In: Statistics and Probability
Data:
Bob wishes to study the relationship between mean annual temperature (Temp) and the mortality rate (SMI) for a type of breast cancer in women at a 5% significance level and collects paired sample data.
SMI 100 96 95 89 89 79 82 72 65 68 53 Temp 50 49 48 47 45 46 44 43 42 40 34
1) Determine predictor variable, x, and response variable, y. What is the response variable?
Group of answer choices
A) y=mean annual temperature (Temp)
B) y=mortality rate (SMI) for a type of breast cancer in women
C) x=mean annual temperature (Temp)
D) x=mortality rate (SMI) for a type of breast cancer in women
2)Calculate r (rounded to the nearest ten-thousandth).
3)Is there sufficient evidence support the claim of linear correlation? Why? (Hint: Complete a linear correlation hypothesis test using Table A-6.)
Group of answer choices
A) There is sufficient evidence to support the claim of linear correlation because the absolute value of r is larger than the critical value from Table A-6 of 0.602.
B) There is not sufficient evidence to support the claim of linear correlation because the absolute value of r is larger than the critical value from Table A-6 of 0.602.
C)There is sufficient evidence to support the claim of linear correlation because the absolute value of r is not larger than the critical value from Table A-6 of 0.735.
D) There is not sufficient evidence to support the claim of linear correlation because the absolute value of r is larger than the critical value from Table A-6 of 0.735.
E) There is sufficient evidence to support the claim of linear correlation because the absolute value of r is not larger than the critical value from Table A-6 of 0.602.
F) There is sufficient evidence to support the claim of linear correlation because the absolute value of r is larger than the critical value from Table A-6 of 0.735.
G) There is not sufficient evidence to support the claim of linear correlation because the absolute value of r is not larger than the critical value from Table A-6 of 0.602.
H) There is not sufficient evidence to support the claim of linear correlation because the absolute value of r is not larger than the critical value from Table A-6 of 0.735.
4)Find the slope of the regression line (rounded to the nearest ten-thousandth).
5)Find the y-intercept of the regression line (rounded to the nearest ten-thousandth).
6) Calculate the best point estimate for the mortality rate (SMI) for a type of breast cancer in women at mean annual temperature (Temp) of 38 (rounded to the nearest ten-thousandth).
7)Construct a prediction interval estimate for SMI at a Temp of 38. What is the value of the lower bound of the interval (rounded to the nearest tenth)?
1)
x = mean annual temperature (Temp)
y = mortality rate (SMI) for a type of breast cancer in women
2)
Correlation coefficient (r) = 0.9 (=CORREL(Xi,Yi)
3)
H0: ρ = 0
HA: ρ not = 0
α = 0.05
rc = 0.602 (Use r table)
∣r∣ > rc = 0.602, Reject H0
There is sufficient evidence to support the claim of linear correlation because the absolute value of r is larger than the critical value from Table A-6 of 0.602.
4)
Excel > Data > Data Analysis > Regression
SUMMARY OUTPUT | ||||||||
Regression Statistics | ||||||||
Multiple R | 0.949879808 | |||||||
R Square | 0.902271649 | |||||||
Adjusted R Square | 0.891412944 | |||||||
Standard Error | 4.892296138 | |||||||
Observations | 11 | |||||||
ANOVA | ||||||||
df | SS | MS | F | Significance F | ||||
Regression | 1 | 1988.770765 | 1988.770765 | 83.09200753 | 7.69148E-06 | |||
Residual | 9 | 215.4110535 | 23.9345615 | |||||
Total | 10 | 2204.181818 | ||||||
Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | Lower 95.0% | Upper 95.0% | |
Intercept | -55.62003454 | 15.0303308 | -3.700519655 | 0.004916604 | -89.62100502 | -21.61906407 | -89.62100502 | -21.61906407 |
Temp | 3.073402418 | 0.337162917 | 9.115481749 | 7.69148E-06 | 2.31068691 | 3.836117926 | 2.31068691 | 3.836117926 |
Y^ = -55.6 + 3.1*X
the slope of the regression line = 3.1
5)
the y-intercept of the regression line = -55.6
6)
If X = 38
Y^ = -55.6 + 3.1*X
Y^ = -55.6 + 3.1*38 = 62.2
7)
X0 | 38 | |
Y^ | 62.2 | |
X bar | 44.3636 | |
MSE | 23.9345615 | |
SSxx | 210.55 | SUM(X-X bar)^2 |
t c | 2.262157163 | Use t table |
95% PI | ||
LOWER | 49.7 | Y^-tc*SQRT(MSE*(1+1/n+(X0-Xbar)^2/SSxx)) |
UPPER | 74.7 | Y^+tc*SQRT(MSE*(1+1/n+(X0-Xbar)^2/SSxx)) |