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Brock University *

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Statistics

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Feb 20, 2024

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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

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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

= 0.271 Z = 3.34 Z 0.05/2 : 1.96 Reject the null hypothesis since there is enough significant evidence to conclude that there’s a difference in patients experiencing pain relief. Question 7 Null hypothesis H O : μ = 0 H 1 : μ not equal to 0 Level of significance α = 0.05 α/2 = 0.025 Degree of freedom = n-1 = 7-1 = 6 T value according to the table = 0.025 N =7 xi x i – x (x i – x)^2 122-120 2 1.14 1.2996 142-145 -3 -3.86 14.8996 135-130 5 4.14 17.1396 158-160 -2 -2.86 8.1796 155-152 3 2.14 4.5796 140-143 3 -3.86 14.8996 130-126 4 3.14 9,8596 = 70.8572 Total x i = 6

Mean = 0.86 Standard deviation = n= 7 sd = 3.4365 T = 0.86 -0 / (3.4365/ square root 7) = 0.662 We fail to reject the null hypothesis as the t test value is less than the critical value. Question 8 a x1 = 120.2. x2 = 131.4. α = 0.05 critical value = 1.96 S1 = 15.4. s2 = 18.9 n1 = 100. n2 = 100 null hypotheses and alternate hypotheses H O : μ1 = μ2 H 1 : μ1 not equal to μ2 Test statistics sp 2 = (n1-1)*s1 2 +(n2-1)*s2 2 / n1+n2-2 = 297.185 SE = square root (sp 2 *(1/n1 +1/n2) = 2.438 Test statistics = 120.2 – 131.4 / 2.438 = -4.5939 Hence -4.59 < 1.96 we reject the null hypotheses.

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Question 9 Null hypothesis: H O : p1=p2=p3 Alternate hypothesis = p1 not equal to p2 not equal to p3 Site 1 Site 2 Site 3 total hypertensive 10 14 12 36 Non hypertensive 68 56 40 164 total 78 70 52 200 Expected frequencies: Site 1 Site 2 Site 3 hypertensive 14.04 12.6 9.36 Non hypertensive 63.96 57.40 42.64 Observed frequencies Expected frequencies Observed frequency – expected frequency (O-E) ^2 (O-E) ^2/E 10 14.04 -4.04 16.3216 1.162 14 12.60 1.4 1.96 0.155 12 9.36 2.64 6.9696 0.744 68 63.96 4.04 16.3216 0.255 56 57.40 -1.4 1.96 0.034 40 42.64 -2.64 6.9696 0.163 = 2.515 Test statistics = 2.515 Degree of freedom = (r-1) (c-1) (2-1) (3-1) = 1*2 = 2 P value 1- 0.7157 = 0.2843 0.2843 > 0.05

Null hypothesis is not rejected. there is not enough evidence to claim that the two variables are dependent at 5% significant level. Question 10 H O : P1 = P2 H 1 : P1 not equal P2. Level of significance = 0.05. 1.96 P1= x1/n1 = 38/100 = 0.38 P2 = x2/n2 = 21/100 = 0.21 P = 38 +21/200 = 0.295 Z = 0.38 – 0.21/ square root (0.295 (1-0.295) (1/100+1/100) = 0.17/0.0644 = 2.636 We reject the hull hypothesis as 2.636 is greater than 1.96.

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