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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: 1848.3

Upper limit: 1818.3

(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

Let's address each part of the question:

(a) Calculating the 90% Confidence Interval for Mean Coffee Sales

We have the following information:

  • Regression equation: y^=2525.68+10.99x\hat{y} = -2525.68 + 10.99x
  • MSE (Mean Square Error): 2421.24
  • Forecasted high (x = 63): We're given x=63x = 63 and want to predict the coffee sales.
  • Additional data: i=131(xixˉ)2n0.0401\sum_{i=1}^{31} \frac{(x_i - \bar{x})^2}{n} \approx 0.0401.
  1. Calculate Predicted Sales at x=63x = 63: y^=2525.68+10.99×63=1664.61\hat{y} = -2525.68 + 10.99 \times 63 = 1664.61

  2. Calculate Standard Error of the Mean Prediction using: SEmean=MSE×(1n+0.0401)SE_{\text{mean}} = \sqrt{\text{MSE} \times \left( \frac{1}{n} + 0.0401 \right)}

    Substituting in the values (with n=31n = 31): SEmean=2421.24×(131+0.0401)11.3SE_{\text{mean}} = \sqrt{2421.24 \times \left( \frac{1}{31} + 0.0401 \right)} \approx 11.3

  3. Find the 90% Confidence Interval: For a 90% confidence level, the critical value t0.05,29t_{0.05, 29} (for 30 degrees of freedom) is approximately 1.699. Margin of Error=t×SEmean=1.699×11.319.2\text{Margin of Error} = t \times SE_{\text{mean}} = 1.699 \times 11.3 \approx 19.2 So, the confidence interval for mean coffee sales is: Lower limit=1664.6119.21645.4\text{Lower limit} = 1664.61 - 19.2 \approx 1645.4 Upper limit=1664.61+19.21683.8\text{Upper limit} = 1664.61 + 19.2 \approx 1683.8

(b) Comparison of Prediction Interval to Confidence Interval

A prediction interval for an individual sale would be wider than a confidence interval for the mean because it must account for both the variability in individual values and the variability in estimating the mean. The answer here is:

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

(c) Comparison of Confidence Intervals at Different Temperatures (52°F vs. 63°F)

Since 52°F is farther from the mean temperature than 63°F, there is more uncertainty in predicting coffee sales at 52°F. This results in a wider confidence interval when x=52x = 52 compared to x=63x = 63. The answer is:

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


Would you like additional details on any part? Here are some related questions to consider:

  1. How would increasing the sample size affect the confidence interval width?
  2. What impact would a higher MSE have on the prediction interval?
  3. Why do we use tt-values rather than zz-values in this scenario?
  4. How is the confidence interval different from the prediction interval conceptually?
  5. How would changing the confidence level (e.g., to 95%) affect the intervals?

Tip: Confidence intervals are always narrower than prediction intervals for the same level of confidence because they estimate the mean rather than individual observations.

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Math Problem Analysis

Mathematical Concepts

Statistics
Regression Analysis
Confidence Interval
Prediction Interval

Formulas

Least-squares regression equation: ŷ = -2525.68 + 10.99x
Standard error of mean prediction: SE_mean = √(MSE × (1/n + 0.0401))
Confidence interval: CI = ŷ ± t × SE_mean

Theorems

Least-squares regression principle
Confidence interval computation
Prediction interval comparison

Suitable Grade Level

College-level (Statistics/Regression Analysis)