Question

3. The data in the table gives the posted speed limits on a sample of roads and the measured speeds of a car on each road. All speeds are in mph Posted 45 50 25 25 30 55 70 70 35 30 45 65 55 Actual 47 47 32 26 38 46 82 65 42 28 45 71 63 3 pts) a. Sketch a scatterplot of the data. You may use your calculator. 9 pts) b. Use your calculator and procedures from this class to test the claim that the data is significantly linearly correlated. Use 0.01 significance.

Answer 1

Construct the scatter plot: Instructions: 1. Import the data in Excel sheet. 2. Go to insert charts → scatter plot. 3. Choose data and click on scatter with markers. 4. Click ok The required output is shown below: Scatter plot y = 0.973 x + 3.712 R2 = 0.871 Y 20 30 40 50 60 70 80

The above scatter plot, it is clear that data points are in increasing direction and hence there is a positive relationship exists between the two variables. b) The null and alternative hypothesis is, H :p=0 H:p=0 Level of significance, a = 0.01 XY 2115 2350 Y Y 2209 2209 800 1024 676 1444 650 1140 2530 5740 4550 1470 840 2025 4615 3465 XX 2025 2500 625 625 900 3025 4900 4900 1225 900 2025 4225 3025 2116 6724 4225 1764 784 2025 5041 3969 Aci Go

From the table, { x = 600, Y = 632, x. Y = 32290, x² = 30900, < y2 = 34210 Use the following formula to work out the correlation coefficient. n.EXY - [X. SY V[n {X? - (EX)*] - [n Y? – (EY)?) 13 - 32290 – 600 - 632 0.9334 V (13 - 30900 – 600] - [13 - 34210 - 6322 The test statistic for the correlation coefficient is, r=0.9334 The degree of freedom is, df =n-2 = 13 -2 = 11 Critical value: The critical value r, for a significance level of a = 0.01 for two tailed test is r = 0.684 The rejection region is, |-|>r, = 0.684 Since the test statistic r = 0.9334>r, = 0.684, so we reject the null hypothesis at 1% level of significance Conclusion: it can be conclude that the data is significantly linearly correlated.