1) Problem #PRAFF5P "PRAFF5P - Confidence Intervals - sigma known - EJS" |
A)
A simple random sample (SRS) of 36 books in the local library is collected. The sample has a mean book length of 284 pages. Assume the distribution for the population is non-normal, and σ = 60 pages. We are going to calculate the 99% confidence interval of the mean number of pages for all books in the library. a) First, what is the number that represents the point estimate for this problem? © STATS4STEM.ORG |
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B)
What conditions must be met before constructing a confidence interval for μ? Check ALL that apply. |
Check All That Apply:
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C)
Now, what is z* for a 99% confidence interval. Hint: Go to the bottom of the t-table. Find the number directly above confidence interval you are looking for. This will usually be in the row marked z, z*, or z∞. For example, z* for a 80% confidence interval is 1.282. |
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Z* is 2.576 for a 99% confidence interval. You can see this if you look at the very bottom of the t-table. Find where it say 99%, then look at the number directly above this number. |
D)
Now, calculate the margin of error for this problem. Round answer to the nearest hundredths. |
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m = z*. |
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E)
Now, lets go ahead and calculate the 99% confidence interval. The inteval can be broken into a lower limit and an upper limit. (Lower Limit, Upper Limit) Where: Lower Limit = Point Estimate - margin of error Upper Limit = Point Estimate + margin of error What is the Lower Limit? Round answer to the nearest hundredths. |
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F)
Now, lets go ahead and calculate the 99% confidence interval. The inteval can be broken into a lower limit and an upper limit. (Lower Limit, Upper Limit) Where: Lower Limit = Point Estimate - margin of error Upper Limit = Point Estimate + margin of error What is the Upper Limit? Round answer to the nearest hundredths. |
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G)
The margin of error for this problem is 25.76. If you wish to hold the confidence level constant at 99%, calculate a new sample size n to reduce the margin of error to just 20. |
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2) Problem #PRAFB56 "PRAFB56 - 150749 - Confidence Intervals - sigma known" |
A)
A simple random sample (SRS) of 35 students taking a national exam has a mean of 82. The standard deviation of the population is assumed to be 15. We are going to calculate the 95% confidence interval of the mean test score of all students across the nation (the population of interest). a) First, what is the number that represent the point estimate for this problem? © STATS4STEM.ORG |
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B)
Now, what is z* for a 95% confidence interval. Link to t-table |
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C)
Now, calculate the margin of error for this problem. Round answer to hundredth. |
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m = z*. |
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D)
Now, lets go ahead and calculate the 95% confidence interval. The inteval can be broken into a lower limit and an upper limit. (Lower Limit, Upper Limit) What is the Lower Limit? Round answer to the nearest hundredths. |
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E)
Now, lets go ahead and calculate the 95% confidence interval. The inteval can be broken into a lower limit and an upper limit. (Lower Limit, Upper Limit) What is the Upper Limit? Round answer to the nearest hundredth. |
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F)
The margin of error for this problem is 4.97. If you wish to hold the confidence level constant at 95%, what would the new sample size need to be to reduce the margin of error to just 3. |
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new.n = 96.04 The new sample size is 97. You must round UP to the nearest whole number. If you round down, your margin of error will be too small. |
3) Problem #PRAFF5H "PRAFF5H - Confidence Intervals - sigma known - EJS" |
A)
A battery manufacturer is interested in knowing how long their batteries last. A simple random sample (SRS) of 50 batteries found that the mean lasting time for the batteries was 76 hours. It is estimated that σ = 5. The battery manufacturer wishes to calculate a 90% confidence interval of the mean lasting time (in hours) of the population. a) First, what is the number that represent the point estimate for this problem? © STATS4STEM.ORG |
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B)
Now, what is z* for a 90% confidence interval. Link to t-table |
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C)
Now, calculate the margin of error for this problem. Round answer to hundredth. |
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m = z*. |
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D)
Now, lets go ahead and calculate the 90% confidence interval. The inteval can be broken into a lower limit and an upper limit. (Lower Limit, Upper Limit) What is the Lower Limit? Round answer to the nearest hundredth. |
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E)
Now, lets go ahead and calculate the 90% confidence interval. The inteval can be broken into a lower limit and an upper limit. (Lower Limit, Upper Limit) What is the Upper Limit? Round answer to the nearest hundredth. |
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F)
The margin of error for this problem is 1.16. If you wish to hold the confidence level constant at 90%, what would the new sample size need to be to reduce the margin of error to just 0.75? |
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n is calculated to be 120.2678. The new sample size is 121. You must round UP to the nearest whole number. If you round down, your margin of error will be too small. |
4) Problem #PRAH9TK "PRAH9TK - 237872 - 237863 - Hypothesis Test - Hypothesis Test Assumptions - EJS" |
When conducting a hypothesis test, what is the first thing that we should assume? © STATS4STEM.ORG |
Multiple Choice:
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5) Problem #PRAH9TB "PRAH9TB - 237863 - Hypothesis Test - 1 Sample Mean - EJS" |
A)
A car manufacturing company claims that it's latest car model exceeds 43 miles per gallon (mpg) on the highway. The head of the legal department for the company informs those in marketing that this claim must be tested before putting this information in marketing materials. As a result, a SRS of 35 cars are tested. The average mpg on the highway was found to be 43.3 for the sample. Past experiments have shown that mpg for individual cars tend to follow a distribution that is normal with a standard deviation σ = 1 mpg. At a significance level of 0.10, has the company obtained evidence to support it's claim? For this problem, what would be considered the alternative hypothesis? A) Ha: μ < 43.3 B) Ha: μ > 43.3 C) Ha: μ < 43 D) Ha: μ > 43 E) Ha: x-bar < 43 F) Ha: x-bar > 43 G) Ha: x-bar < 43.3 H) Ha: x-bar > 43.3 © STATS4STEM.ORG |
Multiple Choice:
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B)
For this problem, what would be considered the null hypothesis? A) H0: μ = 43.3 B) H0: μ < 43.3 C) H0: x-bar = 43 D) H0: x-bar < 43 E) H0: μ = 43 F) H0: μ < 43 G) H0: x-bar = 43.3 H) H0: x-bar < 43.3 |
Multiple Choice:
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C)
For this hypothesis test, the null and alternative hypothesis are: H0: μ = 43 Ha: μ > 43 If we assume that the null is true, what is the mean of the sampling distribution for x-bar? |
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D)
If we assume that the null is true, what is the standard deviation of the sampling distribution for x-bar? Round answer to the nearest hundredth. |
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E)
What is the p-value for this problem? Round answer to the nearest hundredth. |
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F)
If the level of significance for this problem is 0.10, we can conclude to: |
Multiple Choice:
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G)
After a year of advertising that their company's new car had 43 mpg (highway), a lawsuit was filled by a consumer group claiming their was fall advertisement. The consumer group claimed that the car did not have a mpg rating of 43. The consumer group went to court and a judge explained to the group that the court would assume that the car had a mpg rating of 43 until the consumer group could show strong evidence against this claim. Which of the following best describe the null and alternative hypothesis? A) H0: μ = 43 Ha: μ > 43 B) H0: μ = 43 Ha: μ < 43 C) H0: μ = 43 Ha: μ ≠ 43 |
Multiple Choice:
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H)
The lawyers for the car company met with the manufacturing engineer who performed the original hypothesis test to explain the new lawsuit. The engineer quickly realized that the new hypothesis test was: H0: μ = 43 Ha: μ ≠ 43 The engineer then quickly did some mental math in his head and stated - "The p-value for this problems is .08" The lawyer (astonished at the engineers intelligence) asked how he could so quickly calculate a p-value for this hypothesis test. The engineer explained that all one needed to do is multiply the p-value previously calculated for a one-sided test by 2. In other words, to calculate the value of this two-sided test - just multiply .04 (previously calculated) by 2. For clarification, one sided tests include < or > in the alternative hypothesis (HENCE ONE SIDE). Whereas, two sided test include ≠ (HENCE TWO SIDES). ------------------------------------------- The engineer, not sure if the lawyer understood, gave him a simple quiz. He said: Let's say that I wished to test the claim that rulers manufactured by ABC Rulers, Inc was not 12 inches. I begin by conducting a SRS of 27 rules, and calculating a sample mean of 12.1 inches. It is known that individual rulers are normally distributed with a standard deviation of .3 inches. Please calculate the p-value for this situation. The lawyer answer correctly, he said the p-value (rounded to the nearest hundredth) is: |
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6) Problem #PRAH9TS "PRAH9TS - 237878 - If you are a rese..." |
If you are a researcher trying to support a hypothesis, this hypothesis is called the: © STATS4STEM.ORG |
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7) Problem #PRAH9TA "PRAH9TA - 237863 - Hypothesis Test - 1 Sample Mean - EJS" |
A)
Bottles of a particular brand of water is supposed to have 300 milliliters (mL) of nutrients. The amount varies from bottle to bottle because the precision of the machines is not too precise. The distribution is normal with a standard deviation σ = 3 mL. An inspector of the company suspects that the company is under-filling the bottles. The inspector took a sample of six bottles and the results are below in an effort to confirm his suspicion. 299.4, 297.7, 301, 298.9, 300.2, 297 What is the mean of this sample data set? Round answer to the nearest hundredths. © STATS4STEM.ORG |
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Rweb:> x=c(299.4, 297.7, 301, 298.9, 300.2, 297) Rweb:> xbar=mean(x) Rweb:> xbar [1] 299.0333 Answer is 299.03 (ROUNDED) |
B)
When conducting a hypothesis test, what is the first thing that we should assume? |
Multiple Choice:
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C)
For this problem, what would be considered the alternative hypothesis? A) Ha: μ < 299.03 B) Ha: μ > 299.03 C) Ha: μ < 300 D) Ha: μ > 300 E) Ha: x-bar < 300 F) Ha: x-bar > 300 G) Ha: x-bar < 299.03 H) Ha: x-bar > 299.03 |
Multiple Choice:
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D)
For this problem, what would be considered the null hypothesis? A) H0: μ = 299.03 B) H0: μ > 299.03 C) H0: x-bar = 300 D) H0: x-bar > 300 E) H0: μ = 300 F) H0: μ > 300 G) H0: x-bar = 299.03 H) H0: x-bar > 299.03 |
Multiple Choice:
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E)
For this hypothesis test, the null and alternative hypothesis are: H0: μ = 300 Ha: μ < 300 If we assume that the null is true, what is the mean of the sampling distribution for x-bar? |
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F)
If we assume that the null is true, what is the standard deviation of the sampling distribution for x-bar? Round answer to the nearest hundredth. |
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G)
What is the p-value for this problem? Round answer to the nearest hundredth. Link to z-table |
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H)
If the level of significance is 0.05, the most appropriate conclusion would be: |
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8) Problem #PRAH9TR "PRAH9TR - 237877 - As the p-value de..." |
As the p-value decreases, the strength of an argument to reject the null hypothesis: © STATS4STEM.ORG |
Multiple Choice:
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9) Problem #PRAJB74 "PRAJB74 - 240182 - If the p-value is..." |
If the p-value is less than the significance level. The correct conclusion is: © STATS4STEM.ORG |
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10) Problem #PRAJB75 "PRAJB75 - 240183 - As the p-value de..." |
As the p-value decreases. The strength against the null hypothesis: © STATS4STEM.ORG |
Multiple Choice:
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11) Problem #PRAJB76 "PRAJB76 - 240184 - A coffee mug manu..." |
A)
A coffee mug manufacturer calculated a 90% confidence interval for the average weight of a box of mugs to be: (21.3 lbs, 23.6 lbs). From this information, we can conclude: CHECK ALL THAT APPLY © STATS4STEM.ORG |
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We CANNOT conclude
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B)
A coffee mug manufacturer calculated a 95% confidence interval for the amount of coffee held in their new Jumbo coffee mugs to be: (19.3 oz, 21.4 oz). From this information, we can conclude: CHECK ALL THAT APPLY If the manufacturer were to repeat the process of constructing a confidence interval, but was to significantly increase the sample size, the new margin of error would be smaller than that calculated the first time. |
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C)
If one wanted to reduce the margin of error for a 90% confidence interval by half by holding all things constant except for the sample size, one would need to: |
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12) Problem #PRAJB77 "PRAJB77 - 240185 - As your confidenc..." |
As your confidence level increases for confidence intervals. If all other factors are held constant, we can conclude that the margin of error must __________________ . © STATS4STEM.ORG |
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13) Problem #PRAJEHS "PRAJEHS - 237863 - Hypothesis Test - 1 Sample Mean - EJS" |
A)
When conducting a hypothesis test, what is the first thing that we should assume? © STATS4STEM.ORG |
Multiple Choice:
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B)
A metal fabrication company was hired to manufacture metal strips that are 16 inches long. It is suspected that the length of the metal strips do not equal 16 inches. It is known that the the distribution for the manufactured length of individual strips is normal with a standard deviation σ = 1 inch. A quality inspector randomly samples 100 pieces and measures and records the lengths. The sample mean is calculated to be 16.18 inches.
For this problem, what would be considered the null and alternative hypothesis?A) H0: μ = 16.18 HA: μ > 16.18 B) H0: x-bar = 16.18 HA: x-bar ≠ 16.18 C) H0: μ = 16 HA: μ > 16 D) H0: x-bar = 16 HA: x-bar > 16 E) H0: μ = 16 HA: μ ≠ 16 F) Ho: μ = 16.18 HA: μ ≠ 16.18 |
Multiple Choice:
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C)
For this hypothesis test, the sample mean is 16.18 and the null and alternative hypothesis are: If we assume that the null is true, what is the mean of the sampling distribution for x-bar? |
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D)
If we assume that the null is true, what is the standard deviation of the sampling distribution for x-bar? Round answer to the nearest hundredth. |
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E)
What is the p-value for this problem? Round answer to the nearest hundredth. Link to z-table |
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F)
If we assume the level of significance is 0.05, we can: |
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14) Problem #PRAJEZR "PRAJEZR - 237863 - Hypothesis Test - 1 Sample Mean - EJS" |
A)
A laptop advertises that the average battery life is 12 hours. A consumer group tests to see if the average battery life is actually less than the stated battery life. It is known that the the distribution for the battery life of individual laptops is normal with a standard deviation of 2.3 hours. The consumer group randomly samples 36 laptops and they record the battery life of each laptop. The sample mean is calculated to be 11.44 hours.
For this problem, what would be considered the null and alternative hypothesis?A) H0: μ > 12 HA: μ < 12 B) H0: μ = 12 HA: μ < 12 C) H0: x-bar = 11.44 HA: x-bar ≠ 11.44 D) H0: x-bar = 12 HA: x-bar < 12 E) Ho: μ = 11.44 HA: μ < 11.44 F) H0: μ ≠ 12 HA: μ < 12 G) Ho: μ > 11.44 HA: μ < 11.44 © STATS4STEM.ORG |
Multiple Choice:
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B)
For this hypothesis test, the sample mean was calculated as 11.44 hours and the null and alternative hypothesis are: HO: μ = 12 HA: μ < 12 If we assume that the null is true, what is the mean of the sampling distribution for x-bar? |
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C)
If we assume that the null is true, what is the standard deviation of the sampling distribution for x-bar? Round answer to the nearest hundredth. |
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D)
What is the p-value for this problem? Round answer to the nearest hundredth. |
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E)
If we assume the level of significance is 0.05, we can: |
Multiple Choice:
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F)
If we assume the level of significance is 0.10, we can:
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G)
Calculate the margin of error for a confidence interval with a confidence level of 99%. Round answer to the nearest hundredth. Link to t-table |
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15) Problem #PRAJGPU "PRAJGPU - 244483 - A light bulb manu..." |
A light bulb manufacturer randomly samples 50 bulbes to prove the claim that their bulbs last more than 2500 hours. The sample mean was found to be 2612 hours. If the company wished to test their hypothesis at a 0.05 significance level, what would be the probability of a Type I Error? © STATS4STEM.ORG |
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16) Problem #PRAJGQR "PRAJGQR - Problem #244511" |
A)
In the criminal justice system, one is presumed innocent until proven guilty. What would be the correct Null and Alternative Hypothesis? A. Ho: Individual Is Innocent HA: Individual Is Guilty B. Ho: Individual Is Guilty HA: Individual Is Innocent © STATS4STEM.ORG |
Multiple Choice:
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B)
If an innocent person is convicted of a crime, what type of error is this? |
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If a guilty person is found to be "Not Guilty", what type of error is this?
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17) Problem #PRAJGQU "PRAJGQU - 244514 - The Port Authorit..." |
A)
The Port Authority of New York contracts with a rope manufacturer to make custom ropes to secure ships to their docks. The Port Authority requests 1000 ropes to be manufactured. They request that the mean breaking strength for the ropes to be greater than 10,000 lbs. An inexprienced worker at the manufacturing plant mistakenly sets a machine up to produce individual ropes with a breaking strength of 9,995 lbs. and a standard deviation of 25 lbs. Before the Port Authority will take delivery, they ask to inspect a random sample of 100 ropes. The entire lot of 1000 ropes will be rejected if the sample mean of the 100 ropes is found to be less than 10,000 lbs. What is the probability that the ropes are accepted? Round your answer to the nearest hundredth. © STATS4STEM.ORG |
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B)
Before the inspection begin, it is assumed by both parties that the ropes are acceptable. If the ropes pass inspection, this would be considered: |
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C)
A second manufacture is also hired to manufacter 1000 ropes. This manufacturer sets up the machine to produce ropes that have a breaking strength of 10,010 lbs with a standard deviation of 61 lbs. What is the probability that the ropes will be rejected? |
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D)
Before the inspection begin, it is assumed by both parties that the ropes are acceptable. If the ropes fails inspection, this would be considered:
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18) Problem #PRAJGUE "PRAJGUE - 244625 - The newest Nike s..." |
A)
The newest Nike soccer ball has a breaking strength of 600 pounds. The standard deviation of the population is 14 pounds. A researcher (a.k.a. Nacho) selects a sample of 80 soccer balls and finds that the average breaking strength is 598 pounds. Can one reject the claim that the breaking strength is 600 pounds? Find the P-value. Should the null hypothesis be rejected at alpha (α) = 0.01? Which is the correct answer? © STATS4STEM.ORG |
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B)
A year later, this same researcher tests the next installment in the Nike soccer ball franchise. The new ball has a breaking strength of 700 pounds. The standard deviation of the population is 11 pounds. He selects a sample of 100 balls and finds that the average breaking strength is 697pounds. Can one reject the claim that the breaking strength is 700 pounds? Find the P-value. Should the null hypothesis be rejected at α = 0.01? Which is the correct answer? |
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19) Problem #PRAJGU4 "PRAJGU4 - 244646 - Is it possible to..." |
A)
Is it possible to make a Type I Error when HA is true? © STATS4STEM.ORG |
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B)
To make a Type I error, your conclusion must have been to? |
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If HA is true and you fail to reject the null hypothesis, what type of error have you made? |
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20) Problem #PRAJGXU "PRAJGXU - Type Error #1" |
A farmer wants to test the hypothesis: Ho: µ=500 Ha: µ>500 where µ is the amount of corn with pest infestation in a given acre of land. The farmer wants to see if the amount of pests have increased over time. Assume that the actual µ=520, σ=25, n=10, and alpha=.01. Find the probability of Type I Error. Round your answer to the hundredth's place. © STATS4STEM.ORG |
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21) Problem #PRAJGXV "PRAJGXV - Hypothesis Tests 8" |
If you are given the following information, what would be the probability of making a type I error? H0: μ=8000 HA: μ>8000 n= 50 σ=500 μ=8000 α= 0.01 © STATS4STEM.ORG |
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22) Problem #PRAJGXW "PRAJGXW - Hyp Test #10" |
In a hypothesis test, which of the following will cause a decrease in β, the probability of making a type II error? © STATS4STEM.ORG |
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23) Problem #PRAJGXX "PRAJGXX - 244734 - Katy is thin..." |
A)
Katy is thinking of opening Perry's Car Dealership in a suburb of Boston. In order for the dealership to be successful and turn a profit, the average income of people living around the dealership (and the theoretical customers of the shop) must be $100,000. She looks at a random sample of 60 people living near the location where she hopes to open. σ has been shown to be $10,000. If she is using a 0.01 significance level, what is the probability of her making a Type I error? © STATS4STEM.ORG |
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B)
If the average income of a certain area is actually $103,000, we want to find how likely is Katy will make a Type II error? However, before we can calculate the probability of a Type II error, we first need to find x-bar that will be used to as the threshold to reject or fail to reject the null. Calculate this number and then round your answer nearest whole number. |
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C)
What is the probability that Katy makes a Type II error? Round your answer to the nearest hundredth. |
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D)
What is the power? Round your answer to the nearest hundredth. |
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24) Problem #PRAJH4S "PRAJH4S - 245845 - A 95% confidence ..." |
A 95% confidence interval indicates that: © STATS4STEM.ORG |
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25) Problem #PRAFB54 "PRAFB54 - 150747 " |
A transportation researcher calculates a 99% confidence interval on the mean commuting time for individuals living in Queens, NY to be (47.2 min, 58.1 min). Interpret the meaning of the calculated confidence interval. © STATS4STEM.ORG |
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26) Problem #PRAFJ9F "PRAFJ9F - 157578 - 157574 - ILL - Confidence intervals - means - sigma known" |
A)
A study of 40 English composition professors showed that they spent, on average, 12.6 minutes correcting a student's term paper. Find the 90% confident interval of the mean time for all composition papers when σ=2.5 minutes. What is the lower limit of your confidence interval? Answer as a decimal, and round answer to the nearest tenth. © STATS4STEM.ORG |
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B)
What is the upper limit of your confidence interval? Answer as a decimal, and round answer to the nearest tenth. |
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27) Problem #PRAFF7T "PRAFF7T - Confidence Intervals - EJS" |
As we seek to increase our confidence level (while keeping sample size constant), the margin of error: © STATS4STEM.ORG |
Multiple Choice:
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28) Problem #PRAJ739 "PRAJ739 - Confidence Intervals - sigma known - EJS" |
A)
A simple random sample (SRS) of 49 movies in a video store is collected. The sample has a mean length of 90 minutes. Assume the distribution for the population is non-normal, and σ = 20 minutes. We are going to calculate the 99% confidence interval of the mean number of pages for all books in the library. a) First, what is the number that represents the point estimate for this problem? © STATS4STEM.ORG |
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B)
What conditions must be met before constructing a confidence interval for μ? Check ALL that apply. |
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C)
Now, what is z* for a 99% confidence interval. Hint: Go to the bottom of the t-table. Find the number directly above confidence interval you are looking for. This will usually be in the row marked z, z*, or z∞. For example, z* for a 80% confidence interval is 1.282. |
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Z* is 2.576 for a 99% confidence interval. You can see this if you look at the very bottom of the t-table. Find where it say 99%, then look at the number directly above this number. |
D)
Now, calculate the margin of error for this problem. The margin of error is represented by the variable m, and the equation is: m = z*. Round answer to hundredths. |
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E)
Now, lets go ahead and calculate the 99% confidence interval. The inteval can be broken into a lower limit and an upper limit. (Lower Limit, Upper Limit) Where: Lower Limit = Point Estimate - margin of error Upper Limit = Point Estimate + margin of error What is the Lower Limit? Round answer to the nearest hundredths. |
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F)
Now, lets go ahead and calculate the 99% confidence interval. The inteval can be broken into a lower limit and an upper limit. (Lower Limit, Upper Limit) Where: Lower Limit = Point Estimate - margin of error Upper Limit = Point Estimate + margin of error What is the Upper Limit? Round answer to the nearest hundredths. |
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G)
The margin of error for this problem is 7.36. If you wish to hold the confidence level constant at 99%, calculate a new sample size n to reduce the margin of error to just 5. |
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29) Problem #PRAJ8HV "PRAJ8HV - Confidence Intervals - sigma known - EJS" |
A)
A simple random sample (SRS) of 54 adults was taken to see how long it takes to put on 1 shoe and tie the lace. The recorded data showed that the mean time was 11 seconds. Assume the distribution for the population is normal, and σ = 1.5 seconds. We are going to calculate the 95% confidence interval for the mean time in seconds that it takes adults to put on 1 shoe and tie the lace. a) First, what is the number that represents the point estimate , or statistic, for this problem? © STATS4STEM.ORG |
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B)
What conditions must be met before constructing a confidence interval for μ? Check ALL that apply. |
Check All That Apply:
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C)
Now, what is z* for a 95% confidence interval. Link to t-table |
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Z* is 1.960 for a 95% confidence interval. You can see this if you look at the very bottom of the t-table. Find where it say 95%, then look at the number directly above this number. |
D)
Now, calculate the margin of error for this problem. Round answer to nearest hundredths. |
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E)
Now, lets go ahead and calculate the 95% confidence interval. The inteval can be broken into a lower limit and an upper limit. (Lower Limit, Upper Limit) What is the Lower Limit? Round answer to the nearest hundredths. |
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F)
Now, lets go ahead and calculate the upper limit of the 95% confidence interval. Round answer to the nearest hundredths. |
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G)
Which of the following answers is the most appropriate interpretation of the 95% confidence interval? |
Multiple Choice:
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H)
If one holds the confidence level constant at 95%, but wishes to reduce the margin of error to just 0.35, calculate the required sample size n that would be required. |
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30) Problem #PRAKESW "PRAKESW - 272447 - True or False, &n..." |
A)
True or False, Power is the probability of rejecting the null hypothesis when in fact it is false. © STATS4STEM.ORG |
Multiple Choice:
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B)
Power = ? |
Multiple Choice:
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C)
Which of the following cell or cells represent a Type I Error? Check all that apply.
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D)
Which of the previous cell or cells represent a Type II Error? Check all that apply. |
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E)
Which of the previous cell or cells represent a correct conclusion? Check all that apply. |
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31) Problem #PRAP746 "PRAP746 - Problem #385202" |
There are 200 different researchers each studying the social media habits of high school seniors. Each researcher takes a random sample of size 100 from the same population of high school seniors. Each researcher is trying to estimate the mean hours of social media interaction during a school day, and each researcher constructs a 90% confidence interval for the mean. Approximately how many of these 200 confidence intervals will NOT capture the true mean? © STATS4STEM.ORG |
Multiple Choice:
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32) Problem #PRAQJZB "PRAQJZB - Problem #395654" |
A)
A researcher wished to estimate the amount of saturated fat in a bacon, egg, and cheese on biscuit sandwich from Dunkin Donuts. To do this, a researcher randomly sampled 16 sandwiches, and measured the saturated fat of each sandwich. The researcher then calculated the sample mean and found it to be 14.5 grams of saturated fat. From experience, the researcher knows that the standard deviation of the population is 0.6 grams. Using this information, the researcher concluded that the 95% Confidence Interval is: © STATS4STEM.ORG |
Multiple Choice:
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B)
Dunkin Donuts claims that the amount of saturated fat in a bacon, egg, and cheese on biscuit sandwich is 14 grams. The researcher feels that this claim is untrue. Which notation below best represents what the researcher's hypothesis. A) H0: μ = 14 grams Ha: μ < 14 grams B) H0: μ = 14 grams Ha: μ ≠ 14 grams C) H0: μ = 14 grams Ha: μ > 14 grams D) None of the above |
Multiple Choice:
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C)
The researcher assumes a significance level of 0.05. Based on the calculated confidence interval, the researcher may conclude that: |
Multiple Choice:
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33) Problem #PRAQYDW "PRAQYDW - Problem #407576" |
Two history professors are going to take a sample of data from the same population of history students. Professor A will draw a simple random sample from all students taking history. Professor B's sample will consist only of the students in his history class. Both researchers will construct a 99% confidence interval for the mean score on the history mid-year exam using their own sample data. Which researcher can be 99% confident in capturing the true mean of the population of all students taking history? |
Multiple Choice:
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34) Problem #PRAQYTY "PRAQYTY - Problem #407981" |
Which of the following methods could be used to decrease the width of a confidence interval for a mean, if all else stays the same. Check all that apply. |
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35) Problem #PRAQYTZ "PRAQYTZ - Problem #407982" |
When conducting a hypothesis test, there are different errors that may be made, depending on your final conlusion. For example, if you mistakenly conclude to reject the null hypothesis, when in fact you should have failed to reject the null hypothesis, what type of error have you made? |
Multiple Choice:
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36) Problem #PRAQYT2 "PRAQYT2 - Problem #407983" |
A)
A 95% confidence interval for the mean achievement score on a standardized math test for a population of 5th grade students is (82, 86). If the confidence level was increased to 99% and all other factors remain constant, the margin of error would: |
Multiple Choice:
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B)
Confidence intervals have the following general formula: statistic ± (critical value)*(standard deviation of the statistic) For the previous question, the 99% confidence level has a larger width than the 95% confidence interval due to: |
Multiple Choice:
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C)
After reviewing the methodology for calculating the 95% confidence interval, it was reported that the sample size was 24 and the standard deviation of the population was assumed to be 1. From this new information, one can conclude the following critical value was used to calculate the 95% confidence interval: Link to t-table |
Multiple Choice:
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