Question

concrete​ cures, it gains strength. The following data represent the​ 7-day and​ 28-day strength in pounds per square inch​ (psi) of a certain type of concrete. Complete parts​ (a) through​ (f) below. ​7-Day Strength​ (psi), x 2480 3330 2620 3380 3390 ​28-Day Strength​ (psi), y 4120 4850 4190 5020 5220 ​(a) Treating the​ 7-day strength as the explanatory​ variable, x, use technology to determine the estimates of beta 0 and beta 1. beta 0almost equalsb 0equals nothing ​(Round to one decimal place as​ needed.) beta 1almost equalsb 1equals nothing ​(Round to four decimal places as​ needed.) ​28-Day Strength​ (psi), y ​(a) Treating the​ 7-day strength as the explanatory​ variable, x, use technology to determine the estimates of beta 0 and beta 1. beta 0almost equalsb 0equals Round to one decimal place as​ needed.) beta 1almost equalsb 1equals Round to four decimal places as​ needed.) ​(b) Compute the standard error of the​ estimate, s Subscript e. s Subscript eequals ​(Round to one decimal place as​ needed.) ​(c) A normal probability plot suggests that the residuals are normally distributed. Determine s Subscript b 1. Use the answer from part ​(b). s Subscript b 1equals 0.1212   ​(Round to four decimal places as​ needed.) ​(d) A normal probability plot suggests that the residuals are normally distributed. Test whether a linear relation exists between​ 7-day strength and​ 28-day strength at the alphaequals0.05 level of significance. State the null and alternative hypotheses. Choose the correct answer below. A. Upper H 0​: beta 1equals0 Upper H 1​: beta 1not equals0 Your answer is correct.B. Upper H 0​: beta 0equals0 Upper H 1​: beta 0not equals0 C. Upper H 0​: beta 0equals0 Upper H 1​: beta 0greater than0 D. Upper H 0​: beta 1equals0 Upper H 1​: beta 1greater than0 Determine the​ P-value of this hypothesis test. ​P-valueequals Round to three decimal places as​ needed.) What is the conclusion that can be​ drawn? A. Do not reject Upper H 0 and conclude that a linear relation does not exist between the​ 7-day and​ 28-day strength of a certain type of concrete at the alphaequals0.05 level of significance. B. Reject Upper H 0 and conclude that a linear relation does not exist between the​ 7-day and​ 28-day strength of a certain type of concrete at the alphaequals0.05 level of significance. C. Do not reject Upper H 0 and conclude that a linear relation exists between the​ 7-day and​ 28-day strength of a certain type of concrete at the alphaequals0.05 level of significance. D. Reject Upper H 0 and conclude that a linear relation exists between the​ 7-day and​ 28-day strength of a certain type of concrete at the alphaequals0.05 level of significance. Your answer is correct. ​(e) Construct a​ 95% confidence interval about the slope of the true​ least-squares regression line. Lower​ bound: ​(Round to three decimal places as​ needed.) Upper​ bound: ​(Round to three decimal places as​ needed.) ​(f) What is the estimated mean​ 28-day strength of this concrete if the​ 7-day strength is 3000​ psi? A good estimate of the mean​ 28-day strength is 4740.81 psi. ​(Round to two decimal places as​ needed.)

Answer 1

(a) Given that ​ 7-day strength as the explanatory​ (independent) variable (x) and 28-day strength as the predictive (dependent) variable (y). Then, linear regression model is

y =b₀ + b₁*x +e

Using the ordinary least square method, we estimate the coefficient b₀ and b₁.

We do the analysis in excel, get the output.

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.973998
R Square	0.948673
Adjusted R Square	0.931564
Standard Error	130.1327
Observations	5
ANOVA
	df	SS	MS	F	Significance F
Regression	1	938996.4	938996.4	55.44863	0.005013
Residual	3	50803.58	16934.53
Total	4	989800
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%
Intercept (beta0)	1411.308	442.8048	3.1872	0.049818	2.105175	2820.51
X Variable (beta1)	1.075228	0.144396	7.446384	0.005013	0.615695	1.53476

The estimate linear regression equation is

(b) Compute the standard error of the​ estimate

x	y	hat{y}	(y-hat(y))^2
2480	4120	4077.872	1774.731
3330	4850	4991.816	20111.79
2620	4190	4228.404	1474.893
3380	5020	5045.577	654.2056
3390	5220	5056.33	26787.96
		SSE	50803.58

c. A normal probability plot suggests that the residuals are normally distributed. Determine s_b1

	x	(x-bar(x))^2
	2480	313600
	3330	84100
	2620	176400
	3380	115600
	3390	122500
average	3040

(d) A normal probability plot suggests that the residuals are normally distributed. Test whether a linear relation exists between​ 7-day strength and​ 28-day strength at the alpha = 0.05 level of significance. The null and alternative hypothesis is

Using the provide table, we see that t-test value is 7.446 and p-value is 0.005.

The p-value is less than 0.05, reject the null hypothesis and conclude that there is a sufficient evidence that a linear relation exists between​ 7-day strength and​ 28-day strength.

(e) a​ 95% confidence interval about the slope of the true​ least-squares regression line is

lower bound = 0.616

Upper bound = 1.534

(f) the estimated mean​ 28-day strength of this concrete if the​ 7-day strength is 3000​ psi is