Questions: Listed below are the numbers of cricket chirps in 1 minute and the corresponding temperatures in °F. Find the regression equation, letting chirps in 1 minute be the independent (x) variable. Find the best predicted temperature at a time when a cricket chirps 3000 times in 1 minute, using the regression equation. What is wrong with this predicted temperature? Use a significance level of 0.05. Chirps in 1 min: 819, 1014, 811, 874, 1205, 793, 1082, 1199 Temperature (°F): 68.1, 82.6, 75.5, 70.5, 93, 71.2, 82.2, 91.8 The regression equation is ŷ = + + (Round the y-intercept to one decimal place as needed. Round the slope to four decimal places as needed.)

Listed below are the numbers of cricket chirps in 1 minute and the corresponding temperatures in °F. Find the regression equation, letting chirps in 1 minute be the independent (x) variable. Find the best predicted temperature at a time when a cricket chirps 3000 times in 1 minute, using the regression equation. What is wrong with this predicted temperature? Use a significance level of 0.05.

Chirps in 1 min: 819, 1014, 811, 874, 1205, 793, 1082, 1199
Temperature (°F): 68.1, 82.6, 75.5, 70.5, 93, 71.2, 82.2, 91.8

The regression equation is ŷ = + +
(Round the y-intercept to one decimal place as needed. Round the slope to four decimal places as needed.)

Transcript text: Part 1 of 3 Points: 0 of 1 Save Listed below are the numbers of cricket chirps in 1 minute and the corresponding temperatures in ${ }^{\circ} \mathrm{F}$. Find the regression equation, letting chirps in 1 minute be the independent ( $x$ ) variable. Find the best predicted temperature at a time when a cricket chirps 3000 times in 1 minute, using the regression equation. What is wrong with this predicted temperature? Use a significance level of 0.05 . \begin{tabular}{l|cccccccc} Chirps in 1 min & 819 & 1014 & 811 & 874 & 1205 & 793 & 1082 & 1199 \\ \hline Temperature ( $\left.{ }^{\circ} \mathrm{F}\right)$ & 68.1 & 82.6 & 75.5 & 70.5 & 93 & 71.2 & 82.2 & 91.8 \end{tabular} The regression equation is $\hat{y}=\square$ $\square$ $+$ $\square$ (Round the $y$-intercept to one decimal place as needed. Round the slope to four decimal places as needed.)

Solution

Solution Steps

Step 1: Calculate the Means

The means of the chirps ($ \bar{x} $) and temperatures ($ \bar{y} $) are calculated as follows:

\[ \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i = 974.625 \]

\[ \bar{y} = \frac{1}{n} \sum_{i=1}^{n} y_i = 79.3625 \]

Step 2: Calculate the Correlation Coefficient

The correlation coefficient ($ r $) is computed to assess the strength of the linear relationship between chirps and temperature:

\[ r = 0.9536 \]

Step 3: Calculate the Slope ($ \beta $) and Intercept ($ \alpha $)

The numerator for the slope ($ \beta $) is given by:

\[ \text{Numerator for } \beta = \sum_{i=1}^{n} x_i y_i - n \bar{x} \bar{y} = 629913.0 - 8 \cdot 974.625 \cdot 79.3625 = 11123.5875 \]

The denominator for the slope is:

\[ \text{Denominator for } \beta = \sum_{i=1}^{n} x_i^2 - n \bar{x}^2 = 7809753 - 8 \cdot 974.625^2 = 210601.875 \]

Thus, the slope ($ \beta $) is calculated as:

\[ \beta = \frac{11123.5875}{210601.875} = 0.0528 \]

The intercept ($ \alpha $) is calculated using:

\[ \alpha = \bar{y} - \beta \bar{x} = 79.3625 - 0.0528 \cdot 974.625 = 27.8847 \]

Step 4: Formulate the Regression Equation

The regression equation is expressed as:

\[ \hat{y} = 27.8847 + 0.0528x \]

Step 5: Predict Temperature for 3000 Chirps

To predict the temperature when the chirps are 3000, we substitute $ x = 3000 $ into the regression equation:

\[ \hat{y} = 27.8847 + 0.0528 \cdot 3000 = 186.2847 \]

Step 6: Discuss the Reliability of the Prediction

The predicted temperature of $ 186.2847 $ is likely unreliable because it is an extrapolation far beyond the observed data range. This means that the relationship established by the regression may not hold for values outside the range of the original data.

Final Answer

The regression equation is $ \hat{y} = 27.8847 + 0.0528x $ and the predicted temperature for 3000 chirps is $ 186.2847 $. The prediction is unreliable due to extrapolation beyond the observed data range.

\[ \boxed{\hat{y} = 27.8847 + 0.0528x, \text{ predicted temperature } = 186.2847} \]

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