Exercise 5. Assume that we have observed the following values from a normal distribution withknown variance o? = 1 and unknown mean µ.1.23-0.671.161.67-0.242.990.021.170.271.21.Test the hypothesis Ho: µ = 0 against the alternative H1: µ #0 at significance level a =5%.

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Answered: Exercise 5. Assume that we have…

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 sample of size 5 was collected from an unknown population 15 15 16 16 17 1. Use the table of expected z-scores (rounded to the nearest tenth) below to construct the normality plot with expected z-scores on the x-axis and the observations on the y-axis: 6 -1.3 -0.6 -0.2 0.2 0.6 1.3 5 -1.2 -0.5 0 0.5 1.2 -1 19 18 normal O not normal Clear All Draw: 17- 16- 14 T 15- رها 1 2 7 -1.4 -0.8 -0.4 0 0.4 0.8 1.4 8 -1.4 -0.9 -0.5 -0.2 0.2 0.5 0.9 1.4 9 -1.5 -0.9 -0.6 -0.3 0 0.3 0.6 0.9 1.5 2. Based on whether the pattern above is linear or not, in your opinion, was the sample drawn from a normal population or not?
One group of accounting students took a distance learning class, while another group took the same course in a traditional classroom. At α = .10, is there a significant difference in the mean scores listed below? Exam Scores for Accounting Students Statistic Distance Classroom Mean scores x¯1x¯1 = 9.7 x¯2x¯2 = 10.0 Sample std. dev. s1 = 2.7 s2 = 2.9 Number of students n1 = 23 n2 = 23 Specify the decision rule. (Round your answers to 3 decimal places. A negative value should be indicated by a minus sign.) Find the test statistic tcalc. (Round your answer to 4 decimal places. A negative value should be indicated by a minus sign.) Use Excel to find the p-value. (Round your answer to 3 decimal places.)
Let x = age in years of a rural Quebec woman at the time of her first marriage. In the year 1941, the population variance of x was approximately ?2 = 5.1. Suppose a recent study of age at first marriage for a random sample of 41 women in rural Quebec gave a sample variance s2 = 2.9. Use a 5% level of significance to test the claim that the current variance is less than 5.1. Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.) What are the degrees of freedom?
thill An experiment compares two different processes for producing steel plate. The measurements represent the thickness of the plate. The samples gave m = 8, X = 6.701, S1 = 0.108 n = 6, Y = 6.841, S2 = 0.155 %3D %3D Assume that the two populations are normal with equal variances, test Ho: H1 - H2 = 0 against H1 : µ1 - µ2#0 at a 0.05 significance level. 5.
Calculate the missing values in the Stata output (SKIP E!!!). Two-sample t test with A variances X y combined diff Obs Ho: diff = 0 49 30 79 diff = mean(x) Ha: diff |t|) = Std. Dev. 6.5 7.4 6.935314 G E [95% Conf. Intervall 40.33298 39.62126 C F degrees of freedom = 44.86702 Pr{T > t) 42.72811 1.6996 77 Ha: diff > 0 H
Results of a research claim to predict wind speed at a place with a variance of 4 km/hr or less. A random sample of 24 measurements provides a sample variance of s2 = 4.9. Assuming that the population distribution of wind speed is approximately normal, check the validity of the claim at 5% level of significance using chi-square test.
Total fat content (including saturated fat and trans fats) is a key characteristic of cheese. Assume that for a given type of cheese, total fat content (10-1 g/g of cheese) follows a normal distribution with a population mean of 2.85 (unit) and a population variance of 0.01 (unit2). Such an assumption is reasonable because fat content varies inside cheese and the “errors” around the mean total fat content (i.e. the value appearing on the etiquette) are distributed like a bell shape. a) What is the probability that the total fat content of this cheese exceeds 3 (unit)? b) What is the probability that the total fat content varies from 2.8 to 3 (unit)?
You have data drawn from a normal distribution with a known variance of 16. You set up the following NHST: • Ho: data follows a N(2, 4²) • HÃ: data follows a N(µ, 4²) where µ ‡ 2. Test statistic: standardized sample mean z. Significance level set to a = .05. You then collected n = 16 data points with sample mean 1.5. (a) Find the rejection region. Draw a graph indicating the null distribution and the rejection region. (b) Find the z-value and add it to your picture in part (a). (c) Find the p-value for this data and decide whether or not to reject Ho in favor of HA.
Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 4.6 parts/million (ppm). A researcher believes that the current ozone level is at an excess level. The mean of 14 samples is 4.9 ppm with a variance of 1.2 Does the data support the claim at the 0.01 level? Assume the population distribution is approximately normal. Step 2 of 5 : Find the value of the test statistic. Round your answer to three decimal places.
A sample of five cans of standard grade tomatoes in puree was taken at random from a production line immediately after filling, and the solid content of each can was weighed. The company desires to test whether the average weight per can is at the specified level of 22 ounces. Past experience indicates that the distribution of the weights per can is approximately normal. The α risk for the test is to be controlled at .01. For the sample, the mean weight was ?̅ =23.9 and the sample standard deviation was s = 0.822 a. State the null and alternative hypotheses. b. Determine the rejection region – indicate how you did this; don’t just give the answer. c. Compute the test statistic – indicate what formula was used. d. What is your conclusion and why?
You wish to test the following claim (Ha) at a significance level of α=0.002 Ho:μ=81.8 Ha:μ<81.8You believe the population is normally distributed, but you do not know the standard deviation. You obtain a sample of size n=22n=22 with mean M=76.5M=76.5 and a standard deviation of SD=14.6SD=14.6.What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic = IncorrectWhat is the p-value for this sample? (Report answer accurate to four decimal places.) p-value = CorrectThe p-value is... less than (or equal to) αα greater than αα This test statistic leads to a decision to... reject the null accept the null fail to reject the null As such, the final conclusion is that... There is sufficient evidence to warrant rejection of the claim that the population mean is less than 81.8. There is not sufficient evidence to warrant rejection of the claim that the population mean is less than 81.8. The sample data support the claim…

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Exercise 5. Assume that we have observed the following values from a normal distribution with known variance o² = 1 and unknown mean µ. 1.23 -0.67 1.16 1.67 -0.24 2.99 0.02 1.17 0.27 1.21. Test the hypothesis Ho: u = 0 against the alternative H1: u70 at significance level a = 5%.