STATS Practice Test 3

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) (b)

(a)

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 alpha, the correct decision is: ______

(a) (b) (c) (d)

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 alpha, the correct decision is: _____

(a) Ho : p = 0.65 versus Ha : p > 0.65 (b) =(86/125-0.65)/sqrt(0.65*0.35/125) (c) =1-NORM.DIST(0.8907, 0, 1, TRUE) (d) Fail to reject the null hypothesis; We cannot conclude that more than 65% of voters in Phoenix use oxygenated fuels.

A standard painkiller is known to bring relief in 3.5 minutes on average (u). A new painkiller is hypothesized to bring faster relief to patients. A sample of 40 patients is given the new painkillers. The sample yields a mean of 3.2 minutes and a standard deviation of 1.1 minutes. (a) Which of the following pairs of hypotheses are appropriate for this study? ______ (b) Which of the following excel function calculates the test statistic?______ (c) Suppose that the test statistic for this test is −1.7249. Which of the following excel function calculates the p-value of the test?______ (d) Suppose the p-value is calculated to be 0.0423. At the 0.01 significance level alpha, the correct decision is: _______

(a) Ho: μ = 3.5 vs Ha: μ < 3.5 (b) z =(3.2-3.5)/(1.1/sqrt(40)) (c) =NORM.DIST(-1.7249, 0, 1, TRUE) (d) Fail to reject the null hypothesis; the new painkiller does not bring faster relief.

In the past, it is generally agreed that a certain standard treatment yields a mean survival period of 3.5 years for a certain cancer patients. Recently, a new treatment is administered to 49 patients and their duration of survival is recorded. The sample mean and standard deviation of the duration is 3.8 years and 0.7 years, respectively. Suppose we want to test whether the new treatment increases the mean survival period at a 5% level of significance. (a) Set up the null and alternative hypotheses to test the claim. _____ (b) Which of the following excel function calculates the test statistic? _____ (c) Suppose that the test statistic is 3.00. Which of the following excel function calculates the p-value of the test?_____ (d) Suppose that the p-value is calculated to be 0.00135. Do we have enough evidence to conclude that the new treatment increases the mean survival period? _____

(a) Ho: μ = 3.5 vs Ha: μ > 3.5 (b) =(3.8-3.5)/(0.7/sqrt(49)) (c) =1-NORM.DIST(3, 0, 1, TRUE) (d) Yes, because the p value is less than 0.05.

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: 107.1 79.1 68.7 93.7 59.3 86.3 76.7 69.5 64.7 84.9 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) Which of the following excel function calculates the test statistic? ____ (c) Suppose the test statistic is -1.8558. Which of the following excel function calculates the p-value? ______ (d) Suppose the p-value is calculated to be 0.0482. State the conclusion of the test. ______

(a) Ho: μ = 87.5 vs Ha: μ < 87.5 (b) t=(79-87.5)/(14.4838/sqrt(10)) (c) =T.DIST(-1.8558, 9, TRUE) (d) Fail to reject the null hypothesis; the new system is ineffective.

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) only (ii), (iii), (iv) (i) and (v) (i), (iii) and (v) (v) only

(i) and (v)

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? (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

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? =(394-0.22)/sqrt(0.22*0.78/2000) =(394/2000-0.22)/sqrt(0.22*0.78/2000) =(394/2000-0.22)/sqrt(0.197*0.803/2000) =394/2000-0.22/sqrt(0.22*0.78/2000) =(394/2000-0.22)/sqrt(0.22*0.78/394) =(0.22-394/2000)/sqrt(0.22*0.78/2000)

=(394/2000-0.22)/sqrt(0.22*0.78/2000)

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, 790 said that they prepare and file their income taxes before April 15th. Let p 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=0.75 vs Ha: p>0.75.Which of the following excel function calculates the standardized test statistic? =(790/1000-0.75)/sqrt(0.75*0.25/1000) =(790-0.75)/sqrt(0.75*0.25/1000) =(790/1000-0.75)/sqrt(0.75*0.25/790) =(790/1000-0.75)/sqrt(0.79*0.21/1000) =(0.75-790/1000)/sqrt(0.75*0.25/1000)

=(790/1000-0.75)/sqrt(0.75*0.25/1000)

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? =NORM.DIST(-2.483, 0, 1, TRUE) =394/2000 =1-NORM.DIST(-2.483, 0, 1, TRUE) =1-T.DIST(-2.483, 0, 1, TRUE) =(394/2000-0.22)/sqrt(0.22*0.78/2000) =1-NORM.DIST(394/2000, 0, 1, TRUE)

=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=0.05to test the claim. Also suppose that the test statistic equals 0.85. Which formula in Excel finds the p-value? =T.DIST(0.85, 24, TRUE) =1-NORM.DIST(0.85, 0, 1, TRUE) =1 - T.DIST(0.85, 25, TRUE) =NORM.DIST(0.85, 0, 1, TRUE) =T.DIST(0.85, 25, TRUE) =1 - T.DIST(0.85, 24, TRUE)

=1-NORM.DIST(0.85, 0, 1, TRUE)

A sociologist wishes to test Ho: u=42 vs. Ha: u>42. The sociologist takes a sample of size 100 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, 99, TRUE) =2*NORM.DIST(2.34, 0, 1, TRUE) =1-T.DIST(2.34, 99, TRUE) =1-NORM.DIST(2.34, 0, 1, TRUE) =2*(1-T.DIST(2.34, 99, TRUE)) =NORM.DIST(2.34, 0, 1, TRUE)

=1-NORM.DIST(2.34, 0, 1, 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: =2*NORM.DIST(-2.34, 0, 1, TRUE) =1-NORM.DIST(-2.34, 0, 1, TRUE) =T.DIST(-2.34, 9, TRUE) =NORM.DIST(-2.34, 0, 1, TRUE) =2*(T.DIST(-2.34, 9, TRUE)) =1-T.DIST(-2.34, 9, TRUE)

=2*(T.DIST(-2.34, 9, 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(-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 sociologist wishes to test Ho: u=42 vs. Ha: u<42. The sociologist takes a sample of size 100 and calculates a standardized test statistic of -2.34. To calculate a p-value for the test in Excel, the sociologist should use: =2*(1-T.DIST(-2.34, 99, TRUE)) =1-T.DIST(-2.34, 99, TRUE) =NORM.DIST(-2.34, 0, 1, TRUE) =1-NORM.DIST(-2.34, 0, 1, TRUE) =2*NORM.DIST(-2.34, 0, 1, TRUE) =T.DIST(-2.34, 99, TRUE)

=NORM.DIST(-2.34, 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) = 1- T.DIST(-1.22, 10, TRUE) =T.DIST(-1.22, 10, TRUE) =1 - NORM.DIST(-1.22, 0, 1, TRUE) =NORM.DIST(-1.22, 0, 1, TRUE)

=T.DIST(-1.22, 9, 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) = 1- T.DIST(-1.22, 10, TRUE) =T.DIST(-1.22, 10, TRUE) =1 - NORM.DIST(-1.22, 0, 1, TRUE) =NORM.DIST(-1.22, 0, 1, TRUE)

=T.DIST(-1.22, 9, 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)

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? The information is not enough to make a decision. Reject Ho since the P-value is less than the significance level. Fail to reject Ho since the P-value is less than the significance level. Fail to reject Ho since the P-value is greater than .01. Reject Ho since the P-value is greater than .01.

Fail to reject Ho since the P-value is greater than .01.

State the decision rules for the P-value approach. (Select ALL that apply) -if the P-value < α, we accept Ho -If the P-value < critical value, we reject Ho -If the P-value > α, we fail to accept Ho -If the P-value < α, we reject Ho -If the P-value > α, we fail to reject Ho

If the P-value < α, we reject Ho If the P-value > α, we fail to reject Ho

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.

The average room rate in hotels in a certain region is $82.53. A travel agent believes that the average in a particular resort area is more expensive. The null and alternative hypotheses are as follows: H0 : µ = $82.53 Ha : µ > $82.53 If the travel agent performed the hypothesis testing using alpha=5%, interpret what this significance level mean in the context of the problem. -If, in fact the mean room rate in hotels in a certain region is more expensive than $82.53, then there is a 5% chance that the travel agent will conclude that the mean room rate in hotels in that particular region is indeed more expensive than $82.53. -If, in fact the mean room rate in hotels in a certain region is $82.53, then there is a 5% chance that the travel agent will conclude that the mean room rate in hotels in that particular region is not $82.53. -If, in fact the mean room rate in hotels in a certain region is $82.53, then there is a 5% chance that the travel agent will conclude that the mean room rate in hotels in that particular region is more expensive. -If, in fact the mean room rate in hotels in a certain region is $82.53, then there is a 5% chance that the travel agent will conclude that the mean room rate in hotels in that particular region is less expensive. -If, in fact the mean room rate in hotels in a certain region is $82.53, then there is a 5% chance that the travel agent will conclude that the mean room rate in hotels in that particular region is indeed $82.53.

If, in fact the mean room rate in hotels in a certain region is $82.53, then there is a 5% chance that the travel agent will conclude that the mean room rate in hotels in that particular region is more expensive.

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. -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 low or lower than the one obtained. -If, in fact, the mean height of women 20 years of age or older was taller than 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. -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 85% of the samples will result in a sample mean as high or higher than the one obtained. -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.

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.

Tony's Pizzeria's staff can accommodate 65 customers per hour. Tony suspects that his average customer load differs from 65. He performed a hypothesis test Ho : µ = 0.65 versus Ha : µ ≠ 0.65 with α = 0.05 and came up with a P-Value of 0.18. Explain what this​ P-value means. -If, in fact, the mean number of customers per hour is 65 and if we sample the same number of hours repeatedly, then 18% of the samples will result in a sample mean farther from 65 than the sample mean that Tony obtained. -If, in fact, the mean number of customers per hour is 65 and if we sample the same number of hours repeatedly, then 82% of the samples will result in a sample mean farther from 65 than the sample mean that Tony obtained. -If, in fact, the mean number of customers per hour is different from 65 and if we sample the same number of hours repeatedly, then 82% of the samples will result in a sample mean farther from 65 than the sample mean that Tony obtained. -If, in fact, the mean number of customers per hour is different from 65 and if we sample the same number of hours repeatedly, then 18% of the samples will result in a sample mean farther from 65 than the sample mean that Tony obtained.

If, in fact, the mean number of customers per hour is 65 and if we sample the same number of hours repeatedly, then 18% of the samples will result in a sample mean farther from 65 than the sample mean that Tony obtained.

Three years ago, the mean price of an existing single-family home was $243,772. A real estate broker believes that existing home prices in her neighborhood have changed. The null and alternative hypotheses are as follows: Ho: µ = $243,772 Ha: µ ≠ $243,772 If the real estate broker performed the hypothesis testing using alpha=2.5%, interpret what this significance level means in the context of the problem. -If, in fact, the mean price of an existing single-family home was different from $243,772, then there is a 2.5% chance that the real estate broker will conclude that the mean price of an existing single-family home is different from $243,772. -If, in fact, the mean price of an existing single-family home was $243,772, then there is a 2.5% chance that the real estate broker will conclude that the mean price of an existing single-family home is higher than $243,772. -If, in fact, the mean price of an existing single-family home was $243,772, then there is a 2.5% chance that the real estate broker will conclude that the mean price of an existing single-family home is different from $243,772. -If, in fact, the mean price of an existing single-family home was $243,772, then there is a 2.5% chance that the real estate broker will conclude that the mean price of an existing single-family home is equal to $243,772. -If, in fact, the mean price of an existing single-family home was $243,772, then there is a 2.5% chance that the real estate broker will conclude that the mean price of an existing single-family home is lower than $243,772.

If, in fact, the mean price of an existing single-family home was $243,772, then there is a 2.5% chance that the real estate broker will conclude that the mean price of an existing single-family home is different from $243,772.

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. Ho: p=0.78 Ha: p<0.78 Two-tailed test Right-tailed test Left-tailed test We do not have enough information to determine this

Left-tailed test

The null and alternative hypotheses are given. Determine whether the hypothesis test is​ left-tailed, right-tailed, or​ two-tailed. Ho: u≥4 Ha: u<4 Right-tailed test Two-tailed test We do not have enough information to determine this Left-tailed test

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 Yes No way to tell

No

If a null hypothesis is rejected at the significance level alpha=0.005, is it possible that the null hypothesis is not rejected if the same test was done at the significance level alpha= .05 (with everything else staying the same)? -No way to tell -No -Yes

No

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: u=100 vs . Ha: u≠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 .05? -Fail to reject Ho since the P-value is greater than .01. -Reject Ho since the P-value is less than the significance level. -The information is not enough to make a decision. -Fail to reject Ho since the P-value is less than the significance level. -Reject Ho since the P-value is greater than .01

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

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

Which of the following is true about the significance level alpha? The probability of failing to reject the null hypothesis when it is true The probability of rejecting the null hypothesis when it is false The probability of rejecting the null hypothesis when it is true The level of confidence a researcher has on an estimate The probability of failing to reject the null hypothesis when it is false

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-family home 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 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 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 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 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 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.

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 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 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 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 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 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: sigma= 25 Ha: sigma≠ 25 We do not have enough information to determine this Left-tailed test Two-tailed test Right-tailed test

Two-tailed test

A test is made of H0: μ = 20 versus Ha: μ ≠ 20. Suppose the true value of μ is 20, and Ho is rejected. Determine whether the outcome is a Type I error, a Type II error, or a correct decision. -Correct Decision -Type I Error -Type II Error -Both Type I and II Errors -Insufficient information to decide

Type I Error

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)? No No way to tell Yes

Yes

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 Proportion -F-test -T-test for the Mean -Z-test for the Mean -T-Test for Proportion

Z-test for the Mean

On wishes to test Ho: u=880 vs Ha: u<880. A random sample with size 10 produced xbar=850, 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.DIST(850, 9, TRUE) z =(880-850)/(20/sqrt(10)) t =(850-880)/(20/sqrt(10)) =NORM.DIST(850, 880, 20/sqrt(10), TRUE) z =(850-880)/(20/sqrt(10)) t =(880-850)/(20/sqrt(10))

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

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))

In order to test Ho: µ = 10 vs. Ha: µ < 10, a random sample of size 25 is taken from a normally distributed population. From the sample, one obtained xbar=9.70 and s=2. For a test with alpha=0.05, which of the following excel function is correct for its test statistic? t =(9.70-10)/(25/sqrt(2)) =NORM.DIST(9.7, 10, 2/SQRT(25), TRUE) t =(10-9.70)/(0.05/sqrt(25)) t =(9.70-10)/(2/sqrt(25)) z =(9.70-10)/(2/sqrt(25)) z =(10-9.70)/(2/sqrt(25)) =T.DIST(9.7, 24, TRUE)

t =(9.70-10)/(2/sqrt(25))

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 fail to reject Ho when µ > 25

Consider testing Ho: µ ≥ 150 vs. Ha: µ < 150. We commit a type II error if we fail to reject Ho when µ < 150. We commit a type I error if... -we fail to reject Ho when µ < 150 -None of the answer choices -we fail to reject Ho when µ ≥ 150 -we reject Ho when µ < 150we reject Ho when µ ≥ 150

we reject Ho when µ ≥ 150