Exercise 15. For the following four cases, perform statistical tests with significance level a = 5%,testing the hypothesis Ho: μ = 0 against the alternative H₁: 0. Show how you perform eachtest and state the outcome.a) We have observed 100 independent, normally distributed values with known variance o² = 1.The average of the observed values is -0.18.b) We have observed 10 independent samples of a normal distribution with known varianceo² = 4: The observed values are 0.595, 3.923, 0.12, 1.078, 0.578, 1.575, 2.26, -0.031, 1.733,and 1.135.c) We have observed 10 independent samples from a normal distribution. The sample mean is* = -1.405, the exact variance is unknown but the sample variance is s2 = 1.456.

Introduction: A z-test is a type of statistical hypothesis test that is used to determine whether…

Answered: Exercise 15. For the following four…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

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The standard z-value of an arbitrary 16-unit sample from the normal-class population with a mean of 60 and a variance of 100 is 2.0.
A survey was conducted to study the height of people from IBRI and NIZWA. The results are shown here. The researcher is interested to examine is there a significant difference between the average height of people of the two places at a=0.05. Write the hypothesis and evaluate t-test value only. [Assuming the population variances are equal and use 2-decimal places].
Do men score the same on average compared to women on their statistics finals? Final exam scores of eleven randomly selected male statistics students and eleven randomly selected female statistics students are shown below. Male: 59 85 89 66 70 92 75 91 61 65 58 Female: 54 49 81 63 75 58 59 55 75 74 70 Assume both follow a Normal distribution. What can be concluded at the the αα = 0.05 level of significance level of significance? For this study, we should use Select an answer t-test for the difference between two independent population means t-test for the difference between two dependent population means t-test for a population mean z-test for the difference between two population proportions z-test for a population proportion The null and alternative hypotheses would be: H0:H0: Select an answer p1 μ1 Select an answer < = ≠ > Select an answer p2 μ2 (please enter a decimal) H1:H1:…
Please explain
A physical therapist wanted to know whether the mean step pulse of men was less than the mean step pulse of women. She randomly selected 54 men and 70 women to participate in the study. Each subject was required to step up and down a 6-inch platform. The pulse of each subject was then recorded. The following results were obtained. Two sample T for Men vs Women N Mean StDev SE Mean Men Women 98% CI for mu Men - mu Women (- 12.20, - 1.00) T-Test mu Men = mu Women (vs H2 O C. Ho: H1 = H2; Ha: H1 #H2 (b) Identify the P-value and state the researcher's conclusion if the level of significance was a = 0.01. What is the P-value? P-value =
Data on the weights (Ib) of the contents of cans of diet soda versus the contents of cans of the regular version of the soda is summarized to the right. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. Use a 0.01 significance level for both parts. Diet Regular H2 27 27 0.79037 lb 0.80399 lb 0.00449 lb 0.00756 lb a. Test the claim that the contents of cans of diet soda have weights with a mean that is less than the mean for the regular soda. What are the null and alternative hypotheses? O A. Ho: H1 = H2 OB. Ho: H1#H2 Hq: Hy
X is a random variable with mean X = 3, variance o = 2.
A researcher was interested in knowing if mean BMI was different between males and females. He obtained the following test results. Choose the correct interpretation of the results: F test p value = 0.03 Two sample t test, assume equal variance = 0.02 Two sample t test, assume unequal variance = 0.06 Mean BMI is higher in males than females. None Mean BMI is not significantly different between males and females. Mean BMI is significantly different between males and females.
A researcher was interested in knowing if mean BMI was different between males and females. He obtained the following test results. Choose the correct interpretation of the results: F test p value = 0.30 Two sample t test, assume equal variance = 0.02 Two sample t test, assume unequal variance = 0.61 Mean BMI is not significantly different between males and females. None Mean BMI is higher in males than females. Mean BMI is significantly different between males and females.
Assume that the examination marks of students follow normal distri- butivn with unknown mean (4) but known variance (a?) = 0.25. Test the hypothesis Ho : u =8 against H: #8 with the level of significance a = 0.01, the sample mean being equal to 7.8 and the sample size being 50.
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.
The average score on a statistics test was 2.21 with a standard deviation of 0.88. The teacher suspects that the exam was very difficult; Based on the above, you want to adjust the scores so that the mean is 3.37 and the standard deviation is 0.25. Compute the values a and b such that the fit of type aX+b gives the desired mean and variance? Calculate: a= Answer b= Answer TIP: Use expected value and variance properties.

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