Homework - 2
Question 1
CI = x_bar +- (t * s/n**0.5)
Bypass
x_bar = 19
n = 539
s = 10
t = (1 - 0.9) / 2 = 0.05
CI = (19 - (0.05 * 10/539**0.5), 19 + (0.05 * 10/539**0.5))
= (18.978, 19.021)
Angiography
x_bar = 18
n = 847
s = 9
t = (1 - 0.9) / 2 = 0.05
CI = (18 - (0.05 * 9/847**0.5), 18 + (0.05 * 9/847**0.5))
= (17.984, 18.015)
The confidence interval is narrower for Angiography (0.031) by a difference of 0.012.
Question 2
n = 1031
t = 0.05
p = 567 * 0.05 / 1031
= 0.0272
Question 3
CI = x_bar +- (t * s/n**0.5)
x_bar = 5
t = 0.05
n = n
CI = (30, 200)
s = 42.5
30 = 5 - (0.05 * 42.5 / (n**0.5))
-25 * n**0.5 = 2.125
n = 735.766
Question 4
x_bar = 500
a
alpha = 0.05
p = 1 - (410/500) = 0.18
The p-value is greater than the alpha.
b
p = 1 - (410 / 500 + 90) = 0.305
The p-value is greater than the alpha.
c
p = 1 - (410/500) = 0.18
The p-value is greater than the alpha.
Question 5
a
t = (519.5 - 500) / (10 / sqrt(1000))
= 19.5 / 0.316
= 61.71
t = (519.7 - 500) / (10 / sqrt(1000))
= 19.7 / 0.316
= 62.342
b
Jones’ test is statistically significant by a factor of 0.01. The test Smith concluded is not statistically significant by 0.01.
c
Simply stating you ‘reject the hypothesis’ does not convey the significance of the p-value gathered. If the conclusion includes an extreme difference between the p-value found and alpha-value stated, then you can show a test has passed or failed.
Question 6
H0 = Children in higher grades eat healthy food at higher rates.
Creating a confidence interval of 95%.
Children in grade seven eat more healthily than in grade six, and children in grade eight eat better than in grade seven.
Question 7
H0 = There is a significant difference in means between Area 1-Area 2 and Area 1-Area 3.
A t-test should be performed.
There is not a significant difference in Area 1’s means compared to 2 and 3.