Suppose that a random (i.i.d.) sample of nine observations X1, ..., X, is taken from thenormal distribution with unknown mean µ and unknown variance o², and it is desired totest the following hypotheses: Ho µ = 20 vs. Hị : µ + 20. Suppose also that the sampledata are such that X; = 198 and (X; - Xn)² = 72. Carry out a t-test at thelevel of significance (i.e., type I error rate) 0.1 such that each tail of the rejection region hasprobability 0.05. Should the hypothesis Ho be rejected or not?(Maybe) useful facts: The 0.95-quantile of t(8) distribution is 1.860. The value of CDFof t(8) distribution at 2 is 0.960. The 0.95-quantile of t(9) distribution is 1.833. The valueof CDF of t(9) distribution at 2 is 0.962.Instruction: Do not simply provide a Yes/No answer. Please justify your derivation.Include in your answer the following: (1) What is the test statistic? (2) What is the dis-tribution of the test statistic (under Ho)? (3) What is the value of the test statistic in thisparticular data set? (4) What is the rejection region with the desired type I error rate (orp-value)? (5) What comparison leads to your decision of rejecting/not rejection Ho?19

Here given row data for sample statistics Using given data calculate sample statistics and perform…

Answered: level of significance (i.e., type I…

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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For the given scenario, determine the type of error that was made, if any. (Hint: Begin by determining the null and alternative hypotheses.) Researchers state 2.5 hours as the mean amount of television watched by children per night. One parent claims that the mean amount of television watched by children per night is less than 2.5 hours. The parent conducts a hypothesis test and fails to reject the null hypothesis. Assume that in reality, the mean amount of television watched by children per night is 2.3 hours. Was an error made? If so, what type?
Chapter 9, Section 3, Exercise 061 Two intervals are given, A and B, for the same value of the explanatory variable. A: 16.5 to 25.1; B: 16.3 to 25.3; (a) Which interval is the confidence interval for the mean response?A or B Which interval is the prediction interval for the response? A or B (b) What is the predicted value of the response variable for this value of the explanatory variable?Enter the exact answer.The predicted value is=
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Test the claim that the proportion of people who own a pet is smaller than 50% at the 0.05 significance level. The null and alternative hypothesis would be: Но: р 3 0.5 Но:д > 0.5 Но:д 0.5 Ні:р 0.5 Hi:р + 0.5 The test is: left-tailed two-tailed right-tailed Based on a sample of 600 people, 288 owned a pet, the p-value is: |(to 4 decimals) Based on this we: Fail to reject the null hypothesis Reject the null hypothesis
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A Two tailed test, where the null hypothesis is H0: p = 0.95 and the significance level is 10%. When the edge of the rejection region for the lower tail is 0.9399 then what is the value of "?" in P(P-hat > ? | p = 0.95) = 0.05?
It has been reported in the past that the probability that an individual will develop schizophrenia over their lifetime is 0.004. A researcher believes that this proportion has increases since then. She found out that in a random sample of 2500 individuals, 16 developed schizophrenia. Is there evidence to support the claim that the true proportion of people who will develop schizophrenia is greater 0.004? Use hypothesis testing at a = 0.10 level of significance to answer this question. a. State the null and alternative hypotheses. b. What is the probability model (underlying distribution of the sample proportion p)? c. Compute p. d. Compute the test statistic. e. Find the P-value. f. Make the decision whether to reject Ho or not. g. State a conclusion answering the question above. 1
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level of significance (i.e., type I error rate) 0.1 such that each tail of the rejection region has probability 0.05. Should the hypothesis Ho be rejected or not?