Question

Modern medical practice tells us not to encourage babies to become too fat. Is there a positive correlation between the weight x of a 1-year old baby and the weight y of the mature adult (30 years old)? A random sample of medical files produced the following information for 14 females.

x (lb)	20	27	22	26	20	15	25	21	17	24	26	22	18	19
y (lb)	122	123	119	127	130	120	145	130	130	130	130	140	110	115

In this setting we have Σx = 302, Σy = 1771, Σx² = 6690, Σy² = 225,153, and Σxy = 38,391.

What percentage of variation in y is explained by the least-squares model? (Round your answer to one decimal place.)
%

Find or estimate the P-value of the test statistic.

P-value > 0.2500.125 < P-value < 0.250    0.100 < P-value < 0.1250.075 < P-value < 0.1000.050 < P-value < 0.0750.025 < P-value < 0.0500.010 < P-value < 0.0250.005 < P-value < 0.0100.0005 < P-value < 0.005P-value < 0.0005

Conclusion

Reject the null hypothesis. There is sufficient evidence that ρ > 0.Reject the null hypothesis. There is insufficient evidence that ρ > 0.    Fail to reject the null hypothesis. There is sufficient evidence that ρ > 0.Fail to reject the null hypothesis. There is insufficient evidence that ρ > 0.

(e) If a female baby weighs 16 pounds at 1 year, what do you predict she will weigh at 30 years of age? (Round your answer to two decimal places.)
lb

(f) Find S_e. (Round your answer to two decimal places.)
S_e =

(g) Find a 95% confidence interval for weight at age 30 of a female who weighed 16 pounds at 1 year of age. (Round your answers to two decimal places.)

lower limit	lb
upper limit	lb

(h) Test the claim that the slope β of the population least-squares line is positive at the 1% level of significance. (Round your test statistic to three decimal places.)

t =

Find or estimate the P-value of the test statistic.

P-value > 0.2500.125 < P-value < 0.250    0.100 < P-value < 0.1250.075 < P-value < 0.1000.050 < P-value < 0.0750.025 < P-value < 0.0500.010 < P-value < 0.0250.005 < P-value < 0.0100.0005 < P-value < 0.005P-value < 0.0005

Conclusion

Reject the null hypothesis. There is sufficient evidence that β > 0.Reject the null hypothesis. There is insufficient evidence that β > 0.    Fail to reject the null hypothesis. There is sufficient evidence that β > 0.Fail to reject the null hypothesis. There is insufficient evidence that β > 0.

(i) Find an 80% confidence interval for β and interpret its meaning. (Round your answers to three decimal places.)

lower limit
upper limit

Interpretation

For each pound less a female infant weighs at 1 year, the adult weight increases by an amount that falls outside the confidence interval.For each pound less a female infant weighs at 1 year, the adult weight increases by an amount that falls within the confidence interval.    For each pound more a female infant weighs at 1 year, the adult weight increases by an amount that falls within the confidence interval.For each pound more a female infant weighs at 1 year, the adult weight increases by an amount that falls outside the confidence interval.

Answer 1

percentage of variation in y is explained by the least-squares model =r^2*100 =18.0%

test statistic t =

r*(√(n-2)/(1-r2))=

1.621

p value is  0.050 < P-value < 0.075

.Fail to reject the null hypothesis. There is insufficient evidence that ρ > 0.

e)

predicted value =

120.53

f)

Se =√(SSE/(n-2))=

8.76

g)

std error of confidence interval =				s*√(1+1/n+(x0-x̅)2/Sxx)=			9.7832
for 95 % confidence and 12degree of freedom critical t=							2.179
lower limit =		99.21
uppr limit =		141.85

h)

test statistic t =

r*(√(n-2)/(1-r2))=

1.621

p value is  0.050 < P-value < 0.075

i)

std error of slope sb1 =				s/√SSx=		0.6611
for 80 % confidence and -2degree of freedom critical t=						1.3560
80% confidence interval =b1 -/+ t*standard error=					(0.175,1.968)
lower limit =		0.175
uppr limit =		1.968

   For each pound more a female infant weighs at 1 year, the adult weight increases by an amount that falls within the confidence interval