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# Stats.docx

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University of Ottawa

Biology

BIO1109

Carolyn Gordon

Fall

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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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