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# Stats.docx

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School
University of Ottawa
Department
Biology
Course
BIO1109
Professor
Carolyn Gordon
Semester
Fall

Description
Assignment #1 Jordan Malette 6794347 1) a) Ho: u = 0.3961 - 0.1 = 0.2962 Ha: u <0.2962 n=1000 u=0.2962 p^=0.262 SD (p^) = √(PoQo / n) =√ 0.2962)(0.7038)) =0.0144 Z= p^ - u / SD(p^) Z=0.262 - 0.2962 / 0.0144 Zcalc=-2.375 Zcrit=-2.3262 Zcalc < Zcrit therefore there is sufficient evidence to reject the null in favour of the alternative. b) n= (Zα/2 / E)^2 (p^)(q^) n=(2.576 / 0.01)^2 (0.262) (0.738) n=12,830.67 n=12,831 A sample size of 12,831 would be required given the margin of error of +-1% and a confidence interval of 99%. c) Ho: u= 0.3962 Ha: u< 0.3962 S= √ ((p) (1-p)/n) S= √ (0.1176) (0.8824)/17 S= 0.078 T= ((0.1176 – 0.396) / (0.078/√17)) T= -14.7 Since the T value is so high the probably will be practically 0 , indicating that the sample size is tremendously small as we saw by np^= 17 x 0.1176 = 1.992 which is less than 10, which shows that the sample is too small to follow the normal distribution. As the sample is so small this does not allow us to infer anything about the Conservative polls. 2) One-Sample T Test of mu = 26 vs < 26 95% Upper N Mean StDev SE Mean Bound T P 40 24.8125 5.4095 0.8553 26.2536 -1.39 0.086 Since 0.086 > 0.05 we fail to reject the null hypothesis. As shown by the graph it makes sense that we fail to reject the null as that would indicate that it is very likely that the mean value is 26. As a large amount of data lies between 25 and 26 it is quite possible that mean value is in fact 26. As well the data is a little skewed to the right as shown by the graph with the long tail, which would again skew the true mean further to the right. Histogram of BMIfemale 14 12 10 c n 8 u q r 6 F 4 2 0 18 24 30 36 42 BMIfemale 3a) Test and CI for Two Proportions: OWmale, OWfemale Event = 1 Variable X N Sample p OWmale 25 50 0.500000 OWfemale 11 40 0.275000 Difference = p (OWmale) - p (OWfemale) Estimate for difference: 0.225 95% CI for difference: (0.0291568, 0.420843) Test for difference = 0 (vs not = 0): Z = 2.17 P-Value = 0.030 Z= p1^-p2^ / SD(P1^-P2^) P = x1 + x2/ n1 + n2 P = 0.4 Q= 0.6 Sd = √(0.4)(0.6)(1/50+1/40) SD = 0.104 Z = 0.5-0.275 / 0.104 Z= 2.17 Since the Z score is greater than the significance value of 1.96 we have sufficient evidence to reject the null and claim that the amount of Ow Males and Females are not the same. b) P(Z<2.17) = 0.985 P(Z>2.17) – 1.0985 P value = 0.015 x 2 P value = 0.03 The p value would be found by finding the left tail probability of Z
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