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Brock University *

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Statistics

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Feb 20, 2024

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docx

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9

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Week 9 Question 1 Step 1Null hypothesis H O : μ SBP = 120 mmhg Alternate hypotheses H 1 : placebo SBP > μ SBP Level of significance 0.05 Step 2 Because the population size is greater than 30 we will use the z method Step 3 we will reject the hypothesis if μ = 120 mmhg Step 4 Sx= s n Sx = 18.9/ 100 = 1.89 We find z Z = (x – μ )/Sx. = 131.4 120 1.89 = 6.03 Step 5 We reject the null hypothesis as 6.03 > 1.654 Question 2 : Step 1 H O : vitamin level of cancer patients = 3.00 = μ vitamin H 1 : population μ vitamin > cancer patient vitamin level Step 2 : using t methods because the sample size n is less than 30 Step 3: we will reject the null hypothesis is the value is less than or eqal to -1.645 Step 4: Sx= s n = 0.15/ 20 = 0.034 t= x μ Sx = 2.41 3 0.034 = -17.35
we will reject the null hypothesis as the value is less than -1.645. the vitamin levels of cancer patients are not the same as the vitamin level of population. Question 3: Step 1 Null hypothesis H O : placebo data side effect = 0.05 Alternate hypothesis H 1 : placebo data side effect isn’t equal to 0.05 Step 2 : as the value of n is greater than 30 we will use z method Step 3: reject the null hypothesis if greater or equal to 1.960 Step 4 Z = p p 0 p O ( 1 p 0 ) n = 1.376 Step 5 We fail to reject the null hypothesis as the value is less than 1.960. Question 4 Step 1: Null hypothesis H O : P1=0.20 P2= 0.20 P3=0.20 P4 =0.20 Alternate hypothesis H 1 : that the values of the null hypothesis isn’t true Level of significance 5% = 0.05 Step 2 : as its categorial we use the chi square method Step 3: if the value of x 2 is greater than 9.49 we will reject the null hypothesis Step 4: X 2 = ( ( O E ) 2 ) E = 206.9 STEP 5: we reject the null hypothesis as the value is greater than 9.49
Week 10 Question 1 H O : The new compound treatment, and the extent of wound healing is independent of each other. H 1 : The new compound treatment, and the extent of wound healing is dependent of each other. Level of significance = 5% α = 0.05 Treatment 0-25 26-50 51-75 76-100 New compound n = 125 15 37 32 41 Placebo n = 125 36 45 34 10 total 15+36 = 51 37+45 =82 32+34 = 66 41+10 = 51 Frequency table treatment 0-25 25-50 51-75 76-100 New compound (125*51)/250 = 25.5 (125*82)/250 = 41 (125*66)/250 = 33 (125*51)/250 = 25.5 Placebo (125*51)/250 = 25.5 (125*82)/250 = 41 (125*66)/250 = 33 (125*51)/250 = 25.5 E O-E (O-E)^2 (O-E)^2/E 15 25.5 -10.50 110.25 4.3235 37 41 -4 16 0.3902 32 33 -1 1 0.0303 41 25.5 15.50 240.25 9.4216 36 25.5 10.5 110.25 4.3236
45 41 4 16 0.3902 34 33 1 1 0.0303 10 25.5 -15.50 240.25 9.4216 Total = 28.33 Test statistics = 28.33 DF = k-1 (4-1) (2-1) 3*1 =3 According to the table: 7.8177 There are enough evidence to reject the null hypothesis treatment, since the test statistics 28.331 is greater than critical value. Question 6 P1 be the proportion of patients who get relied from new medication. P2 be the proportion of patients who get relied from placebo. H O : P1 = P2 H 1 : P1 not equal P2. Sample proportion of pain relief for two treatments = x1/n1 = 44/44+76 = 0.367 Pooled estimate of the sample proportion = x1 + x2 / n1 + n2 = 44 +21 / 120 + 120
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