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Be sure to submit the HWK 11 Auto grade Quiz which will give you ~20 of your 40 accuracy points.
50 points total: 40 points accuracy, and 10 points completion
Exercise 1 Reconsider the relationship between city air particulate and rates of childhood asthma first discussed in Homework 10. We sample 15 cities for particulate measured in parts-per-million (ppm) of large particulate matter and for the rate of childhood asthma measured in percents.
| variable: | size | mean | variance |
|---|---|---|---|
| x | 15 | 11.42 | 13.05029 |
| y | 15 | 14.51333 | 2.635524 |
- Suppose we sample a new city whose particulate is 13 ppm. If reasonable, create a 95% interval for the predicted rate of childhood asthma in this city. If not reasonable, explain why.
particulate <- c(11.6, 15.9, 15.7, 7.9, 6.3, 13.7, 13.1, 10.8,
6.0, 7.6, 14.8, 7.4, 16.2, 13.1, 11.2)
asthma <- c(14.5, 16.6, 16.5, 12.6, 12.0, 15.8, 15.1, 14.2,
12.2, 13.1, 16.0, 12.9, 16.4, 15.4, 14.4)
asthma_model = lm(asthma ~ particulate)
summary(asthma_model)##
## Call:
## lm(formula = asthma ~ particulate)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.34226 -0.12842 -0.01514 0.12130 0.29164
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 9.41626 0.17344 54.29 < 2e-16 ***
## particulate 0.44633 0.01452 30.73 1.6e-13 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.1963 on 13 degrees of freedom
## Multiple R-squared: 0.9864, Adjusted R-squared: 0.9854
## F-statistic: 944.4 on 1 and 13 DF, p-value: 1.595e-13qt(0.975, df = 13)## [1] 2.160369new_model = data.frame(particulate = 13 )
predict(asthma_model, newdata = new_model, interval = "prediction")## fit lwr upr
## 1 15.21853 14.77771 15.65936nnew_model = data.frame(particulate = 10)
predict(asthma_model, newdata = nnew_model, interval = "confidence")## fit lwr upr
## 1 13.87955 13.76132 13.99777
- Create a 95% confidence interval for the average rate of childhood asthma among all cities with 10 ppm of large particulate. Is this confidence interval wider or narrower than a 95% prediction interval for the rate of childhood asthma in the next city with 10 ppm of large particulate? Explain why you know this without having to build the PI.
- If reasonable, create a 95% interval for the predicted rate of childhood asthma in the next city sampled that has 3 ppm of large particulate. If this is not reasonable, explain why.
Exercise 2. In the paper “Artificial Trees as a Cavity Substrate for Woodpeckers”, scientists provided polystyrene cylinders as an alternative roost. The paper related values of x = ambient temperature (C) and y = cavity depth (cm). A scatterplot in the paper showed a strong linear relationship between x and y. The summary values for x and y are given below:
| Variable | Size | Mean | Variance |
|---|---|---|---|
| Temp (x) | 12 | 10.92 | 137.17 |
| Depth (y) | 12 | 16.36 | 21.28 |
A least-squares linear model for (Depth ~ Temp) was fit, and the intercept was estimated to be 20.12506 with standard error 0.94023. The slope was estimated to be -0.34504 with standard error 0.06008. The MSE for the model is \(2.334^2\).
- Determine the sample correlation (r) from the summaries given.
2*pt(-5.743, df = 10)## [1] 0.0001869496pt(-5.753, df = 10,)## [1] 9.22048e-05pt(-5.753, df = 10, lower.tail = FALSE)## [1] 0.99990782*pt(2.579, df = 10, lower.tail = FALSE)## [1] 0.02746338qt(0.99, df = 10)## [1] 2.763769qt(0.975, df = 10)## [1] 2.228139
- Give the linear regression model with least squares estimates for \(\beta_0\) and \(\beta_1\) relating ambient temperature (x) and hole depth (y).
Determine test statistics and p values for the tests in parts c-f:
- \(H_0: \beta_1=0\) vs \(H_A: \beta_1 \ne 0\)
- \(H_0: \beta_1 \ge 0\) vs \(H_A: \beta_1 < 0\)
- \(H_0: \beta_1 \le 0\) vs \(H_A: \beta_1 > 0\)
- \(H_0: \beta_1=-0.5\) vs \(H_A: \beta_1 \ne -0.5\)
- Compute and interpret a 98% confidence interval for the slope of the regression line \(\beta_1\).
- Construct a 95% prediction interval for the cavity depth of the next hole when ambient temperature is 1 degree Celsius (this temperature value is within the range of those in the original study).