Question

. The following table contains the weights of a group of 12 women, where x represents the woman’s average weight at age 17, and y represents the woman’s weight after menopause.

Weights (in Pounds Weight at Age 17, x ----99-- 123-- 119 --87 --90 --101-- 98-- 121-- 131-- 134--135-- 97

Weight after Menopause, y ----------------121-- 156--145--109--111--137-- 128--142--149--152 --155-- 128

Enter this information on your calculator and then answer the following:

a. What kind of linear relationship seems to exist between weights at age 17 and weight after menopause?

b. Calculate the correlation coefficient, r?

c. Is r statistically significant at the 0.05 level? At the 0.01 level? Explain.

d. Find the equation of the least-squares regression line.

e. Calculate and interpret the coefficient of determination, r^2 .

f. For a woman who weighed 105 pounds at 17 years old, what would be her predicted weight after menopause.

g. What would be the predicted weight of a woman who weighed 180 pounds at age 17?

Answer 1

ANSWER:

1. Bring data in to excel sheet

2.Select both x and y variable and insert scatter plot chart. Select any data point in the excel and add trend line.

(a)

What kind of linear relationship is exists between the variables.

The scatter plot shows the perfect positive linear relaionship between the variables.

(b)

Correlation coefficient r:

USe "=CORREL(Array 1, Array 2) formula in excel to find the r value.

r = 0.94 (Perfect positive)

c)

Is r statistically significant at 0.05 level?

Run a regression analysis in excel followed by below procedures:

1. Go to data tab --> data analysis --> choose Regression

Result:

Highilighted significance F is nothing but the P-value.

P-value = 0.000007

If the p-value is less than the significance level (α=0.05):

• Decision: Reject the null hypothesis.
• Conclusion: "There is sufficient evidence to conclude that there is a significant linear relationship between x and y because the correlation coefficient is significantly different from zero."

d) Equation:

y = 38.536 + 0.877x

e) coefficient of determination :

r-squared = 0.878 (check the summary output)

f) Predict weight 105:

x = 105

y = 38.536 + 0.877x

y = 38.536 + 0.877*(105) = 131

so the predicted weight of women would be 131

g) Predict weight 180:

y = 38.536 + 0.877*(180) = 196

so the predicted weight of women would be 196

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