Should the standard normal (2) distribution or Student's (t) distribution be used for this test?The standard normal (z) distribution should be usedO The Student's t distribution should be usedDetermine the critical value(s) for this hypothesis test. Round the solution(s) to four decimal places. Ifmore than one critical value exists, enter the solutions using a comma-separated list.Determine the test statistic. Round the solution to four decimal places.Determine the appropriate conclusion for this hypothesis test.O The sample data provide sufficient evidence to reject the null hypothesis that the mean weight ofthe spray paint cans is 16 and thus we conclude that the manufacturing line needs inspection.O The sample data do not provide sufficient evidence to reject the alternative hypothesis that themean spray paint can weight is different from 16 and thus we conclude that the manufacturing lineneeds inspection.The sample data do not provide sufficient evidence to reject the null hypothesis that the meanweight of the spray paint cans is 16 and thus we conclude that the manufacturing line does not needinspection.O The sample data provide sufficient evidence to reject the alternative hypothesis that the mean spraypaint can weight is different from 16 and thus we conclude that the manufacturing line does notneed inspection.75°FPartly cloudy Rust-Oleum is a company that manufactures and sells spray paint. Their American Accents line of spraypaints are guaranteed to contain 16 ounces of paint per can. Due to the fact that under-filling cans of spraypaint is bad for customer relations and over-filling spray paint cans is dangerous, Rust-Oleum has a strictquality control process. Every 1.5 hours, a quality control officer randomly selects a sample of 29 spraypaint cans from the manufacturing line and carefully weighs them to ensure that each can of paint contains16 ounces, on average. If the sample mean is found to be sufficiently different than 16 ounces, themanufacturing line is shut down and inspected.The mean weight of a recent sample of 29 spray paint cans was found to be 16.02 ounces with a standarddeviation of 0.11 ounces.Use the critical value method to test the hypothesis that the mean weight of the spray paint cans from thismanufacturing line is different than 16 ounces, using a = 0.1. Assume that the distribution of all spraypaint cans from Rust-Oleum is known to be approximately normally distributed.State the null and alternative hypothesis for this test.Ho: p= 0.094H₁: p> 0.094Determine if this test is left-tailed, right-tailed, or two-tailed.Oright-tailedOleft-tailedtwo-tailedHE75°FPartly cloudy

Hypothesized Population Mean (µ) = 16 The provided sample mean is = 16.02 Sample standard deviation…

Answered: Determine the critical value(s) for…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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You are given the null and alternative hypotheses and the sample information shown below. Complete parts a and b below. Sample 1 n1 = 120 S, = 34 X = 130 Sample 2 n2 = 135 HA: H1- H2#0 S2 = 36 X2 = 150 a. Develop the appropriate decision rule, assuming a significance level of 0.01 is to be used. Select the correct choice below and fill in the answer box(es) to complete your choice. (Round to two decimal places as needed.) O A. Reject Họ if t O B. Reject Họ if t> O C. Reject H, if t< b. Test the null hypothesis and indicate whether the sample information leads you to reject or fail to reject the null hypothesis. Use the test statistic approach. Calculate the value of the test statistic. t= (Round to two decimal places as needed.) Since the test statistic V in the rejection region, the null hypothesis. There is V evidence at a = 0.01 to conclude that the two population means are different.
A business has just completed a marketing campaign. They want to determine if their original market share of 50% has increased as a result of the campaign. A survey conducted found 81 out of 150 people had bought the products. Determine whether the company has increased its market share i. Write down the null and alternative hypotheses for this test.ii. The sample proportion value is missing from the output above. What is this value?iii. Explain why we can use the following test static to conduct this test.? =?̂− ?√??⁄?iv. Use either the ?-value or the ? statistic in the output provided to conduct the test. Has the market share of the company increased?v. The marketing team plan to continue to monitor their market share. In a future study, they want to be able to estimate their market share to within +/−5% and with 90% confidence. What size sample is required? Use the sample proportion from part ii.
Q: Using the data (provided below this question) test the hypotheses that the mean difference in writing and reading scores is zero and that the mean difference in math and science scores is zero. Use the paired-sample procedure. Use a 5% significance level if the problem doesn’t give an alpha. Note: Please only provide the t stat, p(T<=t) value and critical value. No need to explain the steps. School District Writing Reading Math Citizenship Science All Indian Hill 95 98 89 95 91 83 Wyoming 98 96 86 93 87 81 Mason City 96 92 85 94 86 72 Madiera 94 95 88 82 88 69 Mariemont 99 92 74 89 88 68 Sycamore 85 88 80 87 84 68 Forest Hills 93 91 73 85 88 67 Kings Local 92 86 78 82 78 64 Lakota 90 88 73 85 81 64 Loveland 85 93 72 86 86 61 Southwest 92 82 73 78 73 58 Fairf ield 90 86 71 83 77 57 Oak Hills 88 86 75 77 79 57 Three Rivers 87 85 66 84 77 56 Milford 82 86 72 82 76 53 Ross 84 85 66 75 78 52 West Clermont 88 83 63 73 70 48 Reading 88 80 58 76 75…
Floretta, a golfer, claims that her drive distance is not equal to 238 meters, on average. Several of her friends do not believe her, so she decides to do a hypothesis test, at a 5% significance level, to persuade them. She hits 19 drives, collects the proper data, and works through the testing procedure: • Hou = 238; HH #238 • a = 0.05 (significance level) The test statistic is X-Ho Zo == 256-238 14 V19 <= 5.6 • The critical values are -20.025 = -1.96 and 20.025 = 1.96. Conclude whether to reject or not reject Ho. and interpret the results. Select the correct answer below: Reject Ho. At the 5% significance level, the test results are not statistically significant and at best, provide weak evidence against the null hypothesis. Reject Ho. At the 5% significance level, the data provide sufficient evidence to conclude that the mean distance is not equal to 238 meters. Do not reject Ho. At the 5% significance level, the test results are not statistically significant and at best, provide…
Historically, the percentage of U.S. residents who support stricter gun control laws has been 55%. A recent Gallup Poll of 2500 people showed 1723 in favor of stricter gun control laws . Assume the poll was given to a random sample of people. QUESTION Test the claim that the proportion of those favoring stricter gun control has changed from 0.55. Perform a hypothesis test, using a significance level of 0.01. What kind of test will you need to perform? A right, one-tailed z-test A left, one-tailed z-test A two-tailed z- test
You are conducting a study to see if the proportion of women over 40 who regularly have mammograms is significantly different from 0.84. You use a significance level of a = 0.02. Ho:p = 0.84 Ha:p +0.84 You obtain a sample of size n = 700 in which there are 574 successes. What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic What is the p-value for this sample? (Report answer accurate to four decimal places.) p-value = The p-value is... O less than (or equal to) a O greater than a This test statistic leads to a decision to... O Reject the null O Accept the null O Do not reject the null As such, the final conclusion is that... O There is sufficient evidence to warrant rejection of the claim that the proportion of women over 40 who regularly have mammograms is different from 0.84. O There is not sufficient evidence to warrant rejection of the claim that the proportion of women over 40 who regularly have mammograms is different from…
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what do part b and part f say? I'm unsure what all the words say.
Consider the following passage: I ran a regression, with many variables to predict the result of another variable which was the murder rate. One can see that lots of things, can cause the murder rate to increase or decrease. I tried to account for all the important factors, and those factors are the SAT scores, unemployment rate, and international migration per 1,000. The SAT score is, average combined total score participants did on the SAT exam. The unemployment rate is, "a measure of the prevalence of unemployment and it is calculated as a percentage by dividing the number of unemployed individuals by all individuals currently in the labor force." (Wikipedia) International migration per 1,000 is, the number of people who come into a state from other countries per 1,000 people who live in the state. After I run the regression I will look at the t scores and p values and I should hopefully conclude that international migration does not cause crime. Which writing mistakes, if any, did…
The Bureau of Labor Statistics reports that the official unemployment rate for Black people was 10.4% and 4.7% for White people in February 2015. Select all correct answers for this question. O The samples of white and black people are independent. The explanatory variable is the unemployment rate. The response variable is the unemployment rate. The response variable is race.
A psychologist wants to demonstrate the effectiveness of ACT therapy for the treatment of depression. 30 people who were seeking treatment for depression asked to complete the Beck depression inventory (BDI) before the first sessionof therapy and again after one month of therapy. Which statistical test would be best to determine if there is a significant difference between measures? One sample t-test Dependent samples t-test Independent samples t- test ANOVA
Determine the Test statistic. Determine the P-Value. State the final conclusion that addresses the original claim.

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