Question

PLEASE IF POSSIBLE USE PH STAT

Sales of Body	Price of Lens	Gender	Sales of Lens
155	$700	1	122
101	650	1	120
157	725	0	135
180	575	1	95
150	600	0	100
201	750	0	174
99	560	1	118
137	500	0	130
155	675	1	128
165	550	1	166
152	725	0	131
127	750	1	102
217	565	0	165
186	670	0	154
176	600	1	97
123	585	0	129
109	645	0	98
90	575	1	105
176	660	0	120
129	650	1	105

d)        Find a 99% confidence interval for the mean sales of the lens when sales of the camera body are 150 and the price of the lens is $700. Provide an explanation of the meaning of this interval.

e)        Find a 99% prediction interval for the sales of the lens when sales of the camera body are 150 and the price of the lens is $700. Explain its meaning relative to the problem. Compare the width of this interval to that of the interval calculated in the previous part of the problem. Give a reason for any difference in the width of this interval and the one you found in the previous part of the problem.

Answer 1

Sol:

L - 1 1575 631 859 X New Microsoft Excel Worksheet - Microsoft Excel (Product Activation Failed) 0 Х File Home Insert Page Layout Formulas Data Review View Developer OP x * Cut Σ AutoSum και Calibri - 11 A Wrap Text General HH ECopy * Fill Paste BI U H- А Merge & Center +.0 .00 $ - % Conditional Format Cell Insert Delete Format .00.0 Format Painter Sort & Find & Formatting as Table Styles Clear Filter - Select - Clipboard Font Alignment Number Styles Cells Editing - Ofic B A - А B C D E F G H 1 K M N 0 Р Q R S T U V 1 X Y PRED Confidence Prediction 2 Sales of Body price of lens Y Interval Interval use trend use linest 95% prediction 95% confidence 3 155 $ 700.00 637.209 interval: interval: 4 2 101 $ 650.00 621.159 3 Degree of freedom =n-2 5 3 725.00 637.803 95% of the Y-values to there is a 95% 6 4 A 1800 575.00 644.64 07.09 probability that the 7 5 be found for a certain 150ls 5 600 on 1 635 723 154 000.00 023.722 Slope 0.297220389 591.14 Interc. true best-fit line for the 8 6 2011 s X-value will be within 750.00 2017 650 881 130.00 030.00 SE slope 0.482821074 73.9061 SE interc. 7 9 7 99 $ population lies within 560.00 300.00 620.565 RSQ the interval range 0.020618829 73.3916 SEY 10 10 the confidence 8 1371 5 1575 o 500.00 F 0.378952487 around the linear 18 d.f. 11 9 155 $ 675.00 637.209 SS regr 2041.161021 96953.8 SS res interval. regression line. 12 10 165 $ 550.00 640.181 13 11 1521 S 725.00 636.317 SS res is unexplained variation n 20 14 12 127 $ 750.00 628.887 SS regr is explained variation nes Xm Average 15 13 217 $ 565.00 655.637 d.f.=n(data)-k(variables) SSXX use formula=devsq 16 14 186 $ 670.00 646.423 t.95 Use student t test 17 15 176) $ 600.00 643.451 SE 18 16 123 S 585.00 627.698 19 17 109 $ 645.00 623.537 20 18 90 $ 575.00 617.89 21 19 176 $ 660.00 643.451 22 201 129 $ 650.00 629.481 23 24 i (X-X) 99% prediction interval 1. (X - x) 25 X = Y + 1 SE + X = Y +1,SE, 1+ + n n 26 SS. =Y2tVal"SE'SQRT{1+1/count+(X-Xmean)^2/SSxx) XX 27 28 Activate Windows 29 Go to Settings to activate Windows. HE Sheet1 Sheet2 Sheet3 Ready BOI 80% + 12:26 AM Type here to search ge Ac) 90 7/30/2020 SS SE = Vn-2 99% confidence interval =YMVaI'SE'SQRT(1/count+(X-Xmean)^255xx) 9 SS, confidence prediction o

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