The weights (kg) and blood glucose levels (mg/100 ml) of sixteen healthy adult m

Question

The weights (kg) and blood glucose levels (mg/100 ml) of sixteen healthy adult males are listed below: weight glucose weight glucose 64 73 76.2 59.4 82.1 76.7 83.9 64.4 108 104 105 79 101 75.3 82.1 95 93.4 78.9 82.1 73 77.6 109 102 121 107 85 100 104 87 108 102 (a)Find the simple linear regression equation (b) Test Ho: = 0 using both the t test and analysis of variance (c) Test Ho: = 0 (d) Construct a 95% confidence interval for (e) What is the predicted glucose level for a man who weighs 95? Construct the 95% prediction interval for his weight. Let = 0.05 for all tests

Explanation / Answer

Solution:

Required regression output for the given data is summarised as below:

Simple Linear Regression Analysis

Regression Statistics

Multiple R

0.4841

R Square

0.2344

Adjusted R Square

0.1797

Standard Error

9.2761

Observations

16

ANOVA

df

SS

MS

F

Significance F

Regression

1

368.7981

368.7981

4.2861

0.0574

Residual

14

1204.6394

86.0457

Total

15

1573.4375

Coefficients

Standard Error

t Stat

P-value

Lower 95%

Upper 95%

Intercept

61.8769

19.1890

3.2246

0.0061

20.7205

103.0333

Weight

0.5098

0.2462

2.0703

0.0574

-0.0183

1.0378

Part a

From the above given regression output, the simple linear regression equation is given as below:

Y = 0 + 1 X

Y = 61.8769 + 0.5098*X

Blood glucose level = 61.8769 + 0.5098*weight

Where, y-intercept is given as 61.8769 and slope is given as 0.5098.

Part b

Here, we have to use a t test for checking whether a population slope is zero or not.

H0: = 0 versus Ha: 0

We are given

= 0.5098

SE() = 0.2462

Sample size = n = 16

df = n – 1 = 16 – 1 = 15

= 0.05

Test statistic formula is given as below:

t = /SE()

t = 0.5098/0.2462

t = 2.0703

P-value = 0.0574

(By using t-table or excel)

P-value > = 0.05

So, we do not reject the null hypothesis that the population slope is zero.

Population slope is not statistically significant.

There is insufficient evidence to conclude that the population slope is different than zero.

Now, we have to check same hypothesis by using ANOVA F test.

The test statistic value for ANOVA is given as F = 4.2861

P-value = 0.0574

= 0.05

P-value > = 0.05

So, we do not reject the null hypothesis that the population slope is zero.

Population slope is not statistically significant.

Part c

Now, we have to test whether the population correlation coefficient is different from zero or not.

H0: = 0 versus Ha: 0

For checking this test, we have to use t test for population correlation coefficient.

Test statistic formula is given as below:

t = r*sqrt((n – 2)/(1 – r^2))

We are given

r = 0.4841

n = 16

df = n – 2 = 16 – 2 = 14

t = 0.4841*sqrt((16 - 2)/(1 - 0.4841^2))

t = 2.07007

Critical value = 2.144787

(By using t-table or excel)

P-value = 0.05742

= 0.05

P-value > = 0.05

So, we do not reject the null hypothesis that the population correlation coefficient is zero.

There is insufficient evidence to conclude that population correlation coefficient is different from zero.

Population correlation coefficient is not statistically significant.

Part d

Here, we have to find 95% confidence interval for population correlation coefficient . Formula for confidence interval is given as below:

Confidence interval = r -/+ t*sqrt((1 – r^2)/(n – 2))

We are given

Confidence level = 95%

df = 14

So, critical value t = 2.144787 (By using t-table or excel)

Confidence interval = 0.4841 -/+ 2.144787*sqrt((1 - 0.4841^2)/(16 - 2))

Confidence interval = 0.4841 -/+ 2.144787* 0.233857

Confidence interval = 0.4841 -/+ 0.501573

Lower limit = 0.4841 - 0.501573 = -0.01747

Upper limit = 0.4841 + 0.501573 = 0.985673

Confidence interval = (-0.01747, 0.985673)

Above confidence interval contains zero, so we do not reject the null hypothesis.

Part e

We are given X = 95

Y = 61.8769 + 0.5098*X

Y = 61.8769 + 0.5098*95

Y = 110.3079

Predicted blood glucose level = 110.3079 (mg/100ml)

CI = Predicted Y -/+ t*SE

df = 14

Critical value t = 2.144787 (By using t-table or excel)

Confidence interval = 110.3079 -/+ 2.144787*10.50157

Confidence interval = 110.3079 -/+ 22.5236

Lower limit = 110.3079 - 22.5236 = 87.7796

Upper limit = 110.3079 + 22.5236 = 132.8269

Simple Linear Regression Analysis

Regression Statistics

Multiple R

0.4841

R Square

0.2344

Adjusted R Square

0.1797

Standard Error

9.2761

Observations

16

ANOVA

df

SS

MS

F

Significance F

Regression

1

368.7981

368.7981

4.2861

0.0574

Residual

14

1204.6394

86.0457

Total

15

1573.4375

Coefficients

Standard Error

t Stat

P-value

Lower 95%

Upper 95%

Intercept

61.8769

19.1890

3.2246

0.0061

20.7205

103.0333

Weight

0.5098

0.2462

2.0703

0.0574

-0.0183

1.0378

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