Week 15 Quiz

For 342 Palmer penguins, software fits the least-squares line ŷ = −5,780.8314 + 49.6856x, where x is flipper length in mm and ŷ is predicted body mass in g; the observed flippers run from 172 mm to 231 mm. Which sentence correctly interprets the slope?

Select one.

• Each additional gram of body mass predicts about 49.69 mm more flipper length.
• Each additional millimetre of flipper length predicts about 49.69 g more body mass.
• Each additional millimetre of flipper length predicts about 0.8712 g more body mass, the correlation.
• A penguin with a 1 mm flipper is predicted to weigh about 49.69 g.

For 342 Palmer penguins, software fits the least-squares line ŷ = −5,780.8314 + 49.6856x, where x is flipper length in mm and ŷ is predicted body mass in g; the observed flippers run from 172 mm to 231 mm. What body mass does the line predict for a penguin with a 200 mm flipper?

Select one.

• 9,937.12 g
• 4,201.75 g
• 4,653.14 g
• 4,156.29 g

For 342 Palmer penguins, software fits the least-squares line ŷ = −5,780.8314 + 49.6856x, where x is flipper length in mm and ŷ is predicted body mass in g; the observed flippers run from 172 mm to 231 mm. The line predicts 4,156.29 g for a 200 mm flipper. Suppose a field team weighs a new penguin with a 200 mm flipper at 4,500 g. What is its residual?

Select one.

• 343.71 g
• −343.71 g
• 394.28 g
• 4,156.29 g

For the 342 penguins, the regression of body mass on flipper length has r² = 0.759. Which statement correctly interprets r²?

Select one.

• About 75.9% of the 342 penguins lie exactly on the least-squares line.
• Flipper length causes about 75.9% of each penguin's body mass.
• Flipper length accounts for about 75.9% of the variation in body mass.
• Each additional millimetre of flipper predicts 0.759 g more body mass.

For 342 Palmer penguins, software fits the least-squares line ŷ = −5,780.8314 + 49.6856x, where x is flipper length in mm and ŷ is predicted body mass in g; the observed flippers run from 172 mm to 231 mm. True or false: because the intercept is −5,780.83 g, the line gives a trustworthy prediction of −5,780.83 g for a penguin with a 0 mm flipper.

True or false.

• True
• False

Score each face x of a fair die by y = (x − 3.5)², giving 6.25, 2.25, 0.25, 0.25, 2.25, 6.25 for faces 1 to 6. The least-squares line is flat at ŷ ≈ 2.92, and its residuals are about 3.33, −0.67, −2.67, −2.67, −0.67, 3.33. What does this residual plot tell you?

Select one.

• The spread of the residuals grows from left to right, so the spread is not equal.
• The residuals trace a U shape, so a straight line is the wrong form for these data.
• The residuals form a shapeless band around 0, so the straight line fits well.
• The line fits well, because r² for these data is 0.

Software reports a slope of b₁ = 49.6856 g per mm with standard error SE(b₁) = 1.5184 g per mm for the 342 penguins. For the test of H₀: β₁ = 0, what are the t statistic and its degrees of freedom?

Select one.

• t ≈ 32.72 with df = 341
• t ≈ 0.03 with df = 340
• t ≈ 75.44 with df = 340
• t ≈ 32.72 with df = 340

Software reports b₁ = 49.6856 g per mm and SE(b₁) = 1.5184 g per mm on 340 degrees of freedom. Using the conservative critical value t* = 1.976, which is the 95% confidence interval for the population slope β₁?

Select one.

• About 46.69 to 52.69 g per mm
• About 48.17 to 51.20 g per mm
• About 47.71 to 51.66 g per mm
• About 45.77 to 53.60 g per mm

Pair each month's position in the year (1 to 12) with its number of days in a common year. For all twelve months the least-squares slope is about 0.07 days per month position; with February (month 2, 28 days) removed it is about −0.02. What does this show?

Select one.

• February is an extrapolation, because month 2 lies outside the data.
• The slope and the correlation are the same number for these data.
• February is an influential point: removing that one point flips the sign of the slope.
• The calendar data have a strong positive linear relationship.

The penguin study observed birds in the field; it did not assign flipper lengths. The test of the slope of body mass on flipper length gives a p-value far below 0.001. Which conclusion does the study support?

Select one.

• Growing a longer flipper causes a penguin to gain about 49.69 g per mm.
• Flipper length and body mass have a strong positive association, though species may be a lurking variable.
• Body mass causes flipper length, because the p-value is so small.
• The small p-value proves that flipper length is the only cause of body mass.