Using the p-value method, we reject the null hypothesis if the p-value is than or equal to the level of significance. 4) (Consider Ho: u=75 versus H1: µ<75. A random sample of 16 observations taken from thi population produced a sample mean of 73.2. The population is normally distributed with o=6 a) Calculate the test statistic (round to two decimal places) b) Use z-table or technology to get the p-value (round to four decimal places) 5) Consider H u-56 versus Hy ut56 A random sample of 25 observations taken from this

Using the p-value method, we reject the null hypothesis if the p-value is than or equal to the level of significance. 4) (Consider Ho: u=75 versus H1: µ<75. A random sample of 16 observations taken from thi population produced a sample mean of 73.2. The population is normally distributed with o=6 a) Calculate the test statistic (round to two decimal places) b) Use z-table or technology to get the p-value (round to four decimal places) 5) Consider H u-56 versus Hy ut56 A random sample of 25 observations taken from this

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Description: Using the p-value method, we reject the null hypothesis if the p-value isthan or equal to the level of significance.4) (Consider Ho: u=75 versus H1: µ<75. A random sample of 16 observations taken from thispopulation produced a sample mean of 73.2. T

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 research team conducted a study of VARIABILITY in students' test performance. Researchers assume that individual records are normally distributed with unknown mean and variance. They collected a sample of n = 25 test scores with summaries shown below. Sample Mean = X = 75.2 and Sample SD = S = 15 Estimate the POPULATION VARIANCE with confidence C = 0.99 • Show critical values needed • Evaluate the upper and lower confidence limits. UCL = LCL = Solution
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The claim is that for 10 AM body temperatures of males, the mean is less than 98.6°F. The sample size is n=39 and the test statistic is t=−4.041. Use technology to find the P-value. Based on the result, what is the final conclusion? Use a significance level of 0.01. Question content area bottom Part 1 State the null and alternative hypotheses. H0: μ ▼ greater than> less than< not equals≠ equals= enter your response here H1: μ ▼ greater than> less than< less than or equals≤ not equals≠ equals= greater than or equals≥ enter your response here (Type integers or decimals. Do not round. Do not include the degree symbol in your answer.) Part 2 The test statistic is enter your response here. (Round to two decimal places as needed.) Part 3 The P-value is enter your response here. (Round to three decimal places as needed.) Part 4 Based on the P-value, there ▼ is is not sufficient evidence at a significance level of 0.01 to…
Suppose the mean IQ score of people in a certain country is 100. Suppose the director of a college obtains a simple random sample of 39 students from that country and finds the mean IQ is 104.5 with a standard deviation of 13.6. Complete parts (a) through (d) below. (a) Consider the hypotheses Ho: µ= 100 versus H,: µ> 100. Explain what the director is testing. Perform the test at the a = 0.10 level of significance. Write a conclusion for the test. Explain what the director is testing. Choose the correct answer below. O A. The director is testing if the sample provided sufficient evidence that the population mean IQ score is actually not equal to 100. OB. The director is testing if the sample provided sufficient evidence that the population mean IQ score is actually greater than 100. O C. The director is testing if the sample provided sufficient evidence that the population mean IQ score is actually not greater than 100. O D. The director is testing if the sample provided sufficient…
A population has parameters μ = n = 191. 49.2 and o= = 55.3. You intend to draw a random sample of size What is the mean of the distribution of sample means? fx = What is the standard deviation of the distribution of sample means? (Report answer accurate to 2 decimal places.) Ox =
The random variable, x is a water quality parameter, suspended solids with units of parts per million (ppm). A sample size of 60 yields a mean of 59.87 ppm, and a standard deviation of 12.5 ppm. Water quality criteria requires that the suspended solids do not exceed a mean, µ, = 50 ppm . Test the hypotheses H.: u = µ, vs. HA: u > H. (right-tailed test) at a level of significance of 5 % and assume the pop variance is known.
Use the AustStates data. Perform a repeated measures ANOVA using the 9 years as the paired variables and the states as the observations. Based on the Greenhouse-Geisser p-value of p=0.130, sphericity can be assumed for these data. The result of the analysis is F=9.563 on degrees of freedom, for a p-value of You the null hypothesis. 8 and 56, p<0.001, reject 8 and 56, p<0.001, fail to reject 1.0 and 7.3, p=0.016, reject 1.0 and 7.3, p=0.016, fail to reject
A previous study showed to determine whether a sales tax on soda will reduce consumption of soda in the US below the current per-capita level of about 50 gallons of soda per year. Hypotheses Test is Ho: m(mue) =50 vs. Ha: m<50 where m (mue) represents the average annual consumption of soda in communities where the sales tax is implemented. (b) Now supposed sample results give a p-value of 0.41. Interpret this p-value in terms of random chance and in the context of taxes and soda consumption.
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You wish to test the following claim (Ha) at a significance level of a = 0.02. Ho:μι ≥ με Ha:μι < με You believe both populations are normally distributed, but you do not know the standard deviations for either. You obtain a sample of size ni You obtain a sample of size n₂ = = 12 with a mean of M₁ 13 with a mean of M2 test statistic = p-value = = - 10.6. 53 and a standard deviation of SD1 59.5 and a standard deviation of SD₂ = 9.7. [three decimal accuracy] [three decimal accuracy] =
Part 6 of 7 (f) Determine whether to reject H. Use the a=0.10 level of significance. Do not reject the null hypothesis H Part: 6/7 Part 7 of 7 (g) State a conclusion. There is not enough evidence to conclude that the mean lead amount is less than 15 micrograms per liter. X