Exam 7: Relationships in Data

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a. The scatterplot on the right (B) is the appropriate one, because the response variable (temperature) is on the vertical axis.
b. The correlation coefficient is the (positive, because the association is positive) square root of $r^2: \sqrt{.957} ≈ .978$ .
c. The predicted temperature for a cricket chirping at 110 chirps per minute is $35.8 + 0.251 × 110 = 63.41 ^\circ{F}$ . This is not an example of extrapolation because 110 chirps per minute is near the center of the data.
d. The test statistic is 24.89. This is extremely large, so the p-value is essentially 0. The sample data provide overwhelming evidence there is a relationship between chirp frequency and temperature.
e. The critical value for 90% confidence and 30 - 2 = 28 degrees of freedom is t* = 1.701. The 90% confidence interval for the population slope coefficient is therefore 0.25116 $\pm$ 1.701 × 0.01009, which is 0.25116 $\pm$ 0.01716, which is the interval (0.23400, 0.26832). You can be 90% confident the predicted temperature increases between 0.234 and 0.268 degrees for each additional chirp per minute of a cricket.

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a. The slope coefficient is -0.2170. This means that for each additional increase of one degree in mean monthly temperature, the predicted natural gas usage decreases by .217 therms per day.
b. The predicted gas usage in a month with a mean temperature of 50 degrees is $15.37 - 0.2170 \times (50) = 4.52$ therms per day.
c. The correlation coefficient is the negative square root of the $r{}_2$ value, which is $-\sqrt{.864}\approx -.930$ .
d. $r^2 is 86.4\%$ .
e. No. It's common for none, or a very small percentage, of data values to fall exactly on the line. This is not what the $r^2$ value measures.