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A transport department hoped that they could keep costs down by measuring the we

ID: 3044280 • Letter: A

Question

A transport department hoped that they could keep costs down by measuring the weights of big trucks without actually stopping the vehicles and instead using a newly developed "weight-in-motion" scale. To see if the new device was accurate, they conducted a calibration test. They weighed several trucks when stopped (static weight) assuming that this weight was correct Then they weighed the trucks again while they were moving to see how well the new scale could estimate the actual weight. Use the data in the table below to answer parts (a) through () Click the icon to view the data table a) Make a scatterplot for these data. Choose the correct graph below OA. O B. OD. 45 45 25 25 45 b) Describe the direction, form, and strength of the plot. Choose the correct answer below. 1 Data Table o A. There is no evidence of a linear association. 0 B. There is a strong negative linear association. ° C. There is a weak negativ 0 D. There is a weak positive linear association. e linear association. Weight of a Truck sands of Pounds) E. There is a strong positive linear association. Weight-in-Motion Static Weight 25.8 29.8 39.6 24.7 31.3 29.6 29.8 31.2 35.6 40.7 24.7 28.8 40.5 23 32.4 27.8 29.6 31.3 37 34.9 c) Write a few sentences telling what the plot says about the data. Choose the correct answer be 0 A. As the static weight increases it tends to cause the weight-in-motion to increase B. The new scale predicts the static weight fairly well. It may be possible to predict the static ° C. As the static weight increases it tends to cause the weight-in-motion to decrease 0 D. The new scale does not predict the static weight very well. It is probably not possible to pred h to be useful. d) Find the correlation. The correlation is. (Round to three decimal places as needed.)

Explanation / Answer

The statistical software output for this problem is:

Correlation between Weight - in - Motion and Static Weight is:
0.92409822

Hence,

d) Correlation = 0.924

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