, df=5, chi-squared = 12.833

(b, line 1) P=.0975, df=26 Therefore chi-squared is 13.844.

(b, line 2) P = 0.025, df=26 Chi-squared is 41.923.

Remember to show your work graphically.

5. Determine the two x2 - values that divide the area under the curve into a middle 0.90 area and two outside 0.05 areas for a x2 - value with

a. df= 11

b. df = 28

We use the same method here only our lines are at P=0.95 and P=0.05. We’re defining our 90% confidence interval, with 90% of all probabilities within this interval, 5% to the left of it and 5% to the right of it.

(a) for df=11: P=0.95 at chi-squared value of 4.575 and P=0.05 is at chi-squared of 19.675

(b) for df=28: P=0.95 is at chi-squared value of 16.928. If P=0.05, chi-squared = 41.337

I will address questions 6 and 7 a little later. (It’s about 8 PM my time, CST.)

Hopefully I have not introduced any typographical errors in my answer. Please reply if you have any questions.

Best regards,

Jim

JACUSTOMER-i8eog9gm- :

ok, i will wait for it. thank you for the present answers

Customer:

Dear customer:

Customer:

I'm having problems understanding the table of data for question #7. The numbers don't add up and two categorical names are missing.

Customer:

Can you attach the file as a jpeg. (Use the paper clip button) to insert a file in your reply.

Customer:

Hopefully you can scan the table, save it as a jpg, and then attach it.

Customer:

Best regards,

Customer:

Jim

JACUSTOMER-i8eog9gm- :

yes, just a second

Customer:

Alternatively you could:

Customer:

Send me a link if the table is on the web or load the table to www.mediafire.com and send me a link to the table.

Hello, I need my answer ASAP please, how long will it take you to provide me with an answer? Need to know. Thank You!

Customer:

I'm on it. It should be witin the next few minutes.

Customer:

Jim

JACUSTOMER-i8eog9gm- :

ok

JACUSTOMER-i8eog9gm- :

nice

Customer:

Well, I got your table. Thanks. There IS a typo on the table. Note the row for Divorced. 27-60+15=102 in Texas, not 103. But, since the question said 1772 residents, total, that number is, indeed, the total of all residents. So, we'll go with what we have, assuming the 103 is actually 102.

JACUSTOMER-i8eog9gm- :

ok....

Customer:

7. A national survey was conducted to obtain information on the alcohol consumption patterns of Canadian adults by marital status. A random sample of 1772 residents, 18 years old and over, yielded the data displayed in the Table below, for instance, that of the

1772 adults sampled,

1173 are married,

590 abstained, and

411 are married and abstained.

/ 15 marks

Abstain 1-60

Over 60 Total Single 67 213 74 354 Married 411(NNN) NNN-NNNNWidowed 85 51 7 143 Divorced 27 60 15 103 Total Your task is to determine whether there is an association between marital status and alcohol consumption. Determine: (Show all steps) Ho: Marital status and alcohol consumption are statistically independent Ha: Marital status and alcohol consumption are statistically dependent Test for the two assumptions (all expected frequencies are at least 1 and at most 20% of the expected frequencies are less than 5)and indicate in each step what is being performed.

The analysis of this question will also use a Chi-Sq analysis just like #6, but this time we have two variables, marital status and alcohol consumption. We’ll need to use a two-dimensional table.

The table of observed frequencies is given in the problem;

From this we can create a table of Expected Frequencies

Here’s how the cells are calculated:

The value Expected Frequency for Abstain/Single is calculated by taking the Column Total of Observed Frequencies for Abstain (590), multiplying it by the Row Total of Observed Frequencies for Single (354) and dividing this product by the Grand Total of all Observed Frequencies (1772). The result is 117.87.

The rows and columns of Expected Frequency values are totaled. These row and column totals, plus the Grand Total should be exactly what we saw with Observed Frequencies. (Clearly this is easily done in Excel.)

Then the Contributions to Chi-Sq are calculated for each cell.

For each cell, square the difference between Obs Freq and Exp Freq, and divide this square by Expected Frequency.

(I hope you recognize “^2” as the Excel syntax to indicate a square operation. Complete the calculations within the parenthesis an then square the result.)

Notice the Grand Total contribution to Chi-Sq. It is 94.269. This is our Chi-Sq critical value

Our degrees of freedom are (#of rows - 1) x (# XXXXX cols – 1) = (4-1) x (3-1) = 3 x 2 = 6

The question does NOT state the level of confidence we want in our conclusion (90%, 95%, 99%, etc.). So, we’ll see how much confidence we can get with a Chi-Sq of 94.269.

For DF = 6, our critical Chi-Sq is bigger than the Chi-Sq of even the smallest probability, 0.001, (Chi-Sq = 22.458 on the medcalc table. So, we have a lower probability than 0.001 (or 0.1%) that there is no statistical relationship between marital status and alcohol consumption.

We reject the Null Hypothesis and conclude the alternate Hypothesis, “Marital status and alcohol consumption are statistically dependent.” And, we can do so with over 99.9% confidence.

We also check the two assumptions for a two-variable Chi-Square Test:

1 – No cell has an expected frequency less than 1. This is true! We pass! Our lowest is 12.95.

2 – No more than 20% of the cells have expected frequencies less than 5. None do. We pass!

So, Mr. JACUSTOMER-i8eog9gm, I hope these replies effectively answer your questions.

Just for my curiosity, Is this for a college class, an ASQ test prep or what?

Just for your information, all of these problems could have been done in about 10% of the time if we used a statistical software package such as MiniTab or even plain old Excel. But, I assumed you were required to learn the “old fashioned” manual method and that is what I have described for you.

Very best regards,

Jim

PS – Take some time to look these over. When you are satisfied that you understand my explanations, please remember to come back and press the “Accept” button.

Thank you

Customer:

Here's the Observed Frequency Data:

Customer:

Observed Freq

Abstain

1 to 60

Over 60

Total

Single

67

213

74

354

Married

411

633

129

1173

Widowed

85

51

7

143

Divorced

27

60

15

102

Total

590

957

225

1772

Customer:

And the Expected Frequency Data:

Customer:

Expected Freq

Abstain

1 to 60

Over 60

Total

Single

117.87

191.18

44.95

354

Married

390.56

633.50

148.94

1,173

Widowed

47.61

77.23

18.16

143

Divorced

33.96

55.09

12.95

102

Total

590

957

225

1,772

Customer:

That doesn't look too good. Let me try again:

Customer:

Expected Freq

Abstain

1 to 60

Over 60

Total

Single

117.87

191.18

44.95

354

Married

390.56

633.50

148.94

1,173

Widowed

47.61

77.23

18.16

143

Divorced

33.96

55.09

12.95

102

Total

590

957

225

1,772

And the Contribution to Chi-Sq data:

Customer:

Contrib to Chi-Sq

Abstain

1 to 60

Over 60

Total

Single

21.952

2.489

18.776

43.217

Married

1.070

0.000

2.670

3.740

Widowed

29.358

8.908

6.856

45.122

Divorced

1.427

0.438

0.324

2.189

Total

53.807

11.836

28.626

94.269

Customer:

Good Luck,

Customer:

Jim

JACUSTOMER-i8eog9gm- :

Yes, you are right, this is an assignment for marks and i was required to complete it manually. It will take me couple of minutes to look over it. I will press the accept button then. Thank You!

My apologies, Sir. I had worked the problem but I guess I forgot to copy/paste it into the chat window.

6. The table below provides a relative-frequency distribution for the number of years of school completed by Canadian residents, 25 years old and over. /15 marks

Years ReI. Freq.

8 or less = 0.183

9 -11 = 0.153

12 = 0.346

13 - 15 = 0.157

16 or more = 0.161

A random sample of Ontario residents, 25 years old and over, gave the following statistics:

Years Frequency

8 or less = 83

9 -11= 48

12 = 95

13 - 15 = 36

16 or more = 38

Do the data provide sufficient evidence to conclude that the distribution of the number of years of school completed by Ontario residents is different from the national distribution? Show all steps.

To understand if the Ontario distribution is statistically different from the expected national distribution, we will perform a Chi-Squared "Goodness of Fit" This test essentially looks at each category (example: 9 - 11 years of education) and tests whether the observed percentage in that category in Ontario is close to what we'd expect if Ontario followed the national averages.

Here's the Chi-Squared Table with your data and a few calculations:

A B C D E F

Years RelFreq OntCnt ObsFreq ExpCount Contr to chi-sq

1 8 or less 0.183 83 0.277 54.9 14.383

2 9 to 11 0.153 48 0.160 45.9 0.096

3 12 0.346 95 0.317 103.8 0.746

4 13 to 15 0.157 36 0.120 47.1 2.616

5 16 or more 0.161 38 0.127 48.3 2.196

6 Total 1.000 300 1.000 300 20.037

Columns A, B, and C contain the data given in the problem.

Column D is merely taking the column C data and dividing by the total for Column C to calculate the observed frequency or proportion. (It adds up to 100%)

Column E is the Expected Count calculated by (Column B) times the total count (300 observations)

The Contribution to Chi-Squared in column F is calculated as followed:

Contr to Chi-Sq = square of (Obs Count - Exp Count) / Exp Count

For Row 1, this calculation would be:

Contr to Chi-Sq = (83 - 54.9)^2 / 54.9

= (28.1)^2 / 54.9

= 789.61 / 54.9

= 14.383

Do the same calculation for Rows 2 - 5 and sum up Column F to get your critrical Chi-Sq value of 20.037

Degrees of freedom are the number of categories (age groups) minus 1. DF = 4

P-value when DF = 4 and Chi-Sq = 20.037 (from our table)

Our critical Chi-Sq of 20.037 is greater than the Chi-Sq of 18.467 where DF = 4 and P = 0.001 in the medcalc table.

Therefore our probability is less than 0.001 that the Ontario education distribution follows the national average. Our null hypothesis was there is "no difference in the distribution of education in Ontario than in the national average". We can assume less than 0.1% risk (over 99.9% confidence in our conclusion) if we reject this Null and conclude the Alternative Hypothesis: "There IS a difference in the education distributions between Ontario and the national averages."

Clearly, for 95% confidence and even 99% confidence and 99.9% confidence we can say that Ontario averages for years of education are statistically different from the national average.

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