1) Problem #PRAJWVZ "PRAJWVZ - 257167 - Using either the ..." |
A)
Using either the TI-83 or another method, calculate the correct p-value for a test with the following Chi-Square test statistic and degrees of freedom. χ2 = 2.71 df = 1 © STATS4STEM.ORG |
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Hints: |
Click 2nd -> VARS -> X2cdf χ2cdf(2.71, 10000, 1) Note: Use following format for calculating the p-value as chi-squared tests are always right tailed. χ2cdf( χ2 Value, Large Number (10000 or larger) , degrees of freedom ) R or Rweb: pchisq(2.71, 1, lower.tail = FALSE) Note: Use following format for calculating the p-value: pchisq( χ2 Value, df , lower.tail = FALSE ) |
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B)
Using either the TI-83 or another method, calculate the correct p-value for a test with the following Chi-Square test statistic and degrees of freedom.
χ2 = 16.5 df = 6 |
Algebraic Expression:
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Hints: |
Click 2nd -> VARS -> χ2cdf χ2cdf(16.5, 10000, 6) Note: Use following format for calculating the p-value as chi-squared tests are always right tailed. χ2cdf( χ2 Value, Large Number (10000 or larger) , degrees of freedom ) R or Rweb: pchisq(16.5, 6, lower.tail = FALSE) Note: Use following format for calculating the p-value: pchisq( χ2 Value, df , lower.tail = FALSE ) |
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C)
Using either the TI-83 or another method, calculate the correct p-value for a test with the following Chi-Square test statistic and degrees of freedom.
χ2 = 21.4 df = 11 |
Algebraic Expression:
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Hints: |
Click 2nd -> VARS -> χ2cdf χ2cdf(21.4, 10000, 11) Note: Use following format for calculating the p-value as chi-squared tests are always right tailed. χ2cdf( χ2 Value, Large Number (10000 or larger) , degrees of freedom ) R or Rweb pchisq(21.4, 11, lower.tail = FALSE) Note: Use following format for calculating the p-value: pchisq( χ2 Value, df , lower.tail = FALSE ) |
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2) Problem #PRAJVGP "PRAJVGP - 255824 - The following den..." |
The following density curves represent different chi-square density curves for various degrees of freedom (df). True or False: As the degrees of freedom increase for the chi-square distribution, one needs a greater and greater chi-square test statistic to reject the null. © STATS4STEM.ORG |
Multiple Choice:
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3) Problem #PRAJVGM "PRAJVGM - 255822 - Read/Refer to thi..." |
A)
Read/Refer to this chi-square test webpage to answer the following questions. The ________________________is a way of determining whether a set of categorical data came from a claimed discrete distribution or not. The null hypothesis is that they did and the alternate hypothesis is that they didn't. © STATS4STEM.ORG |
Multiple Choice:
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B)
True or False? As with all other tests, certain conditions must be checked before a chi-square test of anything is carried out. |
Multiple Choice:
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C)
True or False: The process for calculating the p-value for test of homogeneity and test of independence are the same. |
Multiple Choice:
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D)
True or False: For a goodness-of-fit test, the degrees of freedom = (r-1)(c-1) |
Multiple Choice:
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E)
True or False: If you're thinking, "homogeneity and independence sound the same!", you're nearly right. The difference is a matter of design. In a test of independence, observational units are collected at random from one population only, whereas for the the test of homogeneity, the data are collected by randomly sampling from each sub-group separately. (Say, 100 blacks, 100 whites, 100 American Indians, and so on.). |
Multiple Choice:
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4) Problem #PRAJWVM "PRAJWVM - 257155 - The following bla..." |
A)
The following bladdercancer data arise from 31 male patients who have been treated for superficial bladder cancer. The data gives the number of recurrent tumors during a particular time after the removal of the primary tumour, along with the size of the original tumor. Is there an association between tumor size and the number of recurrent tumors? Which test would be most appropriate? © STATS4STEM.ORG |
Multiple Choice:
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B)
Define the null and alternative hypothesis. |
Multiple Choice:
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C)
Refer the two way table, what is the expected count when the observed is 15? Round answer to the nearest hundredth. |
Algebraic Expression:
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D)
Refer the two way table, what is the expected count when the observed is 2? Round answer to the nearest hundredth. |
Algebraic Expression:
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E)
What is the degrees of freedom? |
Algebraic Expression:
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F)
What is the chi-squared test statistic? Round answer to the nearest hundredth. |
Algebraic Expression:
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1) Enter observed counts into L1 2) Enter expected counts into L2 3) Highlight L3, then enter formula: (L1 - L2)2/L2 4) Enter 1-VAR STATS L3. 5) χ2 = Σx VALUE |
Answer = 1.01 |
G)
What is the p-value? Round answer to the nearest hundredth.
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Algebraic Expression:
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Hints: |
p-value = x2cdf(1.01, 10000, 3) Note: Use following format for calculating the p-value as chi-squared tests are always right tailed. x2cdf( x2 Value, Large Number (10000 or larger) , degrees of freedom ) |
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H)
Choose the most appropriate conclusion: |
Multiple Choice:
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I)
What condition or conditions have not been satisfied for this test? |
Ungraded Open Response: |
5) Problem #PRAJWY9 "PRAJWY9 - 257268 - This data is from..." |
A)
The data collected reflects the respiratory status of patients recruited for a randomised clinical multicenter trial. In each of two centres, eligible patients were randomly assigned to treatment or placebo. During the treatment, the respiratory status (categorised as poor or good ) was determined at each of four, monthly visits. The trial recruited 111 participants (54 in the treatment group, 57 in the placebo group).The data below is a two table that represents the effect of treatment on respiratory status on the fourth and last monthly visit. Does the data support the claim that the treatment type results in a difference in respiratory status after 4 months? Which of the following tests would be most appropriate? © STATS4STEM.ORG |
Multiple Choice:
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B)
Define the null and alternative hypothesis. A) Ho: There is no association between respiratory status and treatment Ha: There is an association between respiratory status and treatment B) Ho: There is an association between respiratory status and treatment Ha: There is no association between respiratory status and treatment C) p represents the proportion of the population that have a respiratory status rated as good. Ho: Pplacebo = ptreatment Ha: The proportions differ D) p represents the proportion of the population that have a respiratory status rated as good. Ho: Pplacebo = ptreatment Ha: Pplacebo < ptreatment |
Multiple Choice:
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C)
Refer the two way table, what is the expected count when the observed is 32? Round answer to the nearest hundredth. |
Algebraic Expression:
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D)
Refer the two way table, what is the expected count when the observed is 34? Round answer to the nearest hundredth. |
Algebraic Expression:
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E)
What condition or conditions need to be checked before proceeding with the hypothesis test? Click all that apply. |
Check All That Apply:
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F)
Are all conditions met? |
Multiple Choice:
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G)
What is the degrees of freedom for this problem? |
Algebraic Expression:
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Hints: |
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H)
What is the chi-squared test statistic? Round answer to the nearest hundredth. |
Algebraic Expression:
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Hints: |
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1) Enter observed counts into L1 2) Enter expected counts into L2 3) Highlight L3, then enter formula: (L1 - L2)2/L2 4) Enter 1-VAR STATS L3. 5) χ2 = Σx VALUE Using R/Rweb: Use the following code as a guide: obs=c(1, 2, 3) ### Enter your observed counts exp=c(4, 5, 6) ### Enter your expected counts chi.statistic=sum( (obs-exp)^2 / exp ) ### Calculate your chi-squared statistic chi.statistic |
χ2 = 4.06 |
I)
Using either the TI-83 or R/Rweb, calculate the p-value. Round answer to the nearest hundredth.
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Algebraic Expression:
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Hints: |
Click 2nd -> VARS -> X2cdf χ2cdf(4.06, 10000, 1) Note: Use following format for calculating the p-value as chi-squared tests are always right tailed. χ2cdf( χ2 Value, Large Number (10000 or larger) , degrees of freedom ) R/Rweb: pchisq(4.06, 1, lower.tail = FALSE) Note: Use following format for calculating the p-value: pchisq( χ2 Value, df , lower.tail = FALSE ) |
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J)
Assume that the significance level is 0.05, the best conclusion would be: A) Reject the null hypothesis. B) I calculated a p-value of 0.04, which is less than the stated significance level of 0.05. Therefore, we reject the null hypothesis. The data supports the claim that the proportion of individuals that have a respiratory status of "good" differs between the placebo and treatment populations. C) I calculated a p-value of 0.04, which is less than the stated significance level of 0.05. Therefore, we can reject the null hypothesis. D) I calculated a p-value of 0.04, which is less than the stated significance level of 0.05. Therefore, we fail to reject the null hypothesis. The data supports the claim that the proportion of individuals that have a respiratory status of "good" differs between the placebo and treatment populations. E) I calculated a p-value of 0.04, which is less than the stated significance level of 0.05. Therefore, we fail to reject the null hypothesis. The data does not support the claim that the proportion of individuals that have a respiratory status of "good" differs between the placebo and treatment populations. F) I calculated a p-value of 0.04, which is less than the stated significance level of 0.05. Therefore, we fail to reject the null hypothesis. |
Multiple Choice:
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6) Problem #PRAKK76 "PRAKK76 - 277663 - A homeowner just ..." |
A)
A homeowner just installed a new fireplace in his house and had a cord of wood (128 ft3) delivered from a local tree farm. For each cord of wood, the farm advertised the following percentages for each wood type:
When the homeowner received his 1 cord shipment of wood, he organized the wood and recorded how many cubic feet (ft3) of wood he received for each wood type. The data is found below:
The homeowner felt as though the distribution didn't represent the advertised distribution. He went about using a little statistics to prove that the company's advertisement is false. Which of the follow statistical tests would be most appriate in this situation? © STATS4STEM.ORG |
Multiple Choice:
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B)
Define the null and alternative hypothesis. A) Ho: There is no association between wood type and amount of wood received Ha: There is an association between wood type and amount of wood received B) Ho: There is an association between wood type and amount of wood received Ha: There is no association between wood type and amount of wood received C) Ho: The distribution of wood delivered equals the advertised distribution Ha: The distribution of wood delivered does not equal the advertised distribution D) Ho: The distribution of wood delivered does not equal the advertised distribution Ha: The distribution of wood delivered equals the advertised distribution |
Multiple Choice:
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C)
What is the expected count for the apple wood type? Round answer to the nearest hundredth. |
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D)
What is the expected count for the Hemlock wood type? Round answer to the nearest hundredth. |
Algebraic Expression:
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E)
What condition or conditions need to be checked before proceeding with the hypothesis test? Click all that apply. |
Check All That Apply:
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F)
Are all conditions met? |
Multiple Choice:
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G)
What is the degrees of freedom for this problem? |
Algebraic Expression:
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df = 5 - 1 df = 4 |
H)
In order for one to reject the null hypothesis, using a significance level of 0.05, the chi-squared test statistic must found to be _________________ than _________________? Link to Χ2 table. The most appropriate answers for the blanks above are: |
Multiple Choice:
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I)
What is the chi-squared test statistic? Round answer to the nearest hundredth. |
Algebraic Expression:
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1) Enter observed counts into L1 2) Enter expected counts into L2 3) Highlight L3, then enter formula: (L1 - L2)2/L2 4) Enter 1-VAR STATS L3. 5) χ2 = Σx VALUE Using R/Rweb: Use the following code as a guide: obs=c(1, 2, 3) ### Enter your observed counts exp=c(4, 5, 6) ### Enter your expected counts chi.statistic=sum( (obs-exp)^2 / exp ) ### Calculate your chi-squared statistic chi.statistic |
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J)
Using either the TI-83 or R/Rweb, calculate the p-value. Round answer to the nearest hundredth.
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Algebraic Expression:
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Hints: |
Click 2nd -> VARS -> X2cdf χ2cdf(1.79, 10000, 4) Note: Use following format for calculating the p-value as chi-squared tests are always right tailed. χ2cdf( χ2 Value, Large Number (10000 or larger) , degrees of freedom ) R/Rweb: pchisq(1.79, 4, lower.tail = FALSE) Note: Use following format for calculating the p-value: pchisq( χ2 Value, df , lower.tail = FALSE ) |
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K)
Assume that the significance level is 0.05, the homeowner's best conclusion would be: A) I calculated a p-value of 0.77, which is greater than the stated significance level of 0.05. Therefore, we reject the null hypothesis. The data supports the claim that the distribution of wood delivered does not equal the advertised distribution. B) I calculated a p-value of 0.77, which is greater than the stated significance level of 0.05. Therefore, we can fail to reject the null hypothesis. C) I calculated a p-value of 0.77, which is greater than the stated significance level of 0.05. Therefore, we can reject the null hypothesis. D) I calculated a p-value of 0.77, which is greater than the stated significance level of 0.05. Therefore, we fail to reject the null hypothesis. The data fails to supports the claim that the distribution of wood delivered does not equal the advertised distribution. E) I calculated a p-value of 0.77, which is less than the stated significance level of 0.05. Therefore, we fail to reject the null hypothesis. The data proves that the distribution of wood delivered is equal the advertised distribution. F) I calculated a p-value of 0.77, which is greater than the stated significance level of 0.05. Therefore, we reject the null hypothesis. The data supports the claim that the distribution of wood delivered does not equal the advertised distribution. |
Multiple Choice:
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7) Problem #PRARJHG "PRARJHG - 257155 - The following bla..." |
A)
The quine dataset contains data collected from a study of children from Walgett, New South Wales, Australia. The children were classified by Culture, Age, Sex and Learner status and the number of days absent from school in a particular school year was recorded. In particular, this problem will review the relationship between the following two variables:Age - A ge Group: Primary ("F0" ), or forms "F1," "F2" or "F3" .Lrn - Learner Status: factor with levels Average or Slow learner, ("AL" or "SL" ).Review the table below to determine if there is an association between Age Group and Learner Status? Which test would be most appropriate? © STATS4STEM.ORG |
Multiple Choice:
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B)
Define the null and alternative hypothesis. |
Multiple Choice:
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Hints: |
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C)
Refer the two way table, what is the expected count when the observed is 19? Round answer to the nearest hundredth. |
Algebraic Expression:
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D)
Refer the two way table, what is the expected count when the observed is 31? Round answer to the nearest hundredth. |
Algebraic Expression:
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E)
What condition or conditions have not been satisfied for this hypothesis test? |
Ungraded Open Response: |
F)
What is the degrees of freedom? |
Algebraic Expression:
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Hints: |
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G)
Calculate the chi-squared test statistic. Which of the following answers is closest to your answer? |
Multiple Choice:
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Hints: |
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1) Enter observed counts into L1 2) Enter expected counts into L2 3) Highlight L3, then enter formula: (L1 - L2)2/L2 4) Enter 1-VAR STATS L3. 5) χ2 = Σx VALUE |
Answer = 42.71 |
H)
Calculate the p-value. Choose the p-value below that is closest to your calculated answer.
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Multiple Choice:
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Hints: |
p-value = x2cdf(42.71, 10000, 3) Note: Use following format for calculating the p-value as chi-squared tests are always right tailed. x2cdf( x2 Value, Large Number (10000 or larger) , degrees of freedom ) |
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I)
Choose the most appropriate conclusion: |
Multiple Choice:
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