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Topic 7 Homework (Nonadaptive) Question 9 of 16 (1 point)|Question Attempt: 2 of Unlimited

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Question 9 The managers of an outdoor coffee stand in Coast City are examining the relationship between coffee sales and daily temperature. They have bivariate data detailing the stand's coffee sales (denoted by y, in dollars) and the maximum temperature (denoted by x, in degrees Fahrenheit) for each of 31 randomly selected days during the past year. The least-squares regression equation computed from their data is =y−2525.6810.99x. Tomorrow's forecasted high is 63 degrees Fahrenheit. The managers have used the regression equation to predict the stand's coffee sales for tomorrow. They now are interested in both a prediction interval for tomorrow's coffee sales and a confidence interval for the mean coffee sales on days on which the maximum temperature is 63 degrees Fahrenheit. They have computed the following for their data.

mean square error ≈MSE 2421.24 ≈+131−63x2Σ=i131−xix2 0.0401, where x1, x2, ..., x31 denote temperatures in the sample, and x denotes their mean Based on this information, and assuming that the regression assumptions hold, answer the questions below.

(If necessary, consult a list of formulas.)

(a)What is the 90% confidence interval for the mean coffee sales (in dollars) when the maximum temperature is 63 degrees Fahrenheit? (Carry your intermediate computations to at least four decimal places, and round your answer to at least one decimal place.)

Lower limit:

Upper limit:

(b)Consider (but do not actually compute) the 90% prediction interval for an individual value for coffee sales when the maximum temperature is 63 degrees Fahrenheit. How would this prediction interval compare to the confidence interval computed above (assuming that both intervals are computed from the same sample data)?

The prediction interval would have the same center as, but would be narrower than, the confidence interval. The prediction interval would be identical to the confidence interval. The prediction interval would have the same center as, but would be wider than, the confidence interval. The prediction interval would be positioned to the right of the confidence interval. The prediction interval would be positioned to the left of the confidence interval.

(c)For the maximum temperature values in this sample, 52 degrees Fahrenheit is more extreme than 63 degrees Fahrenheit is, that is, 52 is farther from the sample mean maximum temperature than 63 is. How would the 90% confidence interval for the mean coffee sales when the maximum temperature is 63 degrees Fahrenheit compare to the 90% confidence interval for the mean coffee sales when the maximum temperature is 52 degrees Fahrenheit?

The interval computed from a maximum temperature of 63 degrees Fahrenheit would be narrower and have a different center.

The intervals would be identical.

The interval computed from a maximum temperature of 63 degrees Fahrenheit would be wider and have a different center.

The interval computed from a maximum temperature of 63 degrees Fahrenheit would be wider but have the same center.

The interval computed from a maximum temperature of 63 degrees Fahrenheit would be narrower but have the same center.

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Solution

To solve this problem, let's analyze each part step by step, starting with the given least-squares regression equation.

The regression equation provided is: y^=−2525.68+10.99x\hat{y} = -2525.68 + 10.99x where:

  • yy represents coffee sales (in dollars),
  • xx represents the temperature in degrees Fahrenheit.

Tomorrow’s forecasted temperature is 63°F, so we will use this value to find the predicted coffee sales, confidence interval, and address each question part-by-part.

Part (a): 90% Confidence Interval for the Mean Coffee Sales at 63°F

  1. Calculate the predicted mean coffee sales when x=63x = 63: y^=−2525.68+10.99×63\hat{y} = -2525.68 + 10.99 \times 63

  2. Standard Error of the Mean Prediction: The standard error (SE) for predicting the mean is given by: SE=MSE⋅(1n+(63−xˉ)2⋅∑i=131(xi−xˉ)2)\text{SE} = \sqrt{\text{MSE} \cdot \left( \frac{1}{n} + \left(63 - \bar{x}\right)^2 \cdot \sum_{i=1}^{31} \left(x_i - \bar{x}\right)^2 \right)}

    • Given:
      • Mean square error, MSE=2421.24\text{MSE} = 2421.24,
      • ∑i=131(xi−xˉ)2≈0.0401\sum_{i=1}^{31} \left(x_i - \bar{x}\right)^2 \approx 0.0401,
    • Substituting these values will yield the SE.

  3. 90% Confidence Interval Calculation: Use the formula: CI=y^±t⋅SE\text{CI} = \hat{y} \pm t \cdot \text{SE} where tt is the critical value from the tt-distribution for a 90% confidence level with n−2n - 2 degrees of freedom (where n=31n = 31).

Part (b): Comparison of Prediction Interval to Confidence Interval

  • A prediction interval accounts for individual variability and is typically wider than a confidence interval for the mean because it adds more uncertainty for a single observation.
  • The correct answer is:

    "The prediction interval would have the same center as, but would be wider than, the confidence interval."

Part (c): Comparison of Confidence Intervals at 63°F and 52°F

  • Because 52°F is farther from the mean temperature, the confidence interval at 52°F would include more variability, making it wider than the interval at 63°F.
  • Thus, the correct answer is:

    "The interval computed from a maximum temperature of 63 degrees Fahrenheit would be narrower and have a different center."

Would you like a detailed solution for each calculation, or assistance with another part?

Related Questions:

  1. How do we interpret the slope in a regression equation?
  2. What is the difference between a confidence interval and a prediction interval?
  3. How is the mean square error (MSE) calculated in regression analysis?
  4. Why does a prediction interval widen as it moves further from the mean?
  5. How does temperature affect the confidence interval width in this context?

Tip:

In regression, always remember that confidence intervals relate to the mean prediction, while prediction intervals consider individual predictions, inherently making them wider.

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Math Problem Analysis

Mathematical Concepts

Regression Analysis
Confidence Intervals
Prediction Intervals
Least-Squares Method

Formulas

Least-squares regression equation: y = -2525.68 + 10.99x
Standard Error (SE) of the mean: SE = sqrt(MSE * (1/n + (x - mean(x))^2 / sum(x_i - mean(x))^2))
Confidence Interval: CI = ŷ ± t * SE

Theorems

Properties of Linear Regression
T-Distribution

Suitable Grade Level

Undergraduate Statistics or High School Advanced Placement (AP) Statistics

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