Question

What is the relationship between the number of minutes per day a woman spends talking on the phone and the woman's weight? The time on the phone and weight for 8 women are shown in the table below.

Time	29	12	64	90	59	36	53	65
Pounds	99	113	131	148	113	125	115	127

1. Find the correlation coefficient: r=r=   Round to 2 decimal places.
2. The null and alternative hypotheses for correlation are:
   H0:H0: ?rμρ == 0
   H1:H1: ?μρr ≠≠ 0
   The p-value is: (Round to four decimal places)
3. r2r2 = (Round to two decimal places)
4. The equation of the linear regression line is:   
   ˆyy^ = + xx   (Please show your answers to two decimal places)
   
5. Use the model to predict the weight of a woman who spends 46 minutes on the phone.
   Weight = (Please round your answer to the nearest whole number.)

Answer 1

Solution:

Use cor.test function in R to get the correlation coefficient and p value for the hypothesis test for correlation

Time   <- c(29   ,12   ,64,   90,   59,   36,   53,   65)
Pounds <- c(99,   113   ,131,   148   ,113,   125   ,115,   127)  
cor.test(Time,Pounds)

Output:

Pearson's product-moment correlation

data: Time and Pounds
t = 2.8837, df = 6, p-value = 0.02792
alternative hypothesis: true correlation is not equal to 0
95 percent confidence interval:
0.1241646 0.9543008
sample estimates:
cor
0.7621527

Solution-A:

the correlation coefficient: r=0.76

The p-value is: (Round to four decimal places)

p= 0.0279

For this linear regression use lm function in R

linmod <- lm(Pounds~Time)
coefficients(linmod)

summary(linmod)

Output:

coefficients(linmod)
(Intercept) Time
97.8862333 0.4605641
> summary(linmod)

Call:
lm(formula = Pounds ~ Time)

Residuals:
Min 1Q Median 3Q Max
-12.243 -8.487 1.407 8.894 10.534

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 97.8862 8.9269 10.965 3.42e-05 ***
Time 0.4606 0.1597 2.884 0.0279 *
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 10.33 on 6 degrees of freedom
Multiple R-squared: 0.5809,   Adjusted R-squared: 0.511
F-statistic: 8.316 on 1 and 6 DF, p-value: 0.02792

r^2=0.58

y^=97.89+0.46*x

For x=46 ,y^=97.89+0.46*46=119.05=119

t the weight of a woman who spends 46 minutes on the phone=119