Assignment Question Week 7, 8, 9, 10 11
Week 7, Question 2:
(a) Given n = 400, x = 80
So, point estimate p = x/n = 80/400 = 0.2
And q = 1 – p = 0.8
Critical z = 1.96
Margin of error E = 1.96*0.2*0.8400=0.039
Hence 95% confidence interval = (0.2 – 0.039, 0.2 + 0.039) = (0.161, 0.239)
(b) Minimum sample size can be given as,
N = (1.96*0.2*0.80.04)^2 = 61.47
Hence minimum sample size required = 62
Week 8, Question1:
(a) Null Hypothesis: H0: µ = 48400
Alternate Hypothesis: H1: µ > 48400 (claim)
(b) Formula z=x-??/n
(c) Level of significance ? = 0.05
(d) Critical z = 1.645
Rejection region is z > 1.645
(e) z = 50000-484008000100 = 2
(f) z test is in rejection region. So, we can reject the null hypothesis. Hence, sufficient evidence to support the claim that a significant increase in the starting salaries.
Week 9, Question 3:
(a) H0 : Mean process value for all the three process are equal to each other.
Ha : Mean for at least one process is different from others.
(b) DFB = 3 – 1 = 2,
DFW = 11 – 2 = 9 and level of significance ? = 0.05
Critical F = 4.256
Hence, we can reject the null hypothesis if F > 4.256
(c) DFB = 3 – 1 = 2,
DFW = 11 – 2 = 9
MSB = SSB/DFB = 32/2 = 16
MSW = SSW/DFW = 88/9 = 9.778
Hence F test = MSB/MSW
= 16/9.778 = 1.636
(d) Here F test is not greater than 4.256 (critical value). So, we cannot reject the null hypothesis.
Hence, insufficient evidence to support any significant difference in mean for all the three process.
(a) Regression equation is: Total wealth (in 1000’s of dollars) = 45.2159 + 5.3265 * Age
Y = 45.2159 + 5.3265 * x
If age increases by 1 year then wealth tends to increase on average by 5.3265 thousand of dollars.
(b) now x = 50 years
Wealth Y = 45.2159 + 5.3265 * 50 = 311.5409 thousands = $311,541
(c) Coefficient of determination = 0.9115
Hence 91.15% variation in total wealth can be explained by the variation in age.
(d) H0: ? = 0
H1: ? ? 0 (claim)
Standardized test stats is t test.
Level of significance ? = 0.10
Degree of freedom = 6
Critical t = -1.943, 1.943
Rejection region is t < -1.943 or t > 1.943
Test stats t = 5.3265/0.6777 = 7.86
As t test is in rejection region, we can reject the null hypothesis. Hence, there is a significant evidence that slope is different from 0. Hence a significant association occur between age and total wealth.
Week 11: Question 3:
(a) Regression equation can be given by
y = 0.0136 + 0.7992x1 + 0.2280x2 – 0.5796x3
(b) Coefficient od determination is SSR/SST
SST = SSR + SSE = 45.9634 + 2.6218 = 48.5852
So, r2 = 45.9634 / 48.5852 = 0.9460
Here, coefficient of determination is very close to 1. So, strength of relation is very strong between variables.
(c) For degree of freedom 3 and 11, Critical F = 3.587
So, we can reject the null hypothesis of no relation if F > 3.587
Now MSR = SSR/DFR = 45.9634/3 = 15.321
MSE = SSE/DFE = 2.6218/11 = 0.2383
So, F test = MSR/MSE = 15.321 / 0.2383 = 64.29
As F is in rejection region, y is significantly related to the independent variables.
(d) H0: ?3 = 0
H1: ?3 ? 0 (claim)
Level of significance = 0.05
DF = 15 – 4 = 11
Critical t = -2.201, 2.201
We can reject the null hypothesis if t < -2.201 or t > 2.201
Now t test = -0.5796/0.92 = -0.63
Here t-test is within both critical t, so the null hypothesis cannot be rejected. Hence, there is no significant linear association between y and x3.