Consider the following data on x = rainfall volume (m3) and y = runoff volume (m3) for a particular location. x 4 12 14 20 23 30 40 47 55 67 72 83 96 112 127 y 4 10 13 14 15 25 27 46 38 46 53 75 82 99 104 Use the accompanying Minitab output to decide whether there is a useful linear relationship between rainfall and runoff. The regression equation is runoff = -2.07 + 0.850 rainfall Predictor Coef Stdev t-ratio p Constant -2.067 2.412 -0.86 0.407 rainfall 0.85038 0.03708 22.93 0.000 s = 5.321 R-sq = 97.6% R-sq(adj) = 97.4% State the appropriate null and alternative hypotheses. H0: ?1 = 0 Ha: ?1 > 0 H0: ?1 = 0 Ha: ?1 ≠ 0 H0: ?1 = 0 Ha: ?1 < 0 H0: ?1 ≠ 0 Ha: ?1 = 0 Compute the test statistic value and find the P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.) t = P-value = State the conclusion in the problem context. (Use ? = 0.05.) Reject H0. There is a useful linear relationship between runoff and rainfall at the 0.05 level.Reject H0. There is not a useful linear relationship between runoff and rainfall at the 0.05 level. Fail to reject H0. There is not a useful linear relationship between runoff and rainfall at the 0.05 level.Fail to reject H0. There is a useful linear relationship between runoff and rainfall at the 0.05 level. Calculate a 95% confidence interval for the true average change in runoff volume associated with a 1 m3 increase in rainfall volume. (Round your answers to three decimal places.) (_______), (______) m3 For this question, my answers were = [0.742, 0.959]m3 but it came up wrong, any chance you could explain why. I used R programming to evaluate the question; my code written below: # Enter the data x <- c(4, 12, 14, 20, 23, 30, 40, 47, 55, 67, 72, 83, 96, 112, 127) y <- c(4, 10, 13, 14, 15, 25, 27, 46, 38, 46, 53, 75, 82, 99, 104) # Fit the linear regression model model <- lm(y ~ x) # Extract the coefficient and standard error for the slope coef <- coef(model)[2] se <- summary(model)$coefficients[2, 2] # Calculate the confidence interval alpha <- 0.05 t_crit <- qt(1 - alpha / 2, df = length(x) - 2) lower <- coef - t_crit * se upper <- coef + t_crit * se # Print the results round(lower, 3) # Lower bound of CI round(upper, 3) # Upper bound of CI[0.742, 0.959] [0.742, 0.959]m3

Consider the following data on x = rainfall volume (m3) and y = runoff volume (m3) for a particular location. x 4 12 14 20 23 30 40 47 55 67 72 83 96 112 127 y 4 10 13 14 15 25 27 46 38 46 53 75 82 99 104 Use the accompanying Minitab output to decide whether there is a useful linear relationship between rainfall and runoff. The regression equation is runoff = -2.07 + 0.850 rainfall Predictor Coef Stdev t-ratio p Constant -2.067 2.412 -0.86 0.407 rainfall 0.85038 0.03708 22.93 0.000 s = 5.321 R-sq = 97.6% R-sq(adj) = 97.4% State the appropriate null and alternative hypotheses. H0: ?1 = 0 Ha: ?1 > 0 H0: ?1 = 0 Ha: ?1 ≠ 0 H0: ?1 = 0 Ha: ?1 < 0 H0: ?1 ≠ 0 Ha: ?1 = 0 Compute the test statistic value and find the P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.) t = P-value = State the conclusion in the problem context. (Use ? = 0.05.) Reject H0. There is a useful linear relationship between runoff and rainfall at the 0.05 level.Reject H0. There is not a useful linear relationship between runoff and rainfall at the 0.05 level. Fail to reject H0. There is not a useful linear relationship between runoff and rainfall at the 0.05 level.Fail to reject H0. There is a useful linear relationship between runoff and rainfall at the 0.05 level. Calculate a 95% confidence interval for the true average change in runoff volume associated with a 1 m3 increase in rainfall volume. (Round your answers to three decimal places.) (_), () m3 For this question, my answers were = [0.742, 0.959]m3 but it came up wrong, any chance you could explain why. I used R programming to evaluate the question; my code written below: # Enter the data x <- c(4, 12, 14, 20, 23, 30, 40, 47, 55, 67, 72, 83, 96, 112, 127) y <- c(4, 10, 13, 14, 15, 25, 27, 46, 38, 46, 53, 75, 82, 99, 104) # Fit the linear regression model model <- lm(y ~ x) # Extract the coefficient and standard error for the slope coef <- coef(model)[2] se <- summary(model)$coefficients[2, 2] # Calculate the confidence interval alpha <- 0.05 t_crit <- qt(1 - alpha / 2, df = length(x) - 2) lower <- coef - t_crit * se upper <- coef + t_crit * se # Print the results round(lower, 3) # Lower bound of CI round(upper, 3) # Upper bound of CI[0.742, 0.959] [0.742, 0.959]m3

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Related questions

Question

Consider the following data on x = rainfall volume (m³) and y = runoff volume (m³) for a particular location.

x	4	12	14	20	23	30	40	47	55	67	72	83	96	112	127
y	4	10	13	14	15	25	27	46	38	46	53	75	82	99	104

Use the accompanying Minitab output to decide whether there is a useful linear relationship between rainfall and runoff.

The regression equation is runoff = -2.07 + 0.850 rainfall
Predictor	Coef	Stdev	t-ratio	p
Constant	-2.067	2.412	-0.86	0.407
rainfall	0.85038	0.03708	22.93	0.000
s = 5.321 R-sq = 97.6% R-sq(adj) = 97.4%

State the appropriate null and alternative hypotheses.

H₀: ?₁ = 0

H_a: ?₁ > 0

H₀: ?₁ = 0

H_a: ?₁ ≠ 0

H₀: ?₁ = 0

H_a: ?₁ < 0

H₀: ?₁ ≠ 0

H_a: ?₁ = 0

Compute the test statistic value and find the P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.)

t	=
P-value	=

State the conclusion in the problem context. (Use ? = 0.05.)

Reject H₀. There is a useful linear relationship between runoff and rainfall at the 0.05 level.Reject H₀. There is not a useful linear relationship between runoff and rainfall at the 0.05 level. Fail to reject H₀. There is not a useful linear relationship between runoff and rainfall at the 0.05 level.Fail to reject H₀. There is a useful linear relationship between runoff and rainfall at the 0.05 level.

Calculate a 95% confidence interval for the true average change in runoff volume associated with a 1 m³ increase in rainfall volume. (Round your answers to three decimal places.)

(_______), (______) m³

^{For this question, my answers were = [0.742, 0.959]m3 but it came up wrong, any chance you could explain why. I used R programming to evaluate the question; my code written below:}

# Enter the data
x <- c(4, 12, 14, 20, 23, 30, 40, 47, 55, 67, 72, 83, 96, 112, 127)
y <- c(4, 10, 13, 14, 15, 25, 27, 46, 38, 46, 53, 75, 82, 99, 104)

# Fit the linear regression model
model <- lm(y ~ x)

# Extract the coefficient and standard error for the slope
coef <- coef(model)[2]
se <- summary(model)$coefficients[2, 2]

# Calculate the confidence interval
alpha <- 0.05
t_crit <- qt(1 - alpha / 2, df = length(x) - 2)
lower <- coef - t_crit * se
upper <- coef + t_crit * se

# Print the results
round(lower, 3) # Lower bound of CI
round(upper, 3) # Upper bound of CI[0.742, 0.959]
^{[0.742, 0.959]m3}

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x	4	12	14	20	23	30	40	47	55	67	72	83	96	112	127
y	4	10	13	14	15	25	27	46	38	46	53	75	82	99	104

x	4	12	14	20	23	30	40	47	55	67	72	83	96	112	127
y	4	10	13	14	15	25	27	46	38	46	53	75	82	99	104

x	4	12	14	20	23	30	40	47	55	67	72	83	96	112	127
y	4	10	13	14	15	25	27	46	38	46	53	75	82	99	104