The following MINITAB output presents the results of a hypothesis test for a populationmean u.One-Sample Z: XTest of mu = 45 vs > 45The assumed standard deviation = 5.0921Variable N Mean StDevSE95% Lower z PMeanBoundх12246.2598 5.0921 0.461045.50152.730.003Is this a one-tailed or two-tailed test?a.b.What is the null hypothesis?C.What is the P-value?d.Use the output and an appropriate table to compute the P-value for the test of Họ: µ S46 vs. H1: µ > 46.Use the output and an appropriate table to compute a 98% confidence interval for u.e.

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Answered: The following MINITAB output presents…

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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An avionics company uses a new production method to manufacture aircraft altimeters. A single random sample of new altimeters resulted in the errors listed below. Use a 5% significance level to test the claim that the new production method has errors with a standard deviation greater than 32.2 ft, which was the standard deviation for the old production model (o). If it appears that the standard deviation is greater, does the new production method appear to be better or worse than the old method? Should the company take any action? -42 78 -22 -72 -45 15 17 51 -5 -53 -9 -109 Find the upper and lower boundaries of the 95% confidence interval for o. 98 37 32 73 89
The breaking strengths of cables produced by a certain manufacturer have a mean, u, of 1875 pounds, and a standard deviation of 100 pounds. It is claimed that an improvement in the manufacturing process has increased the mean breaking strength. To evaluate this claim, 50 newly manufactured cables are randomly chosen and tested, and their mean breaking strength is found to be 1912 pounds. Can we support, at the 0.1 level of significance, the claim that the mean breaking strength has increased? (Assume that the standard deviation has not changed.) Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places, and round your responses as specified below. (If necessary, consult a list of formulas.) (a) State the null hypothesis H and the alternative hypothesis H,. p H, :0 H, :0 (b) Determine the type of test statistic to use. (Choose one) ▼ D=0 OSO O20 (c) Find the value of the test statistic. (Round to three or more decimal…
In a simple random sample of 46 oral temperatures, the mean was 37.3 degrees Celsius with a standard deviation of 1.1 degrees Celsius. Use this sample to test the claim that the population mean of all oral temperatures is greater than 37 degrees Celsius, with significance level of 0.05. d.f.=45 to.005 = 2.690 to.01 = 2.412 to.025 2.014 to.05 = 1.679 to.10 = 1.301
A simple random sample of 25 filtered 100 mm cigarettes is obtained, and the tar content of each cigarette is measured. The sample has a mean of 19.8 mg and a standard deviation of 3.18 mg. Use a 0.05 significance level to test the claim that the mean tar content of filtered 100 mm cigarettes is less than 21.1 mg, which is the mean for unfiltered king size cigarettes. What do the results suggest, if anything, about the effectiveness of the filters?
Data provided by the National Association of College Employers indicated that the average starting salary for graduates with a Bachelors Degree in Accounting was $37,000 in 2008. In June 2010 a sample of 48 graduating accounting majors provided a sample average starting salary of $38,100, with a sample standard deviation of $5200. Conduct a hypothesis test to determine if it can be assumed that accounting graduates in 2010 had higher average salaries than graduates in 2008, using α = .05.
A random sample of 23 bags of apples (marked as 10 pounds each) received by a large grocery chain tested out as having a mean of 9.2 pounds with a variance of 2.56 pounds. Test whether the true mean of all bags is under 10 pounds. Set up hypotheses. Perform the appropriate test by showing your formula. Interpret the results using a Type I (alpha) error of .05. Also provide the p value here. Also construct a 95% confidence interval around your sample mean (X bar) that should contain the true mean (mu). Also interpret this interval.
5.37 Dental anxiety study. To gauge their fear of going to a den- tist, a random sample of adults completed the Modified Dental Anxiety Scale questionnaire (BMC Oral Health, Vol. 9, 2009). Scores on the scale range from zero (no anx- iety) to 25 (extreme anxiety). The mean score was 11 and the standard deviation was 4. Assume that the distribution of all scores on the Modified Dental Anxiety Scale is ap- proximately normal with µ = 11 and o = 4. a. Suppose you score a 10 on the Modified Dental Anxiety Scale. Find the z-value for your score. b. Find the probability that someone scores between 10 and 15 on the Modified Dental Anxiety Scale. c. Find the probability that someone scores above 20 on the Modified Dental Anxiety Scale.
A survey using a sample size of 11 reveals that the average size of dairy farms in Simcoe County is 1500 hectares with a standard deviation of 150 hectares. As a researcher, you take this information and conduct a hypothesis test and find the calculated value of the test statistic is 2.2. Prior to the study, the following hypotheses were established: но: и - 1400 H1: u > 1400 Use a = 0.01. Based on the test statistic and sample size, what the p-value is so we should the null hypothesis. _(0.0262, .0262, 0.026, .026) Given that the p-value is (greater/less) (greater, larger, >, bigger) than a, we should (accept/reject) _ lаccept) the null hypothesis. 0.062, reject ) 0.0131, accept 0.0026, reject ) 0.0262, accept 0.062, accept
The human resources department of a large investment bank announced that the number of people it interviews monthly has a mean of 115 with a standare deviation of 15.4. The management of the bank suspects that the standard deviation exceeds 15.4. Suppose that the management wants to take a small sample of months and carry out a hypothesis test to see if its suspicions have support. State the null hypothesis H, and the alternative hypothesis H, that is would use for this test. Ho: O H₁: 0 μ 5 OSO ローロ X X a > * 3 р S 20
Scores on a certain "IQ" test for 18-25 year olds are normally distributed. A researcher believes that the average IQ score for students at a certain NJ college is less than 110 points, and so wants to test this hypothesis. The researcher obtain a SRS of 45 student IQ scores from school records and found the mean of the 45 results was 108 with a sample standard deviation of 21. The level of significance (alpha) used for this problem is 0.05/ What is the appropriate test statistic (Student must complete by showing by formula using the ap- propriate values in that formula "showing work" and the final answer and appropriate label)? O T test score = (108-110)/(45/sqrt(21)) = -.2037 O I test score = (108-110)/(21/sqrt(45)) = -.639 T test score = (108-110)/(21/sqrt(45)) = .639 I test score = (110-108)/(21/sqrt(45)) = .639
Activity 4: A manufacturing company of sports equipment has developed a new synthetic fishing line that it claims has a mean breaking strength of 8 kilograms with a standard deviation of 0.5 kilogram. Test the hypothesis that 8 kilograms against the alternative that u # 8 kilograms if a random H1 sample of 50 lines is tested and found to have a mean breaking strength of 7.8 kilograms. Use a 0.01 level of significance,
The average salary in this city is $47,400. Is the average less for single people? 70 randomly selected single people who were surveyed had an average salary of $45,379 and a standard deviation of $13,700. What can be concluded at the αα = 0.05 level of significance? For this study, we should use Correct The null and alternative hypotheses would be: H0:H0: Correct Correct H1:H1: Correct The test statistic = (please show your answer to 3 decimal places.) The p-value = (Please show your answer to 4 decimal places.) The p-value is αα Based on this, we should the null hypothesis. Thus, the final conclusion is that ... The data suggest that the population mean is not significantly less than 47,400 at αα = 0.05, so there is statistically insignificant evidence to conclude that the population mean salary for singles is less than 47,400. The data suggest that the populaton mean is significantly less than 47,400 at αα…

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