2. Using summary statistics in question 6, find the 98% confidence interval for the difference of mean fabricstrengths for the two types using a pooled interval. Interpret your answer. 6. The maintenance cost (y in dollars) and the age of (x in years) of nine cash registers in a department storeis given in the following table.sum71354292181234012635867251Xya. Complete the above table and find the sum of each row.b. Draw the scatter plot. Is there a linear trend?200150с.Find the S, Sw, and Swy.у 1005010d. Find the Pearson's correlation coefficient.

Standard deviation for sample X: Standard Deviation, s: 2.121 Count, N: 9 Sum, Σx: 36 Mean, x̄: 4 Variance, s2: 4.5 s = √4.5 = 2.1213203435596 Standard deviation for sample Y : Standard Deviation, s: 50.316774318092 Count, N: 9 Sum, Σx: 706 Mean, x̄: 78.444 Variance, s2: 2531.777 s = √2531.777 = 50.316 The Data provided is : X (Age) Y (Cost of maintenance) 6 92 7 181 1 23 3 40 6 126 2 35 5 86 4 72 2 51 Total 36 706 Mean 4 78.444 St Dev 2.121 50.316We need to construct the 98%98\% confidence interval for the difference between the population means μ1−μ2\mu_1 - \mu_2, for the case that the population standard deviations are not known. The following information has been provided about each of the samples: Sample Mean 1 (Xˉ1)(\bar X_1) = 44 Sample Standard Deviation 1 (s1)(s_1) = 2.1212.121 Sample Size 1 (N1)(N_1) = 99 Sample Mean 2 (Xˉ2)(\bar X_2)= 78.44478.444 Sample Standard Deviation 2 (s2)(s_2) = 50.13650.136 Sample Size 2 (N2)(N_2) = 99 Based on the information provided, we assume that the population variances are equal, so then the number of degrees of freedom are df=n1+n2−2=9+9−2=16df = n_1 + n_2 -2 = 9 + 9 - 2 = 16 The critical value for α=0.02\alpha = 0.02 and df=16df = 16 degrees of freedom is tc=t1−α/2;n−1=2.583t_c = t_{1-\alpha/2; n-1} = 2.583 The corresponding confidence interval is computed as shown below: Since the population variances are assumed to be equal, we need to compute the pooled standard deviation, as follows: sp=(n1−1)s12+(n2−1)s22n1+n2−2=(9−1)×2.1212+(9−1)×50.13629+9−2=35.483 \begin{array}{ccl} s_p & = & \displaystyle \sqrt{ \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2} } \\ \\ & = & \displaystyle \sqrt{ \frac{(9-1)\times 2.121^2 + (9-1) \times 50.136^2}{9 + 9 - 2} } \\ \\ & = & 35.483 \end{array} Since we assume that the population variances are equal, the standard error is computed as follows: se=sp1n1+1n2=35.483×19+19=16.727 \begin{array}{ccl} se & = & \displaystyle s_p \sqrt{ \frac{1}{n_1} + \frac{1}{n_2} } \\ \\ & = & \displaystyle 35.483 \times \sqrt{ \frac{1}{9} + \frac{1}{9}} \\ \\ & = & 16.727 \end{array} Now, we finally compute the confidence interval: CI=(Xˉ1−Xˉ2−tc×se,Xˉ1−Xˉ2+tc×se)=(4−78.444−2.583×16.727,4−78.444+2.583×16.727)=(−117.658,−31.23) \begin{array}{ccl} CI & = & \displaystyle \left( \bar X_1 - \bar X_2 - t_c \times se, \bar X_1 - \bar X_2 + t_c \times se \right) \\ \\ & = & \displaystyle \left( 4 - 78.444 - 2.583 \times 16.727, 4 - 78.444 + 2.583 \times 16.727 \right) \\ \\ & = & (-117.658, -31.23) \end{array} Therefore, based on the data provided, the 98%98\%98% confidence interval for the difference between the population means is −117.658<μ1−μ2<−31.23, which indicates that we are 98%98\% confident that the true difference between population means is contained by the interval (−117.658,−31.23) (-117.658, -31.23)

Answered: 2. Using summary statistics in question…

Math

Statistics

2. Using summary statistics in question 6, find the 98% confidence interval for the difference of mean fabric strengths for the two types using a pooled interval. Interpret your answer.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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