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# assignment 3 - 3.1 Samples of size 6 were drawn...

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Unformatted text preview: 3.1) Samples of size 6 were drawn independently from two normal populations. α=0.05 Sample 1 12 6 5 Sample 2 7 11 13 3.1.a- Are you allowed to use a pooled variance estimate? 8 5 11 8 5 7 Solution: = 9.37, ℎ ⁄, : = 8.7 , ⎧ ⎪ : = = 5, 2 − : = 0.05 , 2 = 0.025 = = 1.077 ≠1 ) , , ,, . , =1 , ⎨ ⎪ ⎩ = (, , = = 7.15, ⁄, , 1 = ⁄, 1 = , . = ,, 1 = 0.14 7.15 Ftest there is not in rejection region and we fail to reject H0. So we can use a pooled variance estimate. 3.1.b- Test to determine whether the means of the two populations differ. Solution: ℎ = = : : : + + (− − − =0 , ≠0 = 7.83, = 8.5 , = + = 10 = 9.03 )−( 1 = (,, + − ) 1 ), = ⁄, −0.67 − 0 11 9.03 + 66 = 2.228 , = −0.39 ⁄, = −2.228 ttest there is not in rejection region and we fail to reject H0, so means of the two populations don’t differ. 3.1.c- Suppose that the data have been obtained from a matched pairs experiment. Determine whether the population means differ? Solution: Pair Sample 1 Sample 2 Difference = 1 12 7 5 2 6 11 -5 3 5 13 -8 = − 1 = 6 − 1 = 5 , = 5.164 : =0 ℎ : : ≠0 − −0.67 − 0 = = = −0.316 5.164/√6 / ), ⁄ , = 2.571 , = (,, ⁄ 4 8 5 3 5 11 8 3 6 5 7 -2 − , = −2.571 ttest there is not in rejection region and we fail to reject H0. so the population means don’t differ. 2 3.1.d- Discuss about your results in 3.1.b and 3.1.c. Solution: Although two parts (3.1.b and 2.1.c) was solved with different methods, but their results are the same. 3.2) An important statistical measurement in service facilities (e.g., restaurants and banks) is the variability in service times. As an experiment, two bank tellers were observed, and the service times for each of 100 customers were recorded and stored in DA3.2. 3.2.a- Do these data allow us to infer at the 10% significance level that the variance in service times differs between the two tellers? Solution: = 10.95, ℎ ⁄, : = , = 3.35 , ⎧ ⎪ : ⎨ ⎪ ⎩ : = = 99, 2 − , = 0.1 , = 0.05 2 =1 , . , = = (, = 3.27 , , , ) ≠1 = 1.4, , ⁄, , 1 = ⁄, 1 = , . = , , 1 = 0.72 1.4 Ftest there is in rejection region and we reject H 0. Yes. The variance in service times differs between the two tellers. 3.2.b- Construct a 95% confidence interval of the ratio of two population variances. Solution: 1− = 0.95 , = 0.05, = ⁄, , < ⁄, = < < , 1 ⁄, ⁄, , < = (3.27) , ⁄ ⁄ = 1 − ⇒ ⁄, , ⁄, ⁄, 1 = 2.35, 1.4 = ⁄ ⁄ < 1 = 2.35 < , < < , ⁄, < , ⁄, =1− , ⁄, , = 1 − = (3.27)1.4 = 4.56 < 4.56 = 0.95 3.3) Is marriage good for your health? To answer this question, researchers at the University of Toronto Medical School followed 103 couples, each with one spouse who was slightly hypertensive (mild high blood pressure). Participants also completed a questionnaire about their marriage. Three years later, the blood pressure of the previously hypertensive mate was measured. The reduction in blood pressure for those in happy marriage was stored in column 1 of DA3.3. The reduction of those in unhappy marriage 3 was stored in column 2. Can we infer that spouses in happy marriages have a greater reduction in blood pressure than those in unhappy marriage? Solution: Normality investigation of population of both samples: Unhappy 8 Happy 12 10 8 6 4 2 0 -2 -4 R² = 0.9777 6 Xi Xi 4 R² = 0.9816 2 0 -2 -4 -3 -1 1 Zi For independent samples: 3 -3 ℎ ℎ -1 1 Zi 3 First: variance = 10.24, = 6.67 , = : ⎨ ⎪ ⎩ : ≠1 − = 41, 2 − =1 = 1.54, ℎ ⎧ ⎪ = 60, : = 0.074 , , 0.05 < 0.074 < 0.1, ℎ ℎ = 4.92, = 2.33 , = . ℎ = ℎ . Second: Mean ℎ = = : + + (− : : − − =0 , >0 + = 101 = 8.79 )−( 1 + − 1 ) = 2.58 − 0 1 1 8.79 + 61 42 = 4.35 , − ≈ 0 , 0 < 0.1 ℎ ℎ , . Thus we can infer that spouses in happy marriages have a greater reduction in blood pressure than those in unhappy marriage. 4 ...
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