Statistics

Running Head: HYPOTHESIS TESTING 1

HYPOTHESIS TESTING 7

Hypothesis Testing

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Question 1-Exercise 25.1

In a study about train delays in The Netherlands one was interested in whether arrival delays of trains exhibit more variation during rush hours than during quiet hours. The observed arrival

delays during rush hours are modeled as realizations of a random sample from a distribution with variance σ12, and similarly the observed arrival delays during quiet hours correspond to a distribution with variance σ22 . One tests the null hypothesis H0: σ1 =  σ2. What do you choose as the alternative hypothesis?

Solution

Inferential statistics usually does not test the research hypothesis or the testable statement directly. That particularly addresses the null hypothesis .Statistically we test not .The research hypothesis is always the alternaltive hypothesis and the null hypothesis is an opposing statement to the alternative hypothesis.

In regard to our case, our research would be ;is there more variation in arrivals exhibited exhibited by trains during rush hours than in quite hours? .Thus ,the alternative hypothesis is a statement about the belief that reflects arrival delays of trains exhibit more variation during rush hours than during the quite hours. Therefore H1: σ1  >  σ2.

H1: σ1 > σ2

Answer( the first choice)

Question 2-1. Exercise 26.1a

Polygraphs that are used in criminal investigations are supposed to indicate whether a person is lying or telling the truth. However the procedure is not infallible, as is illustrated by the following example. An experienced polygraph examiner was asked to make an overall judgment for each of a total 280 records, of which 140 were from guilty suspects and 140 from innocent suspects. The results are listed in Table 26.2. We view each judgment as a problem of hypothesis testing, with the null hypothesis corresponding to “suspect is innocent” and the alternative hypothesis to “suspect is guilty.”

a.Estimate the probability of a type I error ... that applies to this polygraph method on the basis of Table 26.2.

Solution

The type I error is rejecting the the true null hypothesis. In regard to this situation, it is the same as falsely concluding suspect is guilty . This happened in 9 out of 140 cases as per given information .Thus the probability of type I error is 9/140 =0.064285714.

Question 3- Exercise 26.1b

Estimate the probability of ... a type II error that applies to this polygraph method on the basis of Table 26.2.

Solution

The type II error is accepting the null hypothesis when it is false. In this case , it would to falsely conclude that the suspect is innocent. We have 15 such cases out of the 140 cases.

Therefore the probability of type II error is 15/140 = 0.107142857

Question 4- Exercise 26.2

Consider the testing problem in Exercise 25.11. Compute the probability of committing a type II error if the true value of μ is 1.

Exercise 25.11 – One is given a number t, which is the realization of a random variable T with an N(μ,1) distribution. To test H0: μ = 0 against H1: μ ≠ 0, one uses T as the test statistic. One decides to reject H0 in favor of H1 if |t| ≥ 2.

Solution

In exercise 25.11, we do not reject if |T| ≥ 2 and the probability of type II error would given as; P(|T| <  2)= P(T <2) - P(T< -2),for T is normally distributed with mean of 1and standard deviation of 1.So, P(T >2) - P(T< -2) = P( T-1< 1) - P(T-1 < -3)

=0.8413-0.0013= 0.84

Question 5- Exercise 26.7

A random sample X1,X2 is drawn from a uniform distribution on the interval [0, θ]. We wish to test H0: θ = 1 against H1: θ < 1 by rejecting if X1 + X2 ≤ c. Find the value of c and the critical region that corresponds to a level of significance 0.05. Hint: use Exercise 11.5.

Solution

For the null hypothesis, P(X1 + X2 ≤ c)= 0.05 and assuming uniform distribution , we can find the value of c. For T= X1 + X2 , and density function of f(t)= t. so, we integrate f(t) ,with respect to t ,from zero to c and equate the result to 0.05

c^2/2=0.05 , c = sqrt (0.1) = 0.316227766, c= 0.316(3 d.p) ANSWER .Therefore the corresponding critical region for T = X1 + X2 is [0.0.316]

Question 6; Exercise 27.1a

We perform a t-test for the null hypothesis H0: μ = 10 by means of a dataset consisting of n = 16 elements with sample mean 11 and sample variance 4. We use significance level 0.05.

a. Should we reject the null hypothesis in favor of H1: μ ≠ 10?

Solution

t= (mean -10)/{ stdev./ sqrt.(n)}

t=(11-10)/(2/sqrt.4)= 2 .From the table , t(15, 0.025)= 2.131 but the observed value is 2.The observed value is less than the critical value. So, we do not reject the null hypothesis.

No, do not reject H0 in favor of H1

Question 7- Exercise 27.1b

We perform a t-test for the null hypothesis H0: μ = 10 by means of a dataset consisting of n = 16 elements with sample mean 11 and sample variance 4. We use significance level 0.05.

b. Should we reject the null hypothesis in favor of H1: μ > 10?

Solution

For n =16 , then t(n-1,0.05) = t(15,0.05) = 1.753.But the observed Value is t=2 which is greater than 1.753.Therefore we reject the null hypothesis ,H0: μ = 10.

Answer;

Yes, reject H0 in favor of H1

Question 8:- Exercise 27.3a

Table 27.5 lists the results of tensile adhesion tests on 22 U-700 alloy specimens. The data are loads at failure in MPa. The sample mean is 13.71 and the sample standard deviation is 3.55. You may assume that the data originated from a normal distribution with expectation μ. One is interested in whether the load at failure exceeds 10 MPa. We investigate this by means of a t-test for the null hypothesis H0: μ = 10.

a. What do you choose as the alternative hypothesis?

Solution

In regard to the given information about the test that is “One is interested in whether the load at failure exceeds 10 MPa”, then we can deduce the alternative hypothesis. The alternative hypothesis should be that the load at failure exceeds the 10MPa. So, the alternative hypothesis can be rewritten as H1: μ >10. A

H1: μ >10. Answer-Second choice

Question 9: Exercise 27.3b(1)

Table 27.5 lists the results of tensile adhesion tests on 22 U-700 alloy specimens. The data are loads at failure in MPa. The sample mean is 13.71 and the sample standard deviation is 3.55. You may assume that the data originated from a normal distribution with expectation μ. One is interested in whether the load at failure exceeds 10 MPa. We investigate this by means of a t-test for the null hypothesis H0: μ = 10.

Solution

t= (mean -10)/{ stdev./sqrt.(n)}

t= (13.71-10)/{ 3.55/sqrt(22)} = 4.902

Question 10- Exercise 27.3b(2)

Solution

t(21,0.05)= 1.721 .Since observed value (4.902) >t-critical value(1.721) , we reject the null hypothesis(H0: μ = 10.) in favor of the alternative hypothesis ( H1: μ >10).

Answer;

Reject H0 in favor of H1

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