Circumference Circumference 1856.521739 225.8778297 1500 54 39130435 Mean 2000 M

Question

Circumference Circumference 1856.521739 225.8778297 1500 54 39130435 Mean 2000 Mean 1500 Standard Error 1700 Median 500 Mode 1000 Standard Deviation 1500 Sample Variance 1400 Kurtosis 1500 Skewness 2500 Range 2.651525872 Standard Error 52 Median 50 Mode 12.71627136 Standard Deviat 161.7035573 Sample Varianc 1083.272016 1173478.261 2.040961579 1.370824368 4100 0.725868744 Kurtosis 0.603170931 Skewness 50 Range 33 Minimum 83 Maximum 500 Minimum 000 Maximum 2000 Sum 2500 Count 251 Sum 0.968115661 1500 1500 1600 SUMMARY OUTPUT R Square Adjusted R Squar Standard Error 0.968115661 0.937247934 0.93425974 ANOVA 4 20313E-14 24196481.65 1620040.087 21 77144 78605 Total 259.82 4 656741234 10 11927786 1.57353E-09 71017601 420313E-14 3169 35365 2088.89 72.78748333 92.155 F5 F8 3

Explanation / Answer

1c) From the given output

The regression equation is

Weight = -2629.22 + 82.47*circumference

If circumference is 50

then Weitht = -2629.22 + 82.47 * 50

= 1494.28

14) In this output, F value is two large and p value is too small

It does mean Regression model is significant and useful to predict accurate weight of pumkin.

15) Here Adjusted R2 = 0.9342

It means there is 93.42% of variation is accounted for weight of pumpkin on circumference.

and Correlation coefficient 'r' = 0.9681

It is very close to +1, It means strong positive association between weight of pumpkin anc circumference.

17) Yes we can predict of estimate weight of a pumpkin,since standard error of regression equation is too small. then T stat is higher value. It means regression model is significant.

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