Answered step by step
Verified Expert Solution
Link Copied!

Question

1 Approved Answer

Good Evening I need answers to these questions please, 1.) Part 1? Type I error: A company that manufactures steel wires guarantees that the mean

image text in transcribedimage text in transcribedimage text in transcribedimage text in transcribedimage text in transcribedimage text in transcribedimage text in transcribedimage text in transcribedimage text in transcribedimage text in transcribedimage text in transcribedimage text in transcribedimage text in transcribedimage text in transcribedimage text in transcribed

Good Evening I need answers to these questions please,

1.) Part 1?

image text in transcribedimage text in transcribedimage text in transcribedimage text in transcribedimage text in transcribedimage text in transcribedimage text in transcribedimage text in transcribedimage text in transcribedimage text in transcribedimage text in transcribedimage text in transcribedimage text in transcribedimage text in transcribedimage text in transcribed
Type I error: A company that manufactures steel wires guarantees that the mean breaking strength (in kilonewtons) of the wires is greater than 50. They measure the strengths for a sample of wires and test H0: [1. = 50 versus H1 : p. > 50. If a Type I error is made, What conclusion will be drawn regarding me mean breaking strength? 50- The conclusion will be that the mean breaking strength is less than greater than less than or ureater than A test is made of Ho: u = 20 versus H: u # 20. A sample of size n = 58 is drawn, and x = 18. The population standard deviation is 6 = 8. (a) Compute the value of the test statistic z. (b) Is Ho rejected at the a = 0.05 level? (c) Is Ho rejected at the a = 0.01 level? Part: 0 / 3 Part 1 of 3 (a) Compute the value of the test statistic. Round the answer to at least two decimal places. ZA test of Ho: u = 53 versus H, : u # 53 is performed using a significance level of a = 0.05. The value of the test statistic is Z = -1.97. Part: 0 / 3 Part 1 of 3 Determine whether to reject Ho. Since the test statistic (Choose one) in the critical region, we (Choose one) Ho at X 5 the a = 0.05 level. is not isFacebook: A study showed that two years ago, the mean time spent per visit to Facebook was 19.9 minutes. Assume the standard deviation is o = 6.0 minutes. Suppose that a simple random sample of 103 visits was selected this year and has a sample mean of x = 20.9 minutes. A social scientist is interested to know whether the mean time of Facebook visits has increased. Use the a = 0.01 level of significance and the P-value method with the TI-84 calculator. (a) State the appropriate null and alternate hypotheses. (b) Compute the P-value. (c) State a conclusion. Use the a = 0.01 level of significance. Part: 0 / 4 Part 1 of 4 (a) State the appropriate null and alternate hypotheses. HO: 00 0=0 H1: H This hypothesis test is a (Choose one) V test. right-tailed X 5 left-tailed two-tailedCalibrating a scale: Making sure that the scales used by businesses in the United States are accurate is the responsibility of the National Institute for Standards and Technology (NIST) in Washington, D.C. Suppose that NIST technicians are testing a scale by using a weight known to weigh exactly 1000 grams. The standard deviation for scale reading is known to be 6 = 3.2. They weigh this weight on the scale 58 times and read the result each time. The 58 scale readings have a sample mean of; = 1000.? grams. The calibration point is set too high if the mean scale reading is greater than 1000 grams. The technicians want to perform a hypothesis test to determine whether the calibration point is set too high. Use the at: 0.10 level of signicance and the Pvalue method with the 1184 calculator. Part: 0 I 4 Part 1 of4 State the appropriate null and alternate hypotheses. HoD EIEI EI=EI This hypothesis test is a (Choose one) 1 test. lefttailed righttailed twotai led House prices: Data from the National Association of Realtors indicate that the mean price of a home in Denver, Colorado, in 2.012 was 260.? thousand dollars. A random sample of 71 homes sold in 2013 had a mean price of 286 thousand dollars. l[Ian you conclude that the mean price in 2.013 differs from the mean price in 2012? Assume the population standard deviation is o = 132. Use the a: 0.10I level of signicance and the Pvalue method 1with the TI84 calculator. State the appropriate null and alternate hypotheses. HoD III-=- E|>I E|=I Elatl IJ. (Choose one} 1' test. rial-lt-taI-IEEI - lefttailed twotailed This hypothesis test is a Interpreting calculator display: The following \"Fl34 Plus display presents the results of a hypothesis test. 2 = 1.4T3363499 P = .1405531165 $41.52 n=T What are the null and alternate hypotheses? Good credit: The Fair Isaac lCorporation {ECU} credit score is used by banks and other lenders to determine whether someone is a good credit risk. Scores range from 300 to 850, with a score of T20 or more indicating that a person is a very good credit risk. An economist 1wants to determine whether the mean FICO score is more than the cutoff of 720. She nds that a random sample of 50 people had a mean FICO score of 765 with a standard deviation of 98. Can the economist conclude that the mean FICO score is greater than 720? Use the a: 0.01 level of signicance and the Pvalue memod with the "IT84 Plus calculator. III-:l Elbl E|=l ' Elatl L1 This hypothesis test is a test. righttailed twotailed Weight loss: In a study to determine whether counseling could help people lose weight, a sample of people experienced a groupbased behavioral intervention, which involved weekly meetings with a trained interventionist for a period of six months. The following data are the numbers of pounds lost for 14 people. Assume the population is approximately normal. Perform a hypothesis test to determine whether the mean weight loss is greater than 17 pounds. Use the o: = 0.01 level of signicance and the Pvalue method with the TI84 Plus calculator. 20.3 2?.0 6.1 22.5 19.5 11.2 16.0 19.5 35.9 32.3 10.7 34.3 21.9 17.7 Semldnta'bEmel III-cl El>l E|=l ' Elall |-1 This hypothesis test is a test. righttailed twotailed Commuting to work: A community survey sampled 1923 people in Colorado and asked them how long it took them to commute to work each day. The sample mean oneway commute time was 24.? minutes with a standard deviation of 13 minutes. A transportation engineer claims that the mean commute time is less than 25 minutes. Do the data provide convincing evidence that me engineer's claim is hue? Use the a: 0.1{} level of signicance and the critical value method with the 9 Critical Values for the Student's t Distribution Table. III-:- Eli-I E|=I test E"- H Ianta i lad - This hypothesis test is a righttailed twotail ed College tuition: The mean annual tuition and fees in the 20132014 academic year for a sample of 13 private colleges in California was $36,500 with a standard deviation of $7750. A dotplot shows that it is reasonable to assume that the population is approximately normal. Can you conclude that the mean tuition and fees for private institutions in California is greater than $35,000? Use the o: = l0.01 level of signicance and the critical value method with the 9 Critical Values for the Student's t Distribution Table. (a) State the appropriate null and alternate hypotheses. El-sI El:-I E|=I This hypothesis test is a left ta i led righttailed twotailed In a simple random sample of size 57, there were 45 individuals in the category of interest. Part: 0 / 4 Part 1 of 4 (a) Compute the sample proportion p. Round the answer to at least three decimal places. The sample proportion is X 5Spam: A researcher reported that 71.8% of all email sent in a recent month was spam. A system manager at a large corporation believes that the percentage at his company may be 77%. He examines a random sample of 500 emails received at an email server, and nds that 362 of the messages are spam. l[Ian you conclude that less than 77% of emails are spam? Use both 0'. = 0.05 and a: 0.10 levels of signicance and the critical value method with the table. State the appropriate null and alternate hypotheses. This hypothesis test is a leftta i led righttailed twotailed Curing diabetes: Vertical banded gastroplasty is a surgical procedure that reduces the volume of the stomach in order to produce weight loss. In a recent study, 82 patients with Type 2 diabetes underwent this procedure, and 59 of them experienced a recovery from diabetes. Does this study provide convincing evidence that less than 76% of those with diabetes who undergo this surgery will recover from diabetes? Use the a = 0.05 level of significance and the P-value method with the TI-84 Plus calculator. Part: 0 / 4 Part 1 of 4 (a) State the appropriate null and alternate hypotheses. Ho 00 0=0 P This hypothesis test is a (Choose one) test. X 5 right-tailed left-tailed two-tailedQuit smoking: In a survey of444 HIVpositive smokers, 199 reported that they had used a nicotine patch to try to quit smoking. Can you conclude that less than half of HIVpositive smokers have used a nicotine patch? Use the o.= 0.01 level of signicance and the Pvalue method with the 1184 Plus calculator. III-cl Elhl E|=I 3 El:- P This hypothesis test is a . - righttailed lefttailed twotailed

Step by Step Solution

There are 3 Steps involved in it

Step: 1

blur-text-image

Get Instant Access to Expert-Tailored Solutions

See step-by-step solutions with expert insights and AI powered tools for academic success

Step: 2

blur-text-image

Step: 3

blur-text-image

Ace Your Homework with AI

Get the answers you need in no time with our AI-driven, step-by-step assistance

Get Started

Recommended Textbook for

Intro Stats

Authors: Richard D De Veaux, Paul D Velleman, David E Bock

4th Edition

ISBN: 0321826213, 9780321826213

More Books

Students also viewed these Mathematics questions

Question

Derive Equations 1 for the case /2 < < .

Answered: 1 week ago

Question

Define Decision making

Answered: 1 week ago

Question

What are the major social responsibilities of business managers ?

Answered: 1 week ago

Question

What are the skills of management ?

Answered: 1 week ago