STAT TEST

Which of the following statements are true in hypotheses testing? (i) If we reject when is in fact true, we made Type I error. (ii) If we fail to reject when is in fact true, we made Type II error. (iii) One will reject if P-value is larger than the significance level . (iv) The p-value is the probability that the null hypothesis is true. (v) The significance level 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 Ho when is in fact true, we made Type I error. (ii) If we fail to reject Ho when is in fact false, we made Type II error. (iii) One will reject if P-value is smaller than the significance level (iv) The p-value is the probability that the null hypothesis is true.

(ii) and (iii)

Which of the following statements are NOT true in hypotheses testing? (i) If we reject Ho when 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) (i) only (iii) only (ii) only (i), (ii) and (iii)

(ii) only

Which ones of the following are true about β ? (select ALL that apply) 1-β is the probability that we will reject the null hypothesis when it is false, which is known as the "power of the test" β is the probability of rejecting the null hypothesis when it is true. β is the significance level of a test β is the likelihood of making a Type II error β is the power of the test β is the probability of failing to reject the null hypothesis when it is false.

1-β is the probability that we will reject the null hypothesis when it is false, which is known as the "power of the test" β is the likelihood of making a Type II error β is the probability of failing to reject the null hypothesis when it is false.

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 to test the claim. Which function in Excel finds the test statistic? (Note: 150/200 = 0.75) = NORM.INV(0.75, 0, 1) =NORM.DIST(0.75, 0, 1, TRUE) = (80 - 90) / (0.75/ SQRT(200)) = (0.75 - 0.80)/SQRT(0.80*0.20/200)

= (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 be the proportion of all adults who are in favor of outlawing cigarettes. One is interested in the following hypotheses: Ho:p=0.6 vs Ha:p>0.6 Which of the following excel function calculates the standardized test statistic? =(136/200-0.6)/sqrt(136/200*64/200/200) =(136/200-0.6)/sqrt(0.6*0.4/136) =(136-0.6)/sqrt(0.6*0.4/200) =(136/200-0.6)/sqrt(0.6*0.4/200) =(0.6-136/200)/sqrt(0.6*0.4/200)

=(136/200-0.6)/sqrt(0.6*0.4/200)

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. Which function in Excel finds the test statistic? =NORM.INV((5705-5912)/(300/SQRT(100)) =(5705-5912)/(300/SQRT(100)) =NORM.DIST(5705, 5912, 300, TRUE) =NORM.DIST(5912, 5705, 300, TRUE)

=(5705-5912)/(300/SQRT(100))

In a random sample of 100 adults, 70 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 vs Ha: p>0.6. Which of the following excel function calculates the standardized test statistic? =(70/100-0.6)/sqrt(0.6*0.4/70) =(70/100-0.6)/sqrt(0.6*0.4/100) =(0.6-70/100)/sqrt(0.6*0.4/100) =(70-0.6)/sqrt(0.6*0.4/100) =(70/100-0.6)/sqrt(0.7*0.3/100)

=(70/100-0.6)/sqrt(0.6*0.4/100)

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 hypotheses Ho:p=0.70 vs Ha:p>0.70 Which of the following excel function calculates the standardized test statistic? =(0.7-720/1000)/sqrt(0.7*0.3/1000) =(720/1000-0.7)/sqrt(0.7*0.3/1000) =(720/1000-0.7)/sqrt(0.7*0.3/720) =(720-0.7)/sqrt(0.7*0.3/1000) =(720/1000-0.7)/sqrt(0.72*0.28/1000)

=(720/1000-0.7)/sqrt(0.7*0.3/1000)

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 to test the claim. Which formula in Excel finds the test statistic? =(8.5-8)/(2/SQRT(25)) =1 - ((8.5-8)/(2/SQRT(25))) = NORM.INV(8.5, 8, 2) =8.5-8/2/SQRT(25) =NORM.DIST(8.5, 0, 1, TRUE) =(8-8.5)/(2/SQRT(25))

=(8.5-8)/(2/SQRT(25))

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(-100, 0, 1, TRUE) =2*NORM.DIST(1.52, 0, 1, TRUE) =NORM.DIST(-1.52, 0, 1, TRUE) =2*NORM.DIST(-1.52, 0, 1, TRUE) =1-NORM.DIST(1.52, 0, 1, TRUE) =2*T.DIST(-1.52, 249, TRUE)

=2*NORM.DIST(-1.52, 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? =P.DIST(1.11, 0, 1, TRUE) =1 - NORM.DIST(-1.11, 0, 1, TRUE) =NORM.DIST(-1.11, 0, 1, TRUE) = NORM.DIST(1.11, 0, 1, TRUE) =P.DIST(-1.11, 0, 1, TRUE)

=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=0.10to test the claim. Suppose the test statistic is -1.22. Which function in Excel finds the p-value? =T.DIST(-1.22, 9, TRUE) =T.DIST(-1.22, 10, TRUE) =1 - NORM.DIST(-1.22, 0, 1, TRUE) = 1- T.DIST(-1.22, 10, TRUE) =NORM.DIST(-1.22, 0, 1, TRUE)

=T.DIST(-1.22, 9, TRUE)

Jeffrey, as an eight-year old, established a mean time of 16.43 seconds for swimming the 25-yard freestyle, with a standard deviation of 0.8 seconds. His dad, Frank, thought that Jeffrey could swim the 25-yard freestyle faster using goggles. Frank bought Jeffrey a new pair of expensive goggles and timed Jeffrey for 15 25-yard freestyle swims. For the 15 swims, Jeffrey's mean time was 16 seconds. Frank thought the goggles helped Jeffrey swim faster than the 16.43 seconds. He performed a hypothesis test Ho : µ = 16.43 versus Ha : µ < 16.43 with α = 0.05 and came up with a P-Value of 0.03. Explain what this​ P-value means. If the mean time for swimming the 25-yard freestyle really is 16.43 seconds and if we randomly collect a sample of size n = 15 swims repeatedly, then 3% of the samples will result in a sample mean as high or higher than the one obtained. If the mean time for swimming the 25-yard freestyle really is 16.43 seconds and if we randomly collect a sample of size n = 15 swims repeatedly, then 3% of the samples will result in a sample mean as low or lower than the one obtained. If the mean time for swimming the 25-yard freestyle really is lower than 16.43 seconds and if we randomly collect a sample of size n = 15 swims repeatedly, then 3% of the samples will result in a sample mean as low or lower than the one obtained. If the mean time for swimming the 25-yard freestyle really is higher than 16.43 seconds and if we randomly collect a sample of size n = 15 swims repeatedly, then 3% of the samples will result in a sample mean as low or lower than the one obtained.

If the mean time for swimming the 25-yard freestyle really is 16.43 seconds and if we randomly collect a sample of size n = 15 swims repeatedly, then 3% of the samples will result in a sample mean as low or lower than the one obtained.

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 low or lower than the one obtained. If the mean weight that his players can bench press really is more than 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. 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 far or farther than the one obtained. 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.

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.

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: Ho : p = 0.32 Ha : p < 0.32 If the analyst performed the hypothesis testing using alpha=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 more than 32%. -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 32%. -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 not 32%. -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%. -If, in fact, the proportion of the people who visit the mall actually make a purchase is less than 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%.

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? H0:μ=86.7 Ha:μ<86.7

Left tailed

The null and alternative hypotheses are given. Determine whether the hypothesis test is​ left-tailed, right-tailed, or​ two-tailed.Ho: p=0.78 Ha: p<0.78

Left-tailed test

Which ones of the following are TRUE? (Select ALL that apply) -Alternative hypothesis is a statement of equality or inclusiveness Null hypothesis is a statement of non-equality or non-inclusiveness Null hypothesis is a statement of equality or inclusiveness -β is the likelihood of making a Type II error α is the probability of making a Type I error α is the probability of failing to reject the null hypothesis when it is false. α and the level of confidence add up to 1 Null hypothesis is a statement about the claimα is the power of the test- Alternative hypothesis is a statement of non-equality or non-inclusiveness β is the probability of rejecting the null hypothesis when it is false.

Null hypothesis is a statement of equality or inclusiveness β is the likelihood of making a Type II error α is the probability of making a Type I error α and the level of confidence add up to 1 Alternative hypothesis is a statement of non-equality or non-inclusiveness

In the test of hypothesis Ho: p= 0.56 vs Ha: 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 critical value. Reject Ho since the P-value is less than the significance level. Fail to reject Ho since the P-value is greater than the significance level. Fail to reject Ho since the P-value is greater than the critical value. We do not have enough information to decide.

Reject Ho since the P-value is less than the significance level.

In the test of hypothesis Ho:μ=100 vs Ha:μ>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? Fail to reject since the P-value is greater than .01. Reject since the P-value is greater than .01. Reject since the P-value is less than the significance level. Fail to reject since the P-value is less than the significance level.

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

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 a=0.10 to test the claim. Which distribution should be used? c-distribution p-distribution T-test for the Mean Z-test for the Mean

T-test for the Mean

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 17 blood tests with the mean is 32.7 ml/kg and standard deviation is 4.75 ml/kg. Assume that the population of red blood cell volume is normally distributed. With 5% significant level, test that the mean blood cell volume is different from 28ml/kg? Which test should we use? Z-Test for Proportion T-Test for Proportion Z-test for the Mean F-Test T-test for the Mean

T-test for the Mean

Suppose that the null hypothesis was not 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 increased since then. There is not sufficient evidence to conclude that the mean price of a​ single-family home has decreased from its level two years ago of ​$299,500. 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. There is sufficient evidence to conclude that the mean price of a​ single-family home has increased from its level two years ago of ​$299,500. There is not sufficient evidence to conclude that the mean price of a​ single-family home has increased from its level two years ago of ​$299,500.

There is not sufficient evidence to conclude that the mean price of a​ single-family home has increased 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 not sufficient evidence to conclude that the mean price of a​ single-family home has not changed. There is sufficient evidence to conclude that the mean price of a​ single-family home has not changed. There is sufficient evidence to conclude that the mean price of a​ single-family home has increased. There is not sufficient evidence to conclude that the mean price of a​ single-family home has increased.

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. -There is sufficient evidence to conclude that the standard deviation of monthly cell phone bills is equal to its level three years ago of ​$49.75. -There is not sufficient evidence to conclude that the standard deviation of monthly cell phone bills is less than its level three years ago of ​$49.75. -There is not sufficient evidence to conclude that the standard deviation of monthly cell phone bills is higher than its level three years ago of ​$49.75. -There is sufficient evidence to conclude that the standard deviation of monthly cell phone bills is different from its level three years ago of ​$49.75

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.

A test is made of H0: μ = 20 versus Ha: μ ≠ 20. Suppose the true value of μ is 25, and H0 is not rejected. Determine whether the outcome is a Type I error, a Type II error, or a correct decision.

Type II Error

A Type-II error in a test of hypothesis occurs if?

We do not reject Ho when in fact Ho is false

If a null hypothesis is rejected at the significance level a=0.02 , is it possible that the null hypothesis is not rejected if the same test was done at the significance level a = .002 (with everything else staying the same)?

Yes

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)

if a null hypothesis is rejected at the significance level a=0.005 , is it possible that the null hypothesis is not rejected if the same test was done at the significance level a= .05 (with everything else staying the same)?

no

On wishes to test Ho:u=880 vs Ha:u<880. A random sample with size 10 produced x=850,s=20, Assume that the population follows a normal distribution. For a test with a=0.05, which of the following excel function is correct for its test statistic? t =(880-850)/(20/sqrt(10)) z =(850-880)/(20/sqrt(10)) t =(850-880)/(20/sqrt(10)) =NORM.DIST(850, 880, 20/sqrt(10), TRUE) =T.DIST(850, 9, TRUE) z =(880-850)/(20/sqrt(10

t =(850-880)/(20/sqrt(10))

On wishes to test Ho:μ=880 vs Ha:μ<880. A random sample with size 20 produced x=850,s=20. Assume that the population follows a normal distribution. For a test with a=0.01 , which of the following excel function is correct for its test statistic? t =(880-850)/(20/sqrt(16)) z =(850-880)/(20/sqrt(20)) =NORM.DIST(850, 880, 20/sqrt(20), TRUE) z =(880-850)/(20/sqrt(16)) t =(850-880)/(20/sqrt(20)) =T.DIST(850, 19, TRUE)

t =(850-880)/(20/sqrt(20))

in order to test Ho: µ = 9.5 vs. Ha: µ > 9.5, a random sample of size 15 is taken from a normally distributed population. From the sample, one obtained xbar=9.7 and s=2.9. For a test with alpha=0.10, which of the following excel function is correct for its test statistic? z =(9.7-9.5)/(2.9/sqrt(15)) z =(9.5-9.7)/(2.9/sqrt(15)) t =(9.5-9.7)/(2.9/sqrt(15)) t =(9.7-9.5)/(2.9/sqrt(15)) =1-T.DIST(9.7, 14, TRUE) =1-NORM.DIST(9.7, 9.5, 2.9/sqrt(15), TRUE)

t =(9.7-9.5)/(2.9/sqrt(15))

A Type-I error in a test of hypothesis occurs if

we reject Ho when Ho is in fact is true

Consider testing Ho: µ ≤ 25 vs. Ha: µ > 25. We commit a type I error if we reject Ho when µ ≤ 25. We commit a type II error if we reject Ho when µ > 25 None of the answer choices we reject Ho when µ ≤ 25 we fail to reject Ho when µ > 25 we fail to reject Ho when µ ≤ 25

we reject Ho when µ ≤ 25

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 to test the claim. Which distribution should be used? c-distribution t-distribution z-distribution p-distribution

z-distribution

Which ones of the following are true about the significance level ? a (select ALL that apply) α is the power of the test α is used to be compared to the p value to make a decision about the null hypothesis α is the level of confidence a researcher has on an estimate. α and the level of confidence add up to 1 α is the significance level of a test α is the probability of rejecting the null hypothesis when it is true. α is the probability of failing to reject the null hypothesis when it is false.

α is used to be compared to the p value to make a decision about the null hypothesis α and the level of confidence add up to 1 α is the significance level of a test α is the probability of rejecting the null hypothesis when it is true.