Using Statistics to Test the Relationship between Obesity and Diabetes
Obesity is one of the major causes of diabetes today. Studies show that individuals who are overweight are 85 times more likely to develop diabetes than those whose BMI is less than 22. Obese individuals have increased levels of cytokines, hormones, glycerol, NEFA, and proinflammatory substances that increase their insulin resistance. Reduction in the resistance of insulin leads to the development of type 2 diabetes.
Hypothesis Testing
Null hypothesis: There was no effect within one hour of Advair intervention.
Alternative Hypothesis: There was an effect within the one hour of Advair intervention.
Decision rule: Chi - square>3.84, Reject null hypothesis.
McNemar test: T1 = 990
T2= 990-903 = 87
T3 = 87/(sqrt(990+903) = 87/43.51 = 1.999
T4=1.999*1.999 = 3.998
Decision: Since 3.998 is greater than the critical value (3.84), we reject the null hypothesis.
We conclude thatthere was an effect within one hour of Advair intervention.
Null hypothesis: The median yield this year per tree is equal to 27.
Alternative Hypothesis: The median yield this year per tree is larger than 27.
sign | |
32 | + |
45 | + |
120 | + |
27 | 0 |
63 | + |
24 | - |
57 | + |
27 | 0 |
66 | + |
43 | + |
68 | + |
27 | 0 |
21 | - |
46 | + |
16 | - |
Negative: 3
Zeros: 3
Since we have 3 zeroes, our sample size is 12.
Thus, chi-square critical is 2.
Decision rule: If Chi-square test statistic > 2, we reject the null hypothesis
Test statistic = 9
Decision: Since 9 is greater than the critical value (2), we reject the null hypothesis.
We conclude thatthe median yield this year per tree is larger than 27 (last year's yield). To have this topic explained further by our experts, take our hypothesis testing assignment help.
Probability
a) Probability (got the flu │ you were not vaccinated) = 1556/2272 = 0.6849
b) Probability (got the flu │ you were vaccinated) = 642/1875= 0.3424
c) Relative risk: 0.6849/0.3424 = 2.000173
There is not a positive difference in the Job Satisfaction of these applicants
Alternative Hypothesis: There is a positive difference in the Job Satisfaction of these applicants
Emp Number | Regular hours | Rev hours | Diff | Sign of diff | Rank of Diff | Signed-Rank |
1 | 34 | 38 | -4 | - | 5.5 | -5.5 |
2 | 23 | 23 | 0 | 0 | ||
3 | 75 | 75 | 0 | 0 | ||
4 | 34 | 42 | -8 | - | 11 | -11 |
5 | 52 | 57 | -5 | - | 9 | -9 |
6 | 18 | 18 | 0 | 0 | ||
7 | 26 | 28 | -2 | - | 1.5 | -1.5 |
8 | 43 | 47 | -4 | - | 5.5 | -5.5 |
9 | 72 | 67 | 5 | + | 9 | 9 |
10 | 63 | 65 | -2 | - | 1.5 | -1.5 |
11 | 55 | 58 | -3 | - | 3 | -3 |
12 | 65 | 65 | 0 | 0 | ||
13 | 61 | 66 | -5 | - | 9 | -9 |
14 | 28 | 48 | -20 | - | 14 | -14 |
15 | 68 | 72 | -4 | - | 5.5 | -5.5 |
16 | 17 | 21 | -4 | - | 5.5 | -5.5 |
17 | 9 | 22 | -13 | - | 13 | -13 |
18 | 82 | 71 | 11 | + | 12 | 12 |
Thus, the chi-square critical for the one-tailed test is 26.
Decision rule: If the Chi-square test statistic > 26, we reject the null hypothesis.
Test statistic :
Sum of positive sign ranks = 21
Sum of negative sign ranks = 84
Thus, the test statistic value is 21.
Decision: Since 21 is smaller than the critical value (26), we cannot reject the null hypothesis.
We conclude that there is not a positive difference in the Job Satisfaction of these applicants
Answer 12:
a) Und odds rat = 161*169/112*187 = 1.299
B)
a) Upper 95% Ci = e^ (ln(1.299) + 1.96* sqrt(1/161+1/112+1/187+1/169))
=e^(ln(1.299) +1.96*0.1625)
= e^(0.5801)
= 1.7862
Lower 95% Ci = e^ (ln(1.299) - 1.96* sqrt(1/161+1/112+1/187+1/169))
= 0.9447
ii) Upper 95% Ci = (1.299) + 1.96* sqrt(1/161+1/112+1/187+1/169))
=(1.299) +1.96*0.1625
= 1.6175
Lower 95% Ci = 0.9805
iii) Since the value lies between 95% Ci, the results are not significant.
observed | expected | obs-exp | square | square/exp |
161 | 151.04 | 9.96 | 99.20667 | 0.6568 |
112 | 121.96 | -9.96 | 99.20667 | 0.8134 |
187 | 196.96 | -9.96 | 99.20667 | 0.5037 |
169 | 159.04 | 9.96 | 99.20667 | 0.6238 |
2.5977 |
b) Decision: Since 2.5977 is smaller than the critical value (3.84), we cannot reject the null hypothesis.
c) We conclude that there is no relation between obesity and developing diabetes.
Answer 13:
Null hypothesis: There is no correlation between X and Y.
Alternative Hypothesis: There is a correlation between X and Y.
X | Rank | Y | Rank | d | d^{2} |
45 | 5 | 45 | 7 | -2 | 4 |
34 | 3 | 4 | 2 | 1 | 1 |
76 | 8 | 4 | 2 | 6 | 36 |
54 | 6.5 | 34 | 5.5 | 1 | 1 |
38 | 4 | 22 | 4 | 0 | 0 |
29 | 1.5 | 68 | 8 | -6.5 | 42.25 |
29 | 1.5 | 4 | 2 | -0.5 | 0.25 |
54 | 6.5 | 34 | 5.5 | 1 | 1 |
85.5 |
If t-statistic > +/-2.447, we reject null hypothesis.
Spearman correlation coefficient = 1 – 6*85.5/8*(64-1) = 1-1.0178 = -0.0178
Test statistic = -0.0178*sqrt(8-2)/sqrt(1-(-0.0178^2)) = -0.0178*2.449/0.9998 = -0.0436
Decision: Since -0.0436 is less negative than the critical value (-2.447), we cannot reject the null hypothesis.
We conclude thatthere is no correlation between X and Y.
Answer 14:
Null hypothesis: There is no difference between the two products
Alternative Hypothesis: There is a difference between the two products
Advair | Combined Rank | Symbicort | Combined Rank | |
88 | 1 | 90 | 5 | |
89 | 2 | 90 | 5 | |
90 | 5 | 93 | 8.5 | |
90 | 5 | 93 | 8.5 | |
90 | 5 | 96 | 10 | |
97 | 11.5 | |||
97 | 11.5 | |||
sum | 18 | 60 |
Test statistic:
U = 18 – 5*6/2 = 18-15 = 3
U’ = 5*7 – 3 = 35-3 = 32
Since n and m are less than 20, the test statistic value is 3.
Decision: Since 3 is smaller than the critical value (5), we cannot reject the null hypothesis.
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Regression Analysis
Regression Statistics | ||||||
Multiple R | 0.904179 | |||||
R Square | 0.817541 | |||||
Adjusted R Square | 0.756721 | |||||
Standard Error | 2.980326 | |||||
Observations | 13 | |||||
ANOVA | ||||||
df | SS | MS | F | Significance F | ||
Regression | 3 | 358.1897 | 119.3966 | 13.44201 | 0.001129 | |
Residual | 9 | 79.94109 | 8.882344 | |||
Total | 12 | 438.1308 | ||||
Coefficients | Standard Error | t Stat | P-value | |||
Intercept | 52.8274 | 8.662 | 6.099013 | 0.000179 | ||
A | 1.79 | 0.329 | 5.43 | 0.00 | ||
B | -2.25 | 0.530 | -4.24 | 0.00 | ||
C | -2.24 | 0.894 | -2.51 | 0.03 |
c) Since the given value of % weight of C at 5% is outside the sample range, the model is invalid and we cannot predict % survival.
d) The value of R-square is approximately 82% which tells that 82% of the variation in predicting the survival rate of the original bacteria colony is accounted for by the model and is a good-fit model.
e) Multiple Standard Error of Estimate = sqrt(MSResidual) = sqrt(8.88) = 2.98
Since this is the standard deviation in y values.
Therefore, Multiple Standard Error of Estimate = 2.98 % survival.
From the Empirical rule, for any predicted value of the y variable (% survival), the usual range of values would be +/- 2 * Multiple Standard Error of Estimate OR
Y’ +/- 2 * Multiple Standard Error of Estimate
f)
1. Null hypothesis: None of the independent variables included in the model have a significant impact on the dependent variable.
Alternate hypothesis: Atleast one of the independent variables included in the model have a significant impact on the dependent variable.
2. The critical value of F is 3.86 at 3 and 9 degrees of freedom for a two-tailed test
Decision rule: F-stat >3.86, reject the null hypothesis.
3. Test value: F-stat = MS/MS(Error) = 119.3966/8.88 = 13.44
4. Since 13.44 is greater than the critical value (3.86), we reject the null hypothesis.
We conclude that atleast one of the independent variables included in the model hasa significant impact on the dependent variable.
Analysis of Variance
1. Test value: From ANOVA table t-stat = 5.43
a. Since 5.43 is greater than the critical value (2.179), we reject the null hypothesis.
b. We conclude that % weight A significantly impacts the dependent variable.
% weight B
1. Null hypothesis: % weight B does not significantly impact the dependent variable.
Alternate hypothesis: % weight B significantly impacts the dependent variable.
The critical value of t is 2.179 at 12 degrees of freedom for a two-tailed test
2. Decision rule : t-stat > +/- 2.179, reject null hypothesis.
3. Test value: From ANOVA table t-stat = -4.24
4. Since -4.24 is more negative than the critical value (-2.179), we reject the null hypothesis.
5. We conclude that % weight B significantly impacts the dependent variable.
% weight C
1. Null hypothesis: % weight C does not significantly impact the dependent variable.
Alternate hypothesis: % weight C significantly impacts the dependent variable.
The critical value of t is 2.179 at 12 degrees of freedom for a two-tailed test
2. Decision rule : t-stat > +/- 2.179, reject null hypothesis.
3. Test value: From ANOVA table t-stat = -2.51
4. Since -2.51 is more negative than the critical value (-2.179), we reject the null hypothesis.
5. We conclude that % weight C significantly impacts the dependent variable.
2. Since the values of the correlation among all the independent variables are quite small, there is no evidence of multicollinearity in the model.
3. 82% of the variation in predicting the survival rate of the original bacteria colony is accounted for by the model. It can be considered to be a good-fit model.
The correlation matrix shows that the correlation between % weight A and % survival is the strongest and positive (0.612). The correlation between % survival and % weight B is medium (-0.411) and negative. However, the correlation between % survival and % weight C is weakest and negative (-0.186).
All three predictor variables included in the model significantly impact % survival.