# Statistical Tests, Compose and Solutions

2021-05-12 14:46:33
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The table below lists the days of the week selected by a random sample of 1005 subjects who were asked to identify the best day of the week for family activities. Consider the claim that the days of the week have the same chance of being chosen.

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Sunday Monday Tuesday Wednesday Thursday Friday Saturday

523 20 9 19 11 41 382

Instructions for performing this test in STATDISK can be found in the Statdisk Users Manual.]

Use the Chi-Square Goodness-of-Fit test to see if there is a difference between the choices for best day for family activities. Use a significance level of .05.

Paste results here. H0: there the difference between the choices for the best day for family activities

H1: There is no difference between the choices for the best day for family activities

DF = (r - 1) * (c - 1) = (2 - 1) * (3 - 1) = 2

Er,c = (nr * nc) / n

E1, 1 =1005*523/1005 = 523

E1, 2 =1005*20/1005 = 20

E1, 3 =1005*9/1005 = 9

E1,4 =1005*19/1005 =19

E1,5 =1005*11/1005 =11

E1, 6 =1005*41/1005 = 41

E1, 7 = 1005 *382/1005 =382

X2 = S [ (Or,c - Er,c)2 / Er,c ]

X2 = (523 - 523)2/523 + (20 - 20)2/20 + (9 - 9)2/9

+ (19 - 19)2/19 + (11 - 11)2/11 + (41 - 41)2/41

X2 = 0

What are we trying to show here? We accept the H0 hypothesis that there is a relationship between the choice of the best day and the family activity.

What is the p-value and what does it represent in the context of this problem? P value = 0

State in your own words what the results of this Goodness-of-fit test tells us. The result tells as that there is a relationship between the choice of the best day of the family activities and the days of the week.

Repeat the above procedure using only the weekdays Paste results here. Did you get different results? DF = (r - 1) * (c - 1) = (2 - 1) * (3 - 1) = 2

Er,c = (nr * nc) / n

E1,1= 100*20/100=20

E1,2 =100*9/100=9

E1,3 =100*19/100=19

E1,4=100*11/100=11

E1,5 =100*41/100 =41

X2 = S [ (Or,c - Er,c)2 / Er,c ]

X2 = (20 - 20)2/20 + (9 - 9)2/9 + (19 - 19)2/19

+ (11 - 11)2/11 + (41 - 41)2/41

X2 = 0

There is no difference because it also show that the family activity still have the relationship with the day of the week.

Part II. Chi-Square Goodness of Fit Test (unequal frequencies)

When Antonios Pizza introduced chicken wings to their menu,they asked customers to provide feedback to the following question:

Rate Our Chicken: Did we get it right?

___ Nope___ Almost___ Oh Yes We Did

It would be unrealistic to expect 100% of respondents to select the positive "Oh Yes We Did" response. Instead, one may hope that 40% of customers will select the affirmative response, while 35% will select Nope and 25% will select Almost. Thus, the anticipated (expected) or targeted proportions of all responses may follow this distribution:

Nope = .35

Almost = .25

Oh Yes We Did = .40

Suppose after testing at one location 211 customers provided feedback with the following counts:

Nope = 59

Almost = 19

Oh Yes We Did = 133

Instructions for performing this test in STATDISK can be found in the Statdisk Users Manual.]

Complete the table as necessary.

[Hint: You will need to compute the expected frequencies based on the percentages for the 211 customers. Round to the nearest integer.] Nope Almost Oh Yes We Did

OBSERVED 59 19 133

EXPECTED 21

5 53

Use the Chi-Square Goodness-of-Fit test for Unequal frequencies to see if there is a difference between the observed frequencies and the expected frequencies Use a significance level of .01.

Paste results here. DF = (r - 1) * (c - 1) = (2 - 1) * (3 - 1) = 2

Er,c = (nr * nc) / n

E1,1= 211*80/290=58.

E1,2 =211*24/290=17.

E1,3 =211*186/290=135

E2,1=79*80/290= 22

E2,2=79*24/290 = 7

E2,3 =79*186/290 = 51

X2 = S [ (Or,c - Er,c)2 / Er,c ]

X2 = (59 - 58)2/58 + (19 - 17)2/17 + (133- 135)2/135

+ (21 - 22)2/22 + (5 - 7)2/7 + (53 -51)2/51

X2 = 1/58+4/17+4/135 +1/22 +4/7+4/51

X2 =0.0174 +0.235+0.0296+0.0455+0.571 +0.0784

X2 = 0.977

Chi Sq Value: 0.977000

Prob Dens: 2.876403e-78

Cumulative Probs

Left: 0.000000

Right: 1.000000

99 Degrees Freedom

State the null and alternative hypothesis. H0:there is difference in the observed values

H1: there is no difference in the observed values

What conclusion would you reach, given the result of your Goodness-of-Fit test? [State in your own words following the guidelines for a conclusion statement learned last week.] In my conclusion there is a great difference in the observed values than the expected values on the consumption of cookies.

Part III. Chi-Square Test of Independence

The following contingency table identifies the treatment given for a stress fracture of a foot bone and the outcome of success or failure. Use a 0.05 level of significance to test the claim that success is independent of method of treatment.

The following data was collected related to method of treatment and success/failure

Success Failure

Surgery 54 12

Weight-Bearing Cast 41 51

Non-Weight Bearing Cast 6 Weeks 70 3

Non-Weight Bearing Cast Less Than 6 Weeks 17 5

Hint: Instructions for performing this test in STATDISK can be found in the Stat Disk Users Manual under the heading Chi Square Test of Independence (Contingency Tables).

66 Degrees Freedom

Just looking at the numbers in the table, what is your best guess about the relationship between treatment and success? Are they independent or dependent? The relationship between treatments an success is relatively high based on the numbers provided. This is because they are dependent on each other.

Compute a Chi-Square Test of Independence on this data using a 0.05 level of significance. Paste your results here. DF = (r - 1) * (c - 1) = (2 - 1) * (3 - 1) = 2

Er,c = (nr * nc) / n

E1,1= 66*182/253= 47

E1,2 =66*71/253 = 19

E2,1 =92*182/253 = 66

E2,2=92*71/2253= 26

E3,1=73*182/253= 53

E3,2 =73*71/253= 21

E4,1 =22*182/253=16

E4,2 =22*71/253= 6

Chi Sq Value: 134.730000

Prob Dens: 0.0015272

Cumulative Probs

Left: 0.990136

Right: 0.009864

99 Degrees Freedom

What is the null and alternative hypothesis for this result? HO: there is a relationship between treatment and success of the method of treatment.

H1: There is no relationship between treatment and success of treatment method

What is the p-value for this result? What does this represent? P value = 0.0015272

State your conclusion related to the context of this problem. It indicates that there is a relationship between treatment and the success of the method of treatment.

Part IV. Apply this to your own situation

Using one of the above statistical tests, compose and SOLVE an actual problem from the context of your own personal or professional life.You will need to make up some data and describe which test you will use to analyze the situation. Heres an example:

Example: Do not use this problem!!

State the problem that you are analyzing. Last year, I asked the kids in my neighborhood what kind of cookies they preferred. 50% said chocolate-chip, 20% said oatmeal-raisin, and 30% said sugar cookie. I want to see if this has changed.

Make up some data for the new situation. I asked 50 neighborhood kids what kind of cookie they preferred now and heres what they said:

35 said chocolate-chip

5 said oatmeal-raisin

Determine which type of Chi-Square test you will perform. Since these are unequal frequencies, I will perform a Chi-Square Goodness-of-Fit Test (Unequal Frequencies).

Specify your null and alternative hypotheses. H0: There is no difference this year in the preferences of cookies within the neighborhood kids.

H1: Things have changed.

Setup the test Chocolate-Chip Oatmeal-Raisin Sugar-Cookie

OBSERVED 35 5 10

EXPECTED 25 10 15

Perform the test Paste your STATDISK results here

DF = (r - 1) * (c - 1) = (2 - 1) * (3 - 1) = 2

Er,c = (nr * nc) / n

E1,1= 50*60/100= 30

E1,2 =50*15/100 = 7.5

E1,3 =50*25/100 = 12.5

E2,1=50*60/100= 30

E2,2=50*15/100= 7.5

E2,3 =50*25/100= 12.5

X2 = S [ (Or,c - Er,c)2 / Er,c ]

X2 = (35 - 30)2/30 + (5- 7.5)2/7.5 + (10- 12.5)2/12.5

+ (35 - 30)2/30 + (5- 7.5)2/7.5 + (10- 12.5)2/12.5

X2 = 0.833+0.833+0.833+0.833+0.833+0.833

X2 = 4.998

Chi Sq Value: 4.998000

Prob Dens: 3.422437e-42

Cumulative Probs

Left: 0.000000

Right: 1.000000

95 Degrees Freedom

State your conclusion We have evidence to believe that things have changed because it is less than the level of confidence

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