math 10 final study

Identify the two types of incorrect decisions in a hypothesis test. For each incorrect​ decision, what symbol is used to represent the probability of making that type of​ error? Choose the correct answer below. A. A Type I error is accepting a false null​ hypothesis, whose probability is denoted α. A Type II error is not accepting a true null​ hypothesis, whose probability is denoted β. B. A Type I error is rejecting a true null​ hypothesis, whose probability is denoted α. Type II error is not rejecting a false null​ hypothesis, whose probability is denoted β. C. A Type I error is not accepting a true null​ hypothesis, whose probability is denoted α. A Type II error is accepting a false null​ hypothesis, whose probability is denoted β. D. A Type I error is not rejecting a false null​ hypothesis, whose probability is denoted α. A Type II error is rejecting a true null​ hypothesis, whose probability is denoted β.

B. A Type I error is rejecting a true null​ hypothesis, whose probability is denoted α. A Type II error is not rejecting a false null​ hypothesis, whose probability is denoted β.

The primary concern for performing a hypothesis test to compare the​ means, μ1 and μ2​, of two populations is deciding whether the mean of Population 1 is less than the mean of Population 2. a. Determine the null and alternative hypotheses.​ Note: Always place the mean of Population 1 on the left. b. Classify the hypothesis test as​ two-tailed, left-tailed, or​ right-tailed. Question content area bottom Part 1 a. Complete the hypotheses. H0​: μ1 equals= μ2 Ha​: μ1 less than< μ2 Part 2 b. This hypothesis test is left​-tailed. (left tailed)

If h1 < h2, then left tailed

In a​ poll, 1000 adults in a region were asked about their online vs.​ in-store clothes shopping. One finding was that 29​% of respondents never​ clothes-shop online. Find and interpret a 95​% confidence interval for the proportion of all adults in the region who never​ clothes-shop online. Question content area bottom Part 1 The 95​% confidence interval is from 0.262 to 0.318. ​(Round to three decimal places as​ needed.)

One sample proportion summary confidence interval: With 95​% ​confidence, the true percentage of all adults in the region who never​ clothes-shop online is within the bounds of the confidence interval found in the previous step.

In the following scenario for a hypothesis test for a population​ mean, decide whether the​ z-test is an appropriate method for conducting the hypothesis test. Assume that the population standard deviation is known. Preliminary data analyses reveal that the sample data contain no outliers but that the distribution of the variable under consideration is probably mildly skewed. The sample size is 70. Question content area bottom Part 1 Choose the correct answer below. A. The​ z-test is an appropriate​ method, because the sample contains no outliers. B. The​ z-test is not an appropriate​ method, because the sample size is too small to be useful. C. The​ z-test is not an appropriate​ method, because the sample is not a large sample and the data are highly skewed. D. The​ z-test is an appropriate​ method, because the sample size is sufficiently large that the skewness of the variable does not matter.

Option D

Decide whether applying the​ t-test to perform a hypothesis test for the population mean in question appears reasonable. Explain your answers. A study found that for cardiovascular​ hospitalizations, the mean age of women is 68.5 years. At one​ hospital, a random sample of 15 of its female cardiovascular patients had the ages​ shown, in years. 70.2 62.5 83.2 60.0 73.3 55.4 57.9 50.3 82.3 68.4 56.8 66.7 74.9 70.1 67.3 Question content area bottom Part 1 Choose the correct answer below. A. It is not reasonable to use the​ t-test because the sample is obviously not random. B. It is not reasonable to use the​ t-test because the sample size is too small. C. It is not reasonable to use the​ t-test because the the sample is obviously not normal. D. It is reasonable to use the​ t-test because the sample is probably random and appears to be normally distributed.

Stat -> goodness of fit -> normality test ->shapiro wilk->if the p value is less than the stat, then it's normal

For the provided sample​ mean, sample​ size, and population standard​ deviation, complete parts​ (a) through​ (c) below. x̄ = 29 n = 100 σ = 2 a) Find a 95% confidence interval from __ to __ b) What is the margin of error? c) Interpret the margin of error. Choose the correct answer below. A. We can be​ 95% confident that any possible value of the variable is within the margin of error of the population​ mean, μ. B. We can be​ 95% confident that any possible sample mean is within the margin of error of 29. C. We can be​ 95% confident that any possible value of the variable is within the margin of error of the sample​ mean, 29. D. We can be​ 95% confident that the population​ mean, μ​, is within the margin of error of the sample​ mean, 29.

Statcrunch: (Select) Stat -> T stats -> One Sample ->Summary mean = 29 standard deviation = 2 size = 100 (Select) Perform: Confidence interval for miu (μ) Level = 0.95 a) lower limit and upper limit is shown b) upper limit - lower limit / 2 = 0.396843 c) We can be​ 95% confident that the population​ mean, μ​, is within the margin of error of the sample​ mean, 29.

Suppose a study reported that the average person watched 3.72 hours of television per day. A random sample of 15 people gave the number of hours of television watched per day shown. At the 5​% significance​ level, do the data provide sufficient evidence to conclude that the amount of television watched per day last year by the average person differed from the value reported in the​ study? (Note: x=4.100 hours and s=1.460 ​hours.) Set up the hypotheses for the​ one-mean t-test. H0​: μequals=3.723.72 Ha​: μnot equals≠3.723.72 Part 2 The test statistic is t=1.011.01. ​(Round to two decimal places as​ needed.) Part 3 The​ P-value is 0.3310.331. Do not reject the null hypothesis. The data do not provide sufficient evidence to conclude that the amount of television watched by the average person is different from the value reported in the study.

This is because the P value is less than the T-Stat (NOT critical T) for a One sample T hypothesis test:

Preliminary data analyses indicate that the assumptions for using pooled​ t-procedures are satisfied. Independent random samples of released prisoners in the fraud and firearms offense categories yielded the given information on time​ served, in months. At the 5​% significance​ level, do the data provide sufficient evidence to conclude that the mean time served for fraud is less than that for firearms​ offenses? ​(Note: x1=10.79​, s1=4.19​, x2=19.37​, and s2=4.68​.) Let μ1 be the mean time served for fraud μ2 the mean time served for firearms offenses. What are the correct hypotheses to​ test? A.H0​:μ1=μ2Ha​:μ1>μ2B.H0​:μ1=μ2Ha​:μ1<μ2 C.H0​:μ1≠μ2Ha​:μ1=μ2 D.H0​:μ1>μ2Ha​:μ1=μ2 E.H0​:μ1=μ2Ha​:μ1≠μ2 F. H0​: μ1<μ2 Ha​: μ1=μ2 Part 2 Compute the test statistic. t=negative 4.318−4.318 ​(Round to three decimal places as​ needed.) Part 3 Determine the critical value or values. negative 1.734−1.734 ​(Round to three decimal places as needed. Use a comma to separate answers as​ needed.) Part 4 What is the conclusion of the hypothesis​ test?

To determine the critical value, make sure to check "Show critical value" and the significance is 0.05 T stat is less than critical t, therefore we reject

A sample​ mean, sample​ size, and sample standard deviation are provided below. Use the​ one-mean t-test to perform the required hypothesis test at the 10​% significance level. x=17​, s=5​, n=32​, H0​: μ=20​, Ha​:

Use one sample T summary The test statistic is t= −3.39. ​(Round to two decimal places as​ needed.) Part 2 The​ P-value is0.001. ​(Round to three decimal places as​ needed.) Reject the null hypothesis. The data provide sufficient evidence to conclude that the mean is less than 20. Show critical value: significance: 1 - 0.1 = 0.9

​Cadmium, a heavy​ metal, is toxic to animals.​ Mushrooms, however, are able to absorb and accumulate cadmium at high concentrations. A government set a safety limit for cadmium in dry vegetables at 0.5 parts per million​ (ppm). The cadmium levels in a random sample of one species of edible mushroom are in the accompanying data set. At the 10​% significance​ level, do the data provide sufficient evidence to conclude that the mean cadmium level in this species of mushroom is greater than the​ government's recommended limit of 0.5 ​ppm? Assume that the population standard deviation of cadmium levels in this species of mushroom is 0.32 ppm. Preliminary data analyses indicate that applying the​ z-test is reasonable.​ (Note: The sum of the data is 7.11 ​ppm.) State the hypotheses for the​ one-mean z-test. H0​: μ =0.5ppm Ha​: μ >0.5ppm Compute the value of the test statistic. z=1 Determine the​ P-value. P=0.158

Use statcrunch One sample Z hypothesis test with data note: if P less than Z-Stat (significance level or test statistic), then reject the null hypothesis because there is enough evidence

Find the confidence level and α for a 99​% confidence interval. a) What is the confidence​ level? b) Determine the value of α.

a) 0.99 b) 1-0.99 = 0.01

For the following hypothesis​ test, determine the null and alternative hypotheses.​ Also, classify the hypothesis test as​ two-tailed, left-tailed, or​ right-tailed. The mean length of imprisonment for​ motor-vehicle theft offenders in this country is 20.4 months. A hypothesis test is to be performed to determine whether the mean length of imprisonment for​ motor-vehicle theft offenders in this city differs from the national mean of 20.4 months. a) Choose the correct null and alternative hypotheses below. A. H0​: μ>20.4 months Ha​: μ=20.4 months B. H0​: μ=20.4 months Ha​: μ≠20.4 months Your answer is correct. C. H0​: μ=20.4 months Ha​: μ<20.4 months D. H0​: μ=20.4 months Ha​: μ>20.4 months E. H0​: μ<20.4 months Ha​: μ=20.4 months F. H0​: μ≠20.4 months Ha​: μ=20.4 months b) Which of the following is the correct classification of the hypothesis​ test? A. ​Right-tailed B. ​Two-tailed C. ​Left-tailed

a) option B: H0​: μ=20.4 months Ha​: μ≠20.4months b) Option B, two tailed

A confidence interval for a population mean has a margin of error of 3.8. a. Determine the length of the confidence interval. b. If the sample mean is 46.5​, obtain the confidence interval. c. Construct a graph that illustrates your results.

a. 3.8*2 = 7.6 b. 46.5 - 3.8 = 42.7 (lower bound) 46.5 + 3.8 = 50.3 (upper bound) c. from 41 to 52, x is in the middle and two Es make up the confidence interval

Provided below are summary statistics for independent simple random samples from two populations. Use the pooled​ t-test and the pooled​ t-interval procedure to conduct the required hypothesis test and obtain the specified confidence interval. x1=13​, s1=2.5​, n1=12​, x2=16​, s2=2.2​, n2=12 a.​ Two-tailed test, α=0.10 a.​ First, what are the correct hypotheses for a​ two-tailed test? A. H0​: μ1≠μ2 Ha​: μ1=μ2 B. H0​: μ1=μ2 Ha​: μ1>μ2 C. H0​: μ1<μ2 Ha​: μ1=μ2 D. H0​: μ1>μ2 Ha​: μ1=μ2 E. H0​: μ1=μ2 Ha​: μ1≠μ2 Your answer is correct. F. H0​: μ1=μ2 Ha​: μ1<μ2 b.​Next, compute the test statistic. t=−3.121 ​(Round to three decimal places as​ needed.) Finally, determine the​ P-value. P=0.005 ​(Round to four decimal places as​ needed.) What is the conclusion of the hypothesis​ test? > Since the​ P-value is equal to or less than α​, reject H0. 90​% confidence interval

a. H0​: μ1=μ2 and Ha​: μ1≠μ2 Using statcrunch use the Two sample T summary hypothesis test. SELECT POOLED. alpha = 0.1 P = 0.005, therefore Reject the null hypothesis b. Confidence interval for μ1 - μ2 Level: 0.9. Lower: -4.651 upper: -1.349

Cadmium, a heavy​ metal, is toxic to animals.​ Mushrooms, however, are able to absorb and accumulate cadmium at high concentrations. Some governments have a safety limit for cadmium in dry vegetables at 0.42 part per million​ (ppm). A research team measured the cadmium levels in a random sample of edible​ mushrooms, where the hypothesis test is to decide whether the mean cadmium level in the sample is less than the​ governments' safety limit. The null and alternative hypotheses are H0​: μ=0.42 ​ppm, Ha​: μ<0.42 ppm. Complete parts​ (a) through​ (e) below. a. Explain what it would mean to make a type I error. Choose the correct answer below. A. A type I error would occur if in fact μ=0.42 ​ppm, but the results of the sampling fail to lead to the conclusion that μ<0.42 ppm. B. A type I error would occur if in fact μ=0.42 ​ppm, but the results of the sampling lead to the conclusion that μ<0.42 ppm. Your answer is correct. C. A type I error would occur if in fact μ<0.42 ​ppm, but the results of the sampling lead to the conclusion that μ=0.42 ppm. D. A type I error would occur if in fact μ<0.42 ​ppm, but the results of the sampling fail to lead to the conclusion that μ=0.42 ppm. b. Explain what it would mean to make a type II error. Choose the correct answer below. A. A type II error would occur if in fact μ<0.42 ​ppm, but the results of the sampling fail to lead to the conclusion that μ<0.42 ppm. Your answer is correct. B. A type II error would occur if in fact μ<0.42 ​ppm, but the results of the sampling lead to the conclusion that μ<0.42 ppm. C. A type II error would occur if in fact μ=0.42 ppm​, but the results of the sampling fail to lead to the conclusion that μ<0.42 ppm. D. A type II error would occur if in fact μ=0.42 ppm​, but the results of the sampling lead to the conclusion that μ<0.42 ppm. c. Explain what it would mean to make a correct decision. Choose the correct answer below. A. A correct decision would occur if μ≠0.42 ppm and the results of the sampling lead to the rejection of that​ fact; or if μ=0.42 ppm and the results of the sampling do not lead to that conclusion. B. A correct decision would occur if μ=0.42 ppm and the results of the sampling do not lead to the rejection of that​ fact; or if μ<0.42 ppm and the results of the sampling lead to that conclusion. Your answer is correct. C. A correct decision would occur if μ<0.42 ppm and the results of the sampling do not lead to the rejection of that​ fact; or if μ≠0.42 ppm and the results of the sampling lead to that conclusion. D. A correct decision would occur if μ=0.42 ppm and the results of the sampling lead to the rejection of that​ fact; or if μ<0.42 ppm and the results of the sampling do not lead to that conclusion. d. Now suppose that the results of carrying out the hypothesis test lead to nonrejection of the null hypothesis. Classify that conclusion by error type or as a correct decision if in fact the mean cadmium level in that type of mushrooms equals the safety limit of 0.42 ppm. Choose the correct answer below. A. correct decision because a true null hypothesis is not rejected Your answer is correct. B. type II error C. type I error D. correct decision because a false null hypothesis is rejected e. Now suppose that the results of carrying out the hypothesis test lead to nonrejection of the null hypothesis. Classify that conclusion by error type or as a correct decision if in fact the mean cadmium level in that type of mushrooms is below the safety limit of 0.42 ppm. Choose the correct answer below. A. correct decision because a true null hypothesis is not rejected B. correct decision because a false null hypothesis is rejected C. type II error D. type I error

a. Option B b. Option A c. Option B d. Option A e. Option C

In each​ part, assume that the population standard deviation is known. Decide whether use of the​ z-interval procedure to obtain a confidence interval for the population is reasonable. Explain your answers. a. The sample data contain no​ outliers, the variable under consideration is roughly normally​ distributed, and the sample size is 27 b. The distribution of the variable under consideration is highly​ skewed, and the sample size is 18. c. The sample data contain no​ outliers, the sample size is ​300, and the variable under consideration is far from being normally distributed.

a. The​ z-interval procedure is reasonable since there are no​ outliers, the population is roughly​ normal, and the sample size is over 15. b. The​ z-interval procedure is not reasonable since the sample size is too small for a population that is far from normally distributed. c. The sample data contain no​ outliers, the sample size is ​, and the variable under consideration is far from being normally distributed.

Suppose​ that, in a hypothesis​ test, the null hypothesis is in fact true. a. Is it possible to make a Type I​ error? Explain your answer. b. Is it possible to make a Type II​ error? Explain your answer.

a. ​Yes, it is possible. A Type I error is rejecting the null hypothesis when it is in fact true. b. ​No, it is not possible. A Type II error is not rejecting the null hypothesis when it is in fact false.

A sample​ mean, sample​ size, and population standard deviation are provided below. Use the​ one-mean z-test to perform the required hypothesis test at the 10​% significance level. x=42​, n=13​, σ=7​, H0​: μ=38​, Ha​: μ>38 Question content area bottom Part 1 The test statistic is z=2.06 ​(Round to two decimal places as​ needed.) Part 2 The​ P-value is 0.02 ​(Round to three decimal places as​ needed.) Part 3 Reject the null hypothesis. The data provide sufficient evidence to conclude that the mean is greater than 38.

significance level (alpha) = 1 - confidence interval See question, z test summary

Suppose that the average time spent per day with digital media several years ago was 3 hours and 41 minutes. For last​ year, a random sample of 20 adults in a certain region spent the numbers of hours per day with digital media given in the accompanying table. Preliminary data analyses indicate that the​ t-interval procedure can reasonably be applied. Find and interpret a 90​% confidence interval for last​ year's mean time spent per day with digital media by adults of the region.​ (Note: x=4.69 hr and s=2.23 ​hr.) b. Interpret the 90% confidence interval

stat -> t stat (NOT Z) -> one sample -> with data b. With ​% ​confidence, the mean amount of time spent per day on digital media last year by all adults in the region is between the​ interval's bounds.

Homework:Final Review Question 27, 12.2.79-P Part 3 of 5 HW Score: 50%, 23.5 of 47 points Points: 0 of 1 Save Question content area top Part 1 The number of successes and the sample size for a simple random sample from a population are given below. x=6​, n=24​, H0​: p=0.4​, Ha​: p<0.4​, α=0.05 a. Determine the sample proportion. b. Decide whether using the​ one-proportion z-test is appropriate. c. If​ appropriate, use the​ one-proportion z-test to perform the specified hypothesis test. a. The sample proportion is 0.250.25. ​(Type an integer or a decimal. Do not​ round.) Part 2 b. Is it appropriate to use the​ one-proportion z-test? No You answered No. The correct answer is Yes because both assumptions for a​ one-proportion z-test are satisfied. That​ is, the sample is a simple random sample and both np0 and ​n(1−p0​) are 5 or greater .

use the One sample proportion summary hypothesis test for the Z-stat and P value the p value is bigger than the z stat, the critical z, and therefore we do not reject the null hypothesis

According to a study of political​ prisoners, the mean duration of imprisonment for 39 prisoners with chronic​ post-traumatic stress disorder​ (PTSD) was 31.5 months. Assuming that σ=42 ​months, determine a 90​% confidence interval for the mean duration of​ imprisonment, μ​, of all political prisoners with chronic PTSD. Interpret your answer in words. a. A ​% confidence interval for the population mean is from __ months to __ months. b. Interpret the confidence interval. Select the correct choice below and fill in the answer boxes to complete your choice.

x = 31.5 months n = 39 people σ = 42 months determine 90% confidence interval: a. plug into One sample z summary statcrunch b. We can be ​% confident that the mean duration of​ imprisonment, ​, of all political prisoners with chronic PTSD is somewhere between 16.3 months and 44.3 months.