ch.12 2 O S.I.A. Best Songs V xO Syllabus for Statistics* Ch. 12 HW (1602) - Mat xMy Questions | bartleby x= 6.1 The Standard Norm xBookmarksA webassign.net/web/Student/Assignment-Responses/last?dep=25396979+ Part (b)O Part (c)Calculate the least squares line. Put the equation in the form of: ý = a + bx. (Round your answers to three decimal places.)O Part (d)Find the correlation coefficient r. (Round your answer to four decimal places.)r =Is it significant?O YesNoO Part (e)If the laundry detergent were sold in a 40-ounce size, find the estimated cost. (Use your equation from part (c). Round your answer to two decimal places.)$O Part (f)If the laundry detergent were sold in an 70-ounce size, find the estimated cost. (Use your equation from part (c). Round your answer to two decimal places.)$O Part (g)+ Part (h)O Part (i) O S.I.A. Best Songs V xOSyllabus for Statistics w x* Ch. 12 HW (1602) - Mat xMy Questions | bartleby x= 6.1 The Standard Norm x* BookmarksA webassign.net/web/Student/Assignment-Responses/last?dep=25396979DThe cost of a leading liquid laundry detergent in different sizes is given below.Size (ounces)Cost ($)163.89324.79645.6920010.49O Part (a)Using "size" as the independent variable and "cost" as the dependent variable, draw a scatter plot.Cost ($)Cost ($)81086644Size (oz)200Size (oz)2005010015050100150Cost ($)Cost ($)8.10아6.8644Size (oz)20050100150Şize (oz)20050100150

Hi! Thank you for the question, As per the honor code, we are allowed to answer three sub-parts at a time so we are answering the first three as you have not mentioned which of these you are looking for. Please re-submit the question separately for the remaining sub-parts. From the given information, Consider, the data table can be expressed as: Size Cost 16 3.89 32 4.79 64 5.69 200 10.49a. The correct scatter plot can be shown below: Since, Cost is dependent variable and size is taken to be independent variable. Hence, the correct option is a) c. The data can be expressed as also: Size =x Cost =y x-x y-y x-x2 x-xy-y 16 3.89 -62 -2.325 3844 144.15 32 4.79 -46 -1.425 2116 65.55 64 5.69 -14 -0.525 196 7.35 200 10.49 122 4.275 14884 521.55 ∑x=312 ∑y=24.86 ∑x-x2=21040 ∑x-xy-y =738.6 The regression line can be computed as: Considering, Y=α+βx whereas, α=intercept and β =slope Thus, the slope coefficient can be computed as: β=∑x-xy-y∑x-x2 =738.621040 =0.0135105 The intercept is - α=∑y-β∑xn =24.86-0.0135105×3124 =3.476844 Hence, the required regression equation is y⏞=3.477+0.035x d. The coefficient of correlation can be computed as: Size =x Cost =y x2 y2 xy 16 3.89 256 15.1321 62.24 32 4.79 1024 22.9441 153.28 64 5.69 4096 32.3761 364.16 200 10.49 40000 110.0401 2098 ∑x=78 ∑y=6.215 ∑x2=45376 ∑y2=180.4924 ∑xy=2677.68 r=n∑xy-∑x×∑yn∑x2-∑x2×n∑y2-∑y2 =4×2677.68-78×6.2154×45376-782×4×180.4924-6.2152 =0.9340 Hence, the correlation coefficient is 0.9340. The significance test of correlation coefficient can be performed as: The null and alternative hypothesis are: Ho:ρ=0H1:ρ≠0 Lets significance be 0.05 Test statistics: t=rn-21-r2 =0.9340×4-21-0.93402 =3.6971 P-value can be computed as: p-value =2×Prt>3.6971 =0.034347 Since, p-value is less than assume level of significance that is, p-value =0.034347<0,05 as a result null hypothesis gets rejected that is coefficient of correlation is significant