Stats Exam 3 Practice
The recommended daily allowance of iron for females aged 19-50 is 18 mg/day. A dietitian believes that elderly women (on average) get less than 18 mg/day. The dietitian uses hypothesis testing to check this belief. In this scenario, a Type I error would be: (a) Deciding that elderly women get less than the recommended allowance when they don't. (b) Deciding that elderly women get at least the recommended allowance when they don't.
(a)
The target temperature for a hot beverage the moment it is dispensed from a vending machine is 170°F. A consumer organization believes that a particular brand of machine dispenses beverages whose temperature exceeds 170°F. In this scenario, a Type 2 error would be: (a) Deciding that the beverages do not exceed 170°F when they do. (b) Deciding that the beverages exceed 170°F when they don't.
(a)
The target temperature for a hot beverage the moment it is dispensed from a vending machine is 170°F. A consumer organization believes that a particular brand of machine dispenses beverages whose temperature exceeds 170°F. In this scenario, a Type I error would be: (a) Deciding that the beverages do not exceed 170°F when they do. (b) Deciding that the beverages exceed 170°F when they don't.
(b)
Which of the following statements are true in hypothesis testing? (i) If we reject Ho when Ho is in fact true, we made Type I error. (ii) If we reject Ho when Ho is in fact false, we made Type II error. (iii) One will reject Ho if P-value is smaller than the significance level alpha.
(i) and (iii)
Which of the following statements are true in hypotheses testing? (i) If we reject Ho when Ho is in fact true, we made Type I error. (ii) If we fail to reject Ho when Ho is in fact true, we made Type II error. (iii) One will reject Ho if P-value is larger than the significance level alpha. (iv) The p-value is the probability that the null hypothesis is true. (v) The significance level alpha is the probability of making a Type I error.
(i) and (v)
Which of the following statements are true in hypotheses testing? (i) If we fail to reject Ha when Ha is in fact true, we made Type I error. (ii) If we fail to reject Ho when Ho is in fact false, we made Type II error. (iii) One will reject Ho if P-value is smaller than the significance level alpha. (iv) The p-value is the probability that the null hypothesis is true.
(ii) and (iii)
Which of the following statements are true in hypotheses testing? (i) If we fail to reject Ha when Ha is in fact true, we made Type I error. (ii) If we fail to reject Ho when Ho is in fact false, we made Type II error. (iii) One will reject Ho if P-value is smaller than the significance level α (iv) The p-value is the probability that the null hypothesis is true.
(ii) and (iii)
A forensic anthropologist claims that 80% of female skeletons have a sub-pubic angle less than the often-cited 90 degrees. She randomly selects 200 skeletons and finds that 150 have an angle less than 90 degrees. Use alpha=0.10 to test the claim. Which function in Excel finds the test statistic? (Note: 150/200 = 0.75)
= (0.75 - 0.80)/SQRT(0.80*0.20/200)
In a random sample of 200 adults, 136 say they are in favor of outlawing cigarettes in certain areas. Let p be the proportion of all adults who are in favor of outlawing cigarettes. One is interested in the following hypotheses: H0: p=0.6 vs Ha: p>0.6 Which of the following excel function calculates the standardized test statistic?
=(136/200-0.6)/sqrt(0.6*0.4/200)
In a random sample of 200 adults, 136 say they are in favor of outlawing cigarettes in certain areas. Let p be the proportion of all adults who are in favor of outlawing cigarettes. One is interested in the following hypotheses: Ho: p = 0.6 Ha: p > 0.6 Which of the following excel function calculates the standardized test statistic?
=(136/200-0.6)/sqrt(0.6*0.4/200)
A fashion company releases a new line of prom dresses that are all under $100. They claim this is a bargain because the average cost of a prom dress is over $300. A customer decides to test this and takes a random sample of 25 dresses. She finds the mean is $250 with a standard deviation of $10. Assume the population is normally distributed and use α =0.05 to test the claim. Which function in Excel finds the test statistic?
=(250-300)/(10/SQRT(25))
Over the past few years, the proportion of American adults who use tobacco products has been 22%. A researcher wishes to test whether the proportion is greater now. He takes a random sample of 2000 adults and finds that 394 use tobacco products. Test the researcher's claim at a 10% level of significance. Which of the following excel function calculates the standardized test statistic?
=(394/2000-0.22)/sqrt(0.22*0.78/2000)
In a random survey of 1000 people in the United States, 720 said that they prepare and file their income taxes before April 15th. Let � be the true proportion of people in the United States prepare and file their income taxes before April 15th. One wants to test the following hypothesesHo:p=0.70 vs Ha:p>0.70. Which of the following excel function calculates the standardized test statistic?
=(720/1000-0.7)/sqrt(0.7*0.3/1000)
Over the past few years, the proportion of American adults who use tobacco products has been 22%. A researcher wishes to test whether the proportion is greater now. He takes a random sample of 2000 adults and finds that 394 use tobacco products. Test the researcher's claim at a 10% level of significance. Suppose the test statistic is -2.483.Which function in Excel finds the P-value?
=1-NORM.DIST(-2.483, 0, 1, TRUE)
An English professor is studying the use of semicolons over time. She estimates that in the Georgian era, authors used more than 8 semicolons per page. It is well known in her field that the standard deviation of semicolons in this era is 2. She randomly selects 25 pages from different books and finds the average amount of semicolons is 8.5. Assume the population is normally distributed and use alpha=0.05 to test the claim. Also suppose that the test statistic equals 0.85. Which formula in Excel finds the p-value?
=1-NORM.DIST(0.85, 0, 1, TRUE)
In the test of hypothesis H0:μ=100vs Ha:μ≠100. A sample of size 250 yields the standard test statistic z=1.00. Which of the following excel function calculates the P-value for this test?
=2*NORM.DIST(-1, 0, 1, TRUE)
In the test of hypothesis Ho:u=100 vs Ha:u≠100. A sample of size 250 yields the standard test statistic z=1.00. Which of the following excel functions calculates the P-value for this test?
=2*NORM.DIST(-1, 0, 1, TRUE)
In the test of hypothesis Ho:u=100 vs Ha:u≠100. A sample of size 250 yields the standard test statistic z=1.52. Which of the following excel function calculates the P-value for this test?
=2*NORM.DIST(-1.52, 0, 1, TRUE)
In the test of hypothesis Ho:u=100 vs Ho:u≠100. A sample of size 250 yields the standard test statistic z=1.52. Which of the following excel function calculates the P-value for this test?
=2*NORM.DIST(-1.52, 0, 1, TRUE)
A diabetic claims that the average cost of insulin per year for a Type 1 diabetic is $5,705. She takes a sample of 100 Type 1 diabetics and finds their average cost is $5, 912 with a standard deviation of $300. Use alpha=0.05 to test the claim. Suppose the test statistic is 1.65. Which function in Excel finds the p-value?
=2*NORM.DIST(-1.65, 0, 1, TRUE)
A diabetic claims that the average cost of insulin per year for a Type 1 diabetic is $5,705. She takes a sample of 100 Type 1 diabetics and finds their average cost is $5, 912 with a standard deviation of $300. Use α=0.05 to test the claim. Suppose the test statistic is 1.65.Which function in Excel finds the p-value?
=2*NORM.DIST(-1.65, 0, 1, TRUE)
A forensic anthropologist claims that 80% of female skeletons have a sub-pubic angle less than the often-cited 90 degrees. She randomly selects 200 skeletons and finds that 150 have an angle less than 90 degrees. Use alpha=0.10 to test the claim. Suppose the test statistic is -2.1. Which function in Excel finds the p-value?
=2*NORM.DIST(-2.1, 0, 1, TRUE)
A veterinarian reads that 15% of dogs are allergic to chicken. However, he claims that the actual proportion is less than this. He randomly selects 15 of his customers and asks them if their dog is allergic to chicken. 3 of them say yes. Assume the population is normally distributed and use alpha=0.01. Suppose the test statistic is -1.11. Which function in Excel finds the p-value?
=NORM.DIST(-1.11, 0, 1, TRUE)
A veterinarian reads that 15% of dogs are allergic to chicken. However, he claims that the actual proportion is less than this. He randomly selects 15 of his customers and asks them if their dog is allergic to chicken. 3 of them say yes. Assume the population is normally distributed and use α =0.01 Suppose the test statistic is -1.11. Which function in Excel finds the p-value?
=NORM.DIST(-1.11, 0, 1, TRUE)
A nutritionist claims that the average amount of sugar in a 16 oz soda is at least 50 g. He randomly samples 10 sodas and finds they contain an average of 54 g of sugar with a standard deviation of 3 g. Assume the population is normally distributed and use alpha=0.10 to test the claim. Suppose the test statistic is -1.22. Which function in Excel finds the p-value?
=T.DIST(-1.22, 9, TRUE)
A sociologist wishes to test H0:μ=42 vs. Ha:μ<42. The sociologist takes a sample of size 10 and calculates a standardized test statistic of -2.34. To calculate a p-value for the test in Excel, the sociologist should use:
=T.DIST(-2.34, 9, TRUE)
A sociologist wishes to test Ho:u=42 vs. Ha:u<42. The sociologist takes a sample of size 10 and calculates a standardized test statistic of -2.34. To calculate a p-value for the test in Excel, the sociologist should use:
=T.DIST(-2.34, 9, TRUE)
Some have argued that throwing darts at the stock pages to decide which companies to invest in could be a successful stock-picking strategy. Suppose a researcher decides to test this theory and randomly chooses 50 companies to invest in. After 1 year, 26 of the companies were considered winners; that is, they outperformed other companies in the same investment class. To assess whether the dart-picking strategy resulted in a majority of winners, the researcher tested the following: Ho : p = 0.50 versus Ha : p > 0.50 We obtained a P-value of 0.3886. Explain what this P-value means.
About 39 in 100 samples will give a sample proportion as high or higher than the one obtained if the population proportion really is 0.5.
A random sample of 125 registered voters in Phoenix is asked if they favor the use of oxygenated fuels year-round to reduce air pollution and 86 voters responded positively. The city office claims that more than 65% of voters in Phoenix use oxygenated fuels. Test this claim at a 5% level of significance. (a) Determine the suitable null and alternative hypotheses. (b) Which of the following excel function calculates the test statistic z? (c) Suppose that the test statistic for this test is z=0.8907. Which of the following excel function calculates the p-value of the test? (d) Suppose the p-value is calculated to be 0.1865. At the 0.05 significance level α, the correct decision is
Answer 1: Ho : p = 0.65 versus Ha : p > 0.65 Answer 2: =(86/125-0.65)/sqrt(0.65*0.35/125) Answer 3: =1-NORM.DIST(0.8907, 0, 1, TRUE) Answer 4: Fail to reject the null hypothesis; We cannot conclude that more than 65% of voters in Phoenix use oxygenated fuels.
A moving company claims that more than 90% of its customers are happy with its service. To test this claim the company surveys 500 customers and finds that 465 of them are actually happy with the service. Note that 465/500 = 0.93. (a) Set up the null and alternative hypotheses to test for the company's claim. (b) Which of the following excel function calculates the test statistic z? (c) Suppose that the test statistic for this test is z=2.2361. Which of the following excel function calculates the p-value of the test? (d) Suppose that the P-value for this test is 0.0127. At the significance level of 0.01, will you conclude that the company's claim is not plausible? Choose your decision with the correct reason.
Answer 1: Ho : p = 0.90 versus Ha : p > 0.90 Answer 2: =(0.93-0.90)/sqrt(0.90*0.10/500) Answer 3: =1-NORM.DIST(2.2361, 0, 1, TRUE) Answer 4: Yes, since p-value is greater than 0.01
The mean waiting time at the drive-through of a fast food restaurant from the time an order is placed to the time the order is received is 87.5 seconds. A manager devises a new drive-through system that she believes will decrease the wait time. As a test, she initiates the new system at her restaurant and measures the wait time for 10 randomly selected orders. The wait times are provided in the table below: CHART Based on this sample, the average wait time is 79 seconds and the standard deviation is 14.4838 seconds. Suppose the wait time is normally distributed. Is the new system effective? Use 1% as the level of significance. (a) Set up the null and alternative hypotheses to test whether the new system is effective. (b)...
Answer 1: Ho: μ = 87.5 vs Ha: μ < 87.5 Answer 2: t=(79-87.5)/(14.4838/sqrt(10)) Answer 3: =T.DIST(-1.8558, 9, TRUE) Answer 4:Fail to reject the null hypothesis; the new system is ineffective.
A test is made of H0: μ = 20 versus Ha: μ ≠ 20. Suppose the true value of μ is 25, and H0 is rejected. Determine whether the outcome is a Type I error, a Type II error, or a correct decision.
Correct Decision
In the test of hypothesis H0:μ=100 vs . Ha:μ≠100, we already have the P-value for this test: P-value=0.025. Which of the following (with the reason) will be correct with significance level at .01?
Fail to reject H0 since the P-value is greater than the significance level .01
In the test of hypothesis Ho:u=100 vs . Ha:u>100, we already have the p-value for this test: P-value=0.0228. Which of the following (with the reason) will be correct with a significance level at .01?
Fail to reject Ho since the P-value is greater than .01.
To test Ho : µ = 103 versus Ha : µ ≠ 103, a simple random sample of size n = 35 is obtained. We obtained a P-value of 0.004. Explain what this P-value means.
If 1000 random samples of size n = 35 are obtained, about 4 samples are expected to result in a mean as extreme or more extreme than the one observed if µ = 103.
A college football coach records the mean weight that his players can bench press as 275 pounds, with a standard deviation of 55 pounds. Three of his players thought that the mean weight was more than that amount. They asked 30 of their teammates for their estimated maximum lift on the bench press exercise. He performed a hypothesis test Ho : µ = 275 versus Ha : µ > 275 with α = 0.025 and came up with a P-Value of 0.1331. Explain what this P-value means.
If the mean weight that his players can bench press really is 275 pounds and if we randomly collect a sample of size n = 30 players repeatedly, then approximately 13% of the samples will result in a sample mean as high or higher than the one obtained.
Several years ago, the mean height of women 20 years of age or older was 63.7 inches. Suppose that a random sample of 45 women who are 20 years of age or older today results in a mean height of 64.8 inches. Suppose a researcher wants to assess whether women are taller today. She performed a hypothesis test Ho : µ = 63.7 versus Ha : µ > 63.7 with α = 0.10 and came up with a P-Value of 0.15. Explain what this P-value means.
If, in fact, the mean height of women 20 years of age or older was 63.7 inches and if we sample the same number of women repeatedly, then 15% of the samples will result in a sample mean as high or higher than the one obtained.
Suppose a consumer group suspects that the proportion of households that have three cell phones is 45%. A cell phone company has reason to believe that the proportion is not 45%. Before they start a big advertising campaign, they conduct a hypothesis test. Their marketing people survey 150 households with the result that 47 of the households have three cell phones. The null and alternative hypotheses are as follows: H0 : p = 0.45 Ha : p ≠ 0.45 If the marketing people performed the hypothesis testing using alpha =8%, interpret what this significance level means in the context of the problem.
If, in fact, the proportion of households that have three cell phones is 45%, then there is an 8% chance that the marketing people will conclude that the proportion of households that have three cell phones is not 45%.
Suppose a consumer group suspects that the proportion of households that have three cell phones is 45%. A cell phone company has reason to believe that the proportion is not 45%. Before they start a big advertising campaign, they conduct a hypothesis test. Their marketing people survey 150 households with the result that 47 of the households have three cell phones. The null and alternative hypotheses are as follows: H0 : p = 0.45 and Ha : p ≠ 0.45. If the marketing people performed the hypothesis testing using �=8%, interpret what this significance level means in the context of the problem.
If, in fact, the proportion of households that have three cell phones is 45%, then there is an 8% chance that the marketing people will conclude that the proportion of households that have three cell phones is not 45%.
In a clinical trial, 23 out of 838 patients taking a prescription drug daily complained of flu-like symptoms. Suppose that it is known that 2.3% of patients taking competing drugs complain of flu-like symptoms. A researcher claims that more than 2.3% of this drug's users experience flu-like symptoms as a side effect. The null and alternative hypotheses are as follows: H0 : p = 0.023 Ha : p > 0.023 If the researcher performed the hypothesis testing using alpha=1%, interpret what this significance level means in the context of the problem.
If, in fact, the proportion of patients taking this drug who experience flu-like symptoms is 2.3%, then there is a 1% chance that the researcher will conclude that the proportion of patients taking this drug who experience flu-like symptoms is more than 2.3%.
A market research analyst claims that 32% of the people who visit the mall actually make a purchase. You think that less than 32% buy something and decide to test the claim. You stand by the exit door of the mall starting at noon and ask 82 people as they are leaving whether they bought anything. You find that only 20 people made a purchase. The null and alternative hypotheses are as follows: H0 : p = 0.32 Ha : p < 0.32 If the analyst performed the hypothesis testing using α=12%, interpret what this significance level means in the context of the problem.
If, in fact, the proportion of the people who visit the mall actually make a purchase is 32%, then there is a 12% chance that the analyst will conclude that the proportion of the people who visit the mall actually make a purchase is less than 32%.
The null and alternative hypotheses are given. Determine whether the hypothesis test is left-tailed, right-tailed, or two-tailed (Ho: μ ≥ 4 Ha: μ < 4)
Left-Tailed Test
The null and alternative hypotheses are given. Determine whether the hypothesis test is left-tailed, right-tailed, or two-tailed. Ho:o=10 and Ha:o<10
Left-tailed test
If a null hypothesis is rejected at the significance level alpha=0.001, is it possible that the null hypothesis is not rejected if the same test was done at the significance level alpha= .01 (with everything else staying the same)?
No
In the test of hypothesis Ho:p=0.56 vs . Ho:p≠0.56, we already have the p-value for this test: P-value = 0.00000187. Which of the following (with the reason) will be correct with a significance level of 0.05?
Reject Ho since the P-value is less than the significance level.
In the test of hypothesis Ho:u=100 vs. Ha:u>100, we already have the p-value for this test: P-value=0.0228. Which of the following (with the reason) will be correct with significance level at .05?
Reject Ho since the P-value is less than the significance level.
The null and alternative hypotheses are given. Determine whether the hypothesis test is left-tailed, right-tailed, or two-tailed. Ho: p≤ 0.45 Ha: p > 0.45
Right tailed test
The null and alternative hypotheses are given. Determine whether the hypothesis test is left-tailed, right-tailed, or two-tailed. Ho:μ≤4 Ha: Mew>4
Right-tailed test
A fashion company releases a new line of prom dresses that are all under $100. They claim this is a bargain because the average cost of a prom dress is over $300. A customer decides to test this, and takes a random sample of 25 dresses. She finds the mean is $250 with a standard deviation of $10. Assume the population is normally distributed and use α=0.05 to test the claim. Which distribution should be used to test the claim?
T-test for the Mean
A nutritionist claims that the average amount of sugar in a 16 oz soda is at least 50 g. He randomly samples 10 sodas and finds they contain an average of 54 g of sugar with a standard deviation of 3 g. Assume the population is normally distributed and use alpha=0.10 to test the claim. Which distribution should be used?
T-test for the Mean
Which one of the following is true about β?
The probability of failing to reject the null hypothesis when it is false.
Which one of the following is true about the significance level alpha?
The probability of rejecting the null hypothesis when it is true.
Suppose that the null hypothesis was rejected. State the conclusion based on the results of the test. According to the Federal Housing Finance Board, the mean price of a single-familyhome two years ago was $299,500. A real estate broker believes that because of the recent credit crunch, the mean price has decreased since then.
There is sufficient evidence to conclude that the mean price of a single-family home has decreased from its level two years ago of $299,500.
Suppose the null hypothesis is rejected. State the conclusion based on the results of the test. Three years ago, the mean price of a single-family home was $243,732. A real estate broker believes that the mean price has increased since then.
There is sufficient evidence to conclude that the mean price of a single-family home has increased
Suppose that the null hypothesis is rejected. State the conclusion based on the results of the test. According to the report, the standard deviation of monthly cell phone bills was $49.75 three years ago. A researcher suspects that the standard deviation of monthly cell phone bills is higher today.
There is sufficient evidence to conclude that the standard deviation of monthly cell phone bills is higher than its level three years ago of $49.75.
The null and alternative hypotheses are given. Determine whether the hypothesis test is left-tailed, right-tailed, or two-tailed. (Ho: p=0.1 and Ha: p ≠ 0.1)
Two tailed test
If a null hypothesis is rejected at the significance level alpha=0.02, is it possible that the null hypothesis is not rejected if the same test was done at the significance level alpha = .002 (with everything else staying the same)?
Yes
If a null hypothesis is rejected at the significance level α=0.05, is it possible that the null hypothesis is not rejected if the same test was done at the significance level α = .025 (with everything else staying the same)?
Yes
If the null hypothesis is rejected at the significance level alpha=0.05, is it possible that the null hypothesis is not rejected if the same test was done at the significance level alpha=0.01 (with everything else staying the same)?
Yes, it is possible
A diabetic claims that the average cost of insulin per year for a Type 1 diabetic is $5,705. She takes a sample of 100 Type 1 diabetics and finds their average cost is $5, 800 with a standard deviation of $575. Use Alpha=0.05 to test the claim. Which distribution should be used?
Z-distribution
Total blood volume (in ml) per body weight is important in medical research. For healthy adults, the red blood cell volume mean is about 28 ml/kg. Red blood cell that is too low or too high can indicate a medical problem. Suppose that Roger has 7 blood tests and the mean is 32.7 ml/kg. Suppose that red blood cell volume follows normal distribution with the population standard deviation is known to be 4.75 ml/kg. With 5% significant level, test that the mean blood cell volume is different from 28ml/kg? Which test should we use?
Z-test for the Mean
On wishes to test H0: μ=880 vs Ha: μ<880 A random sample with size 10 produced x¯=850, s=20. Assume that the population follows a normal distribution. For a test with α=0.05, which of the following excel function is correct for its test statistic?
t =(850-880)/(20/sqrt(10))
On wishes to test Ho: u=880 vs Ha: u<880. A random sample with size 20 produced x¯=850, s=20. Assume that the population follows a normal distribution. For a test with alpha=0.01, which of the following excel function is correct for its test statistic?
t =(850-880)/(20/sqrt(20))
On wishes to test H0: u=880 vs Ha: u<880. A random sample with size 16 produced X¯=865, s=20. Assume that the population follows a normal distribution. For a test with alpha=0.05, which of the following excel function is correct for its test statistic?
t =(865-880)/(20/sqrt(16))
A Type-II error in a test of hypothesis occurs if _____________.
we do not reject Ho when in fact Ho is false
Consider testing Ho: µ = 10 vs. Ha: µ ≠ 10. We commit a type I error if we reject Ho when µ = 10. We commit a type II error if...
we fail to reject Ho when µ ≠ 10
A Type-I error in a test of hypothesis occurs if _____________.
we reject Howhen in fact Ho is true
A diabetic claims that the average cost of insulin per year for a Type 1 diabetic is $5,705. She takes a sample of 100 Type 1 diabetics and finds their average cost is $5, 800 with a standard deviation of $575. Use α =0.05 to test the claim. Which distribution should be used?
z-distribution