Math  /  Data & Statistics

QuestionListed below are the numbers of cricket chirps in 1 minute and the corresponding temperatures in F{ }^{\circ} \mathrm{F}. Find the regression equation, letting chirps in 1 minute be the independent ( xx ) 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 & 973 & 752 & 1048 & 973 & 848 & 1071 & 846 & 1128 \\ \hline Temperature ( F{ }^{\circ} \mathrm{F} ) & 77.1 & 66 & 86.6 & 83.5 & 73.5 & 85.8 & 76.5 & 83.9 \end{tabular}
The regression equation is y^=\hat{y}= \square ++ \square xx. (Round the yy-intercept to one decimal place as needed. Round the slope to four decimal places as needed.)

Studdy Solution

STEP 1

1. The relationship between the number of cricket chirps and temperature can be modeled using linear regression.
2. The independent variable x x is the number of chirps in 1 minute.
3. The dependent variable y y is the temperature in degrees Fahrenheit.
4. We are using a significance level of 0.05 to evaluate the regression model.

STEP 2

1. Calculate the regression equation.
2. Use the regression equation to predict the temperature for 3000 chirps.
3. Evaluate the prediction and discuss any potential issues.

STEP 3

Calculate the means of x x and y y .
xˉ=973+752+1048+973+848+1071+846+11288 \bar{x} = \frac{973 + 752 + 1048 + 973 + 848 + 1071 + 846 + 1128}{8}
yˉ=77.1+66+86.6+83.5+73.5+85.8+76.5+83.98 \bar{y} = \frac{77.1 + 66 + 86.6 + 83.5 + 73.5 + 85.8 + 76.5 + 83.9}{8}

STEP 4

Calculate the slope b1 b_1 using the formula:
b1=(xixˉ)(yiyˉ)(xixˉ)2 b_1 = \frac{\sum{(x_i - \bar{x})(y_i - \bar{y})}}{\sum{(x_i - \bar{x})^2}}

STEP 5

Calculate the y-intercept b0 b_0 using the formula:
b0=yˉb1xˉ b_0 = \bar{y} - b_1 \bar{x}

STEP 6

Substitute the values of b0 b_0 and b1 b_1 into the regression equation:
y^=b0+b1x \hat{y} = b_0 + b_1 x

STEP 7

Predict the temperature when the number of chirps is 3000:
y^=b0+b1×3000 \hat{y} = b_0 + b_1 \times 3000

STEP 8

Discuss the prediction:
- Evaluate if the prediction is reasonable given the range of data. - Discuss any extrapolation concerns since 3000 chirps is outside the observed data range.
The regression equation is:
y^=b0+b1x \hat{y} = b_0 + b_1 x
The predicted temperature for 3000 chirps is:
y^=b0+b1×3000 \hat{y} = b_0 + b_1 \times 3000
Note: The actual numerical values for b0 b_0 and b1 b_1 need to be calculated using the provided data.

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