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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A researcher claims that the amounts of acetaminophen in a certain brand of cold tablets have a standard deviation different from the s = 3.3 mg claimed by the manufacturer. Assuming that a hypothesis test is conducted to test this claim, identify the null hypothesis H0 and the alternative hypothesis Ha. Which is correct? H0: s = 3.3 Ha: s ≠ 3.3 H0: s ≠ 3.3 Ha: s = 3.3 Ho: s > 3.3 Ha: s < 3.3 H0: s 3.3 Ha: s > 3.3
Below are the SPSS results of a one-sample t-test. Use the SPSS results to answer One-Sample Statistics N Mean Std. Deviation Std. Error Mean IncrFed$, 11 prgms 1139 5.576 2.453 ( a ) One-Sample Test Test Value = 5 95% Confidence Interval of the Difference t df Sig. (2-tailed) Mean Difference Lower Upper IncrFed$, 11 prgms (b) (c) 0 5.576 0.4333 0.7186 Replace “a,” “b,” and “c” in the tables above with the correct values. (Note that the population mean is 5.) Round to the third decimal place. At the .05 level of significance (two-tailed test), do we reject the null hypothesis?
A random sample of 663 teens who said they did not typically eat fast food reported an average daily calorie intake of 2258, with standard deviation 1519. A random sample of 413 teens who said they did typically eat fast food reported an average daily calorie intake of 2637, with standard deviation 1138. The two populations are not terribly skewed. Based on the above description, which of the following conditions are met for conducting a hypothesis test for the equality of two means? Choose one or more that apply. Select one or more: a. Both samples are random b. Samples are independent of one another c. Both samples are normal enough given the sample size d. Two population means need to be the same
The following is data on endotoxinconcentration in settled dust both for a sample of urbanhomes and for a sample of farm homes: U: 6.0 5.0 11.0 33.0 4.0 5.0 80.0 18.0 35.0 17.0 23.0F: 4.0 14.0 11.0 9.0 9.0 8.0 4.0 20.0 5.0 8.9 21.0 9.2 3.0 2.0 0.3 a. Determine the value of the sample standard deviationfor each sample, interpret these values, and thencontrast variability in the two samples. [Hint:Σxi = 237.0 for the urban sample and = 128.4 forthe farm sample, and Σxi2 = 10,079 for the urbansample and 1617.94 for the farm sample.] b. Compute the fourth spread for each sample and compare. Do the fourth spreads convey the same messageabout variability that the standard deviations do?Explain. c. The authors of the cited article also provided endotoxin concentrations in dust bag dust: U: 34.0 49.0 13.0 33.0 24.0 24.0 35.0 104.0 34.0 40.0 38.0 1.0F: 2.0 64.0 6.0 17.0 35.0 11.0 17.0 13.0 5.0 27.0 23.0 28.0 10.0 13.0 0.2
The following is a random sample of the price of regular unleaded gasoline from 10 gas stations in the area. 1.85 1.99 1.94 2.25 2.39 2.19 2.19 1.95 2.18 2.09 Suppose that the population standard deviation of gasoline prices is not known. Give a 95% confidence interval for the mean gasoline prices.
Consider the following hypotheses for a normal random sample: H0:μ=102, H1:μ≠102 Complete the following computer output. Variable Sample size Sample mean Sample std. dev. 95% Confidence Interval Lower-Bound of the mean μ 95%Confidence Interval Upper-Bound of the mean μ Value of test statistic p-value X N=16 x¯=98.33 s=8.61
Consider the following hypothesis test : H0:μ =12.5 Ha:μ ‡12.5 **THAT A sample of size 18 provided a sample mean (12.9) and sample standard deviation is (0.8). *What is the rejection rule ,using critical vale approach? What is your conclusion at α=0.05?
The security department of a factory wants to knowwhether the true average time required by the night guard to walk his round is 30 minutes. If, in a ran-dom sample of 32 rounds, the night guard averaged 30.8 minutes with a standard deviation of 1.5 minutes, deter-mine whether this is sufficient evidence to reject the null hypothesis μ = 30 minutes in favor of the alternative hypothesis μ Z 30 minutes. Use the four steps in the ini-tial part of Section 1 and the 0.01 level of significance.
A sports statiscian claims that the mean winning times for Boston Marathon women's division open champions is at least 2.68 hours. The mean winning time of a sample of 30 randomly selected Boston Marathon women's open division is 2.60 hours. Assume the population standard deviation is 0.32 hour. Alpha = 0.05, can you reject the claim?

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