Question 10.1.2
When the House value = $230,000
Substitute amount of rental for y and the value of house for x
Amount of Rental = 0.0244 * (230,000) + 5363.9 =10,975.9
When House Value of House= $400,000. Amount of Rental = 0.0244 * (400,000) + 5363.9 =15,123.9
10,975.9 is closer to true income than 15,123.9. The prediction is done using both interpolation an extrapolation. Interpolation refers to the prediction within the range of data values while extrapolation is the prediction of outside the range of data values. The assumptions of the linear regression model may not hold outside the data range and hence increased chances of outliers. When the House value is $230,000, the x-value is within the range of values hence increased chances of accuracy as compared to House value = 400,000 which is outside the range of values.
Question 10.1.4
Regression equation is y= 1.6606 x + 69.739
Where y is Parental care (%) and x is Health Expenditure (in % of GDP)
Prenatal Care (%) = 1.6606 (Health Expenditure in %) + 69.739
When x = 5.0% of GDP on health expenditure
Prenatal Care (%) = 1.6606*(5) + 69.739 = 78.042%
When x= 12.0% of GDP.
Prenatal Care (%) = 1.6606*(12) + 69.739 = 89.6662%
78.042% is closer to true percentage as compared to 89.6662%. Prediction through extrapolation comes with increased risk of inaccuracies because the assumptions of the linear regression model may not hold outside the data range. When the Health Expenditure is 5%), the x-value is within the range of data values hence increased chances of accuracy as compared when the Health Expenditure is 12 % which is outside the range of data values.
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Question 10.2.2
|
Value |
Rental |
|
| Value |
1 |
|
| Rental |
0.764716 |
1 |
Correlation = 0.764716. The two variables are strongly positively correlated implying that an increase in one variable causes a simultaneous increase in the other.
The coefficient of determination, r 2 = correlation 2
= 0.764716* 0.764716 = 0.584790181
Coefficient determination of 0.5848 implies that 58.48% of the variance in Y variable (value of House) is predictable from X variable (amount of rental costs).
Question 10.2.4
|
Health Expenditure (% of GDP) |
Prenatal Care (%) |
|
| Health Expenditure (% of GDP) |
1 |
|
| Prenatal Care (%) |
0.171505231 |
1 |
Correlation = 0.171505231 . This means that Health Expenditure (% of GDP) is weakly positively correlated with Parental Care (%), an increase in one variable causes a simultaneous increase in the other.
The coefficient of determination, r 2 = correlation 2
= 0.171505231* 0.171505231 = 0.029414044 = 0.0294
Coefficient determination of 0.0294 implies that 2.94% of the variance in Y variable ( Prenatal Care (% ) is predictable by X variable (Health Expenditure (% of GDP).
Question 10.3.2
The Pearson’s correlation coefficient is 0.764716
|
Value |
Rental |
|
| Value |
1 |
|
| Rental |
0.764716 |
1 |
The null hypothesis: r coefficient is NOT significantly different from zero, (r= 0)
Alternative hypothesis: r coefficient is significantly different from zero (r > 0)
Read the critical value from Tables for critical value: Pearson’s coefficient
The degrees of freedom = n- 2 = 48 – 2 = 46
Level of significance = 0.05
The critical value for 1-tailed is = 0.243
Since the correlation coefficient r > than the critical value (0.7647> 0.243), we reject the null hypothesis and conclude that the correlation coefficient is positive at 5% level of significance.
Question 10.3.4
|
Health Expenditure (% of GDP) |
Prenatal Care (%) |
|
| Health Expenditure (% of GDP) |
1 |
|
| Prenatal Care (%) |
0.171505231 |
1 |
The null hypothesis, H0: r coefficient is NOT significantly different from zero, (r= 0)
Alternative hypothesis, H1: r coefficient is significantly different from zero (r 0)
Read the critical value from Tables for critical value: Pearson’s coefficient
The degrees of freedom = n-2 = 15-2 = 13
Level of significance = 0.05
The critical value for 2-tailed is = 0.514 .Since the correlation coefficient r < than the critical value (0.1715< 0.514), we fail to reject the null hypothesis and conclude that the correlation coefficient is not significant at 5% level of significance.
Question 11.1.2
Null hypothesis, H0: There is no association between the variables.
Alternative hypothesis, H1: There exists an association between the variables.
Chi-square =
df = (No. of rows-1)(No. of columns-1) = 3*2 = 6
|
Observed |
|||||
|
Period |
Row | ||||
| Activity | Morning | Noon | Afternoon | Evening | Total |
| Travel |
6 |
6 |
14 |
13 |
39 |
| Feed |
28 |
4 |
0 |
56 |
88 |
| Social |
38 |
5 |
9 |
10 |
62 |
| Column Total |
72 |
15 |
23 |
79 |
189 |
Calculate expected value = (row total * column total)/ Overall Total
|
Expected |
||||
| Activity | Morning | Noon | Afternoon | Evening |
| Travel |
14.85714 |
3.095238095 |
4.746031746 |
16.3015873 |
| Feed |
33.52381 |
6.984126984 |
10.70899471 |
36.78306878 |
| Social |
23.61905 |
4.920634921 |
7.544973545 |
25.91534392 |
| Observed | Expected | (Observed - Expected) | (Observed - Expected)^2 | /Expected |
|
6 |
14.85714 |
-8.857142857 |
78.44897959 |
5.28021978 |
|
28 |
33.52381 |
-5.523809524 |
30.51247166 |
0.91017316 |
|
38 |
23.61905 |
14.38095238 |
206.8117914 |
8.756144393 |
|
6 |
3.095238 |
2.904761905 |
8.437641723 |
2.726007326 |
|
4 |
6.984127 |
-2.984126984 |
8.905013857 |
1.275036075 |
|
5 |
4.920635 |
0.079365079 |
0.006298816 |
0.001280082 |
|
14 |
4.746032 |
9.253968254 |
85.63592845 |
18.04369061 |
|
0 |
10.70899 |
-10.70899471 |
114.6825677 |
10.70899471 |
|
9 |
7.544974 |
1.455026455 |
2.117101985 |
0.280597668 |
|
13 |
16.30159 |
-3.301587302 |
10.90047871 |
0.668675909 |
|
56 |
36.78307 |
19.21693122 |
369.2904454 |
10.03968558 |
|
10 |
25.91534 |
-15.91534392 |
253.2981719 |
9.774061759 |
| Total |
68.46456706 |
Critical value at 0.05 level of significance = 12.59
The Chi-square test statistic > critical value, we therefore reject the null hypothesis and conclude that there exists sufficient statistical evidence to conclude that that the activity and time period are dependent for dolphins.
Question 11.1.4
Null hypothesis, H0: There is no association between the variables.
Alternative hypothesis, H1: There exists an association between the variables.
Chi-square =
df = (No. of rows-1)(No. of columns-1) = 4*3 = 12
|
Age Group |
Row Total | |||||
| Education | 25-34 | 35-44 | 45-54 | 55-64 | >64 | |
| Did not complete HS |
5416 |
5030 |
5777 |
7606 |
13746 |
37575 |
| Competed HS |
16431 |
1855 |
9435 |
8795 |
7558 |
44074 |
| College 1-3 years |
8555 |
5576 |
3124 |
2524 |
2503 |
22282 |
| College 4 or more years |
9771 |
7596 |
3904 |
3109 |
2483 |
26863 |
| Column Total |
40173 |
20057 |
22240 |
22034 |
26290 |
130794 |
|
Expected |
|||||
|
Age Group |
|||||
| Education | 25-34 | 35-44 | 45-54 | 55-64 | >64 |
| Did not complete HS |
11541.05292 |
5762.051585 |
6389.192165 |
6330.011698 |
7552.6916 |
| Competed HS |
13537.20203 |
6758.660321 |
7494.271603 |
7424.855238 |
8859.0108 |
| College 1-3 years |
6843.852057 |
3416.900424 |
3788.795205 |
3753.701148 |
4478.7512 |
| College 4 or more years |
8250.893 |
4119.387671 |
4567.741028 |
4525.431916 |
5399.5464 |
| Observed | Expected | (Observed - Expected) | (Observed - Expected)^2 | /Expected |
|
5416 |
11541.05292 |
-6125.052915 |
37516273.21 |
3250.680288 |
|
16431 |
13537.20203 |
2893.797972 |
8374066.705 |
618.5965673 |
|
8555 |
6843.852057 |
1711.147943 |
2928027.281 |
427.8332227 |
|
9771 |
8250.893 |
1520.107 |
2310725.292 |
280.0576001 |
|
5030 |
5762.051585 |
-732.0515849 |
535899.523 |
93.00498531 |
|
1855 |
6758.660321 |
-4903.660321 |
24045884.54 |
3557.788585 |
|
5576 |
3416.900424 |
2159.099576 |
4661710.981 |
1364.309872 |
|
7596 |
4119.387671 |
3476.612329 |
12086833.29 |
2934.133482 |
|
5777 |
6389.192165 |
-612.1921648 |
374779.2466 |
58.658315 |
|
9435 |
7494.271603 |
1940.728397 |
3766426.712 |
502.5740875 |
|
3124 |
3788.795205 |
-664.7952047 |
441952.6642 |
116.6472824 |
|
3904 |
4567.741028 |
-663.7410279 |
440552.1521 |
96.44858353 |
|
7606 |
6330.011698 |
1275.988302 |
1628146.147 |
257.2106064 |
|
8795 |
7424.855238 |
1370.144762 |
1877296.669 |
252.8394977 |
|
2524 |
3753.701148 |
-1229.701148 |
1512164.914 |
402.8463787 |
|
3109 |
4525.431916 |
-1416.431916 |
2006279.372 |
443.3343401 |
|
13746 |
7552.691637 |
6193.308363 |
38357068.48 |
5078.595859 |
|
7558 |
8859.010811 |
-1301.010811 |
1692629.13 |
191.0629941 |
|
2503 |
4478.751166 |
-1975.751166 |
3903592.67 |
871.5806092 |
|
2483 |
5399.546386 |
-2916.546386 |
8506242.821 |
1575.362487 |
|
22373.56564 |
Critical value at 0.05 level of significance = 21.03
The Chi-square test statistic > critical value, we therefore reject the null hypothesis and conclude that there exists sufficient statistical evidence to conclude educational attainment and age are dependent
Question 11.2.4
Null hypothesis, H0: There variables are in the same proportion
Alternative hypothesis, H1: The variables are not in the same proportion
Chi-square =
df = (k-1) = 3
| Age |
14-May |
15-29 | 30-49 | 50-69 | Total |
| Cardiovascular Frequency |
8 |
16 |
56 |
433 |
513 |
| All Cause Proportion |
0.1 |
0.12 |
0.26 |
0.52 |
|
| Expected = Total* All Cause Proportion |
51.3 |
61.56 |
133.38 |
266.76 |
| Observed | Expected | (Observed -Expected) | (Observed -Expected)^2 | /Expected |
|
8 |
51.3 |
-43.3 |
1874.89 |
36.54756335 |
|
16 |
61.56 |
-45.56 |
2075.7136 |
33.71854451 |
|
56 |
133.38 |
-77.38 |
5987.6644 |
44.89177088 |
|
433 |
266.76 |
166.24 |
27635.7376 |
103.5977568 |
| Total |
218.7556355 |
Critical value at 0.05 level of significance = 7.81
The Chi-square test statistic > critical value, we therefore reject the null hypothesis and conclude that there exists sufficient statistical evidence to conclude that deaths from cardiovascular disease are not in the same proportion as all deaths for the different age.
Question 11.2.6
Null hypothesis, H0: Reasons for choosing the care are equally likely
Alternative hypothesis, H1: Reasons for choosing the care are not equally likely
Chi-square =
|
Safety |
Reliability |
Cost |
Performance |
Comfort |
Looks |
|
84 |
62 |
46 |
34 |
47 |
27 |
Expected = 84 + 62 + 46 + 34 + 47 + 27 = 300/6 = 50
Chi-square =
Chi-square = = 42.2
Degrees of Freedom = k-1
= 6-1 = 5
Critical value at 0.05 level of significance = 11.07
The Chi-square test statistic > critical value, we therefore reject the null hypothesis and conclude that there exists sufficient statistical evidence to conclude that the reasons for choosing the care are NOT equally likely.
