# 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.

Week 10, question 2:

(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.

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