1) Problem #PRAJBNX "PRAJBNX - 239650 - The average numbe..." |
The average number of calories consumed per day is 2000. The distribution is normal and 68% lies in the middle of the distribution between 1850 and 2150. What is the notation for this distribution? © STATS4STEM.ORG |
Multiple Choice:
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2150 - 2000 = 150 Therefore, we now know that the standard deviation is 150. |
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2) Problem #PRAE8BJ "PRAE8BJ - 147072 - A population has ..." |
A)
A population has a mean of 50 and a standard deviation of 6. (a) When n=16, what are the mean of the sampling distribution of x-bar? Round answer to the nearest hundredth. © STATS4STEM.ORG |
Algebraic Expression:
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Hints: |
Link to Rweb Link to code |
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B)
A population has a mean of 50 and a standard deviation of 6. (b) When n=16, what is the standard deviation of the sampling distribution of x-bar? Round answer to the nearest hundredth. |
Algebraic Expression:
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=6/4 =1.5 |
C)
A population has a mean of 50 and a standard deviation of 6. (c) When n=20, what is the mean of the sampling distribution of x-bar? Round answer to the nearest hundredth. |
Algebraic Expression:
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Hints: |
Link to Rweb Link to code |
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D)
A population has a mean of 50 and a standard deviation of 6. (d) When n=20, what is the standard deviation of the sampling distribution of x-bar? Round answer the nearest hundredth. |
Algebraic Expression:
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=6/4.4721 =1.34 |
3) Problem #PRAE9HX "PRAE9HX - 148231 - A ruler manufactu..." |
A)
A ruler manufacturing company has a machine that manufactures rulers that are normally distributed with a mean of 12 inches long and standard deviation of .01 inches. Find the probability that a randomly selected ruler will have a length greater than 12.005 inches. Answer as a decimal rounded to the nearest hundredth. Link to z-table © STATS4STEM.ORG |
Algebraic Expression:
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B)
Refering to the previous problem, the manufacturing engineer decides to randomly sample 5 rulers to ensure the product is conforming to quality standards. It is expected that the means of the samples will vary from sample to sample. This sampling distribution of the sample means will have which distribution? |
Multiple Choice:
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C)
What is the probability that a random sample of 5 rulers has a mean greater than 12.005? Answer as a decimal rounded to the nearest hundredths. |
Algebraic Expression:
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Therefore, |
z = 0.005/0.00447 z = 1.12 |
Therefore, P(x-bar > 12.005) = 1 - 0.8686 = 0.13 (ROUNDED) |
D)
Will the probability of getting a mean greater 12.005 increase or decrease if the engineer decides to increase the sample size from 5 to 10? |
Multiple Choice:
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E)
What is the probability that a random sample of 10 rulers has a mean between 12.003 and 12.006? Answer as a decimal rounded to the nearest hundredths. |
Algebraic Expression:
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z12.003 = (12.003 - 12) / .003162 = 0.9486 |
From z-table: P(x-bar < 12.006) - P(x-bar < 12.003) = 1.90 - 0.95 = .1424 |
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4) Problem #PRAE9HW "PRAE9HW - 148230 - The length of tim..." |
A)
The length of time it takes a person to play a certain video game is normally distributed with a mean of 20 minutes and standard deviation of 3 minutes. Find the probability it takes a random individual less than 16 minutes. Round answer to the nearest hundredth. Link to z-table. © STATS4STEM.ORG |
Algebraic Expression:
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B)
Now, what is the probability that a sample of 2 individuals will have an average time less than 16 minutes. Round answer to the nearest hundredth. |
Algebraic Expression:
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Therefore, |
z = -4/2.1213 z = -1.89 |
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C)
Now, what is the probability that a sample of 10 individuals will have an average time greater than 21.4 minutes. Round answer to the nearest hundredth. |
Algebraic Expression:
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Therefore, |
z = 1.4/0.949 z = 1.48 |
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D)
Now, what is the probability that a sample of 8 individuals will have an average time between 19 and 22 minutes. Answer as a decimal rounded to the nearest hundredth. |
Algebraic Expression:
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Therefore, |
z19 = (19 - 20)/(3/√8) |
z19 = -1/1.061 = -0.94 |
P(19 < X-BAR < 22) = 0.9706 - 0.1736 = 0.797 or 0.80 (ROUNDED) |
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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H)
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 |
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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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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