The following MINITAB output (first shown in Exercise 16 in Section 6.1) presents theresults of a hypothesis test for a population mean u.One-Sample Z: XTest of mu = 73.5 vs not = 73.5The assumed standard deviation = 2.3634Variable N Mean StDev SE Mean95% CIх145 73.2461 2.36340.1963 (72.8614, 73.6308) –1.29 0.196a.Can Họ be rejected at the 5% level? How can you tell?b. Someone asks you whether the null hypothesis H,: H = 73 versus H1: H# 73 can berejected at the 5% level. Can you answer without doing any calculations? How?

From the given information, Test of hypothesis: H0: µ =73 H0: µ ≠73

Answered: The following MINITAB output (first…

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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Carry out the hypothesis test for the mean of a normally distributed population given the following information: Ho: µ ≥2.5 Ha: µ ≠2.5 Sample size = 30 mean = 1.8, variance =1.7, α = .05
There is a claim that playing e-sport increase food intake. The following table (Table 2) shows data on food intake (in Kcal) for 15 men on the 2 hours playing e-sport each night and for 15 men on the 4 hours playing e-sport each night. Conduct a hypothesis test with a = 0.05 to determine there is a significant difference in mean food intake for two different group. Table 2 2-hour playing e-sport 3586 4470 3068 5338 2221 4791 4435 3099 3187 3901 3868 3869 4878 3632 4518 4965 3918 1987 4993 5220 3653 3510 3338 4100 5792 4547 3319 3336 4304 4057 4-hour playing e-sport
The following MINITAB output presents the results of a hypothesis test for the difference Hx - Hy between two population means. Paired T for X - Y Mean StDev SE Mean 12 134.233 68.376 19.739 12 100.601 94.583 27.304 Difference 12 33.6316 59.5113 17.1794 95% lower bound for mean difference: 2.7793 T-Test of mean difference = 0 (vs > 0): T-Value = 1.96 P-Value = 0.038 Is this a one-tailed or two-tailed test? What is the null hypothesis? Can H, be rejected at the 1% level? How can you tell? a. b. C. d. Use the output and an appropriate table to compute a 98% confidence interval for µx Hy.
A sample of size 115 will be drawn from a population with mean 48 and standard deviation 12. Use the TI-83 Plus/TI-84 Plus calculator. Part 1 of 2 (a) Find the probability that x will be less than 45. Round the answer to at least four decimal places. The probability that x will be less than 45 is 0.0037 Part: 1/ 2 Part 2 of 2 (b) Find the 25 th percentile of x. Round the answer to at least two decimal places. The 25 th percentile is Skip Part Check Save For Later Subm © 2022 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Cente ......... .........a ... MacBook Air DII 80 F10 F7 F9 esc F4 F6 F2 F3 F1 @ 23 $ % & 1 2 3 4 6. 8.
Consider a dataset containing the heights of 38 females who said they prefer to sit in the front of the classroom yields a sample average of 62 inches and a sample standard deviation of 2.5 inches. Suppose that the average height for the general population of college females is 65 inches. An analyst wants to test the claim that the true average height of females who prefer to sit in the front of the classroom is lower than it is for the general population. In the hypothesis for the above claim (at significance level of 10%), the appropriate test, test statistic (name and value), and pvalue is the approximate Z-test with z-statistic = 62 and pvalue = 0. the approximate Z-test with z-statistic = 7.40 and pvalue not calculable because normality is not given. = 0. the approximate Z-test with z-statistic = -7.40 and pvalue = 0.
In a lake pollution study, the concentration of lead in the upper sedimentary layer of a lake bottom is measured from 25 sediment samples. The sample mean and the standard deviation of the measurements are found to be 0.38 and 0.06, respectively. Suppose Ho : u = 0.34 H1: u# 0.34 (a) State Type I and Type Il errors. (b) Conduct a hypothesis test at 0.01 level of significance by doing the seven-step classical approach. (please show all seven steps, formulas, calculations and the curve)
3.3.15 Use z scores to compare the given values. Based on sample data, newborn males have weights with a mean of 3293.7 g and a standard deviation of 551.3 g. Newborn females have weights with a mean of 3000.6 g and a standard deviation of 590.5 g Who has the weight that is more extreme relative to the group from which they came; a male who weighs 1700 g or a female who weighs 1700 g? V has the weight that is more extreme. Since the z score for the male is z= and the z score for the female isz= the (Round to two decimal places.)
The state's education secretary claims that the average cost of one year's tuition for all private high schools in the state is $2350. A sample of 30 private high schools is selected, and the average tuition is $2315. The standard deviation for the population is $38. At α = 0.05, is there enough evidence to reject the claim that the average cost of tuition is equal to $2350?
7.1
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.
8.2 Use technology to find the P-value for the hypothesis test described below. The claim is that for a smartphone carrier's data speeds at airports, the mean is μ=17.00 Mbps. The sample size is n=21 and the test statistic is t=−1.798
9.9

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