contestada

Measurements of the sodium content in samples of two brands of chocolate bar yield the following results (in grams):

Brand A: 34.36 31.26 37.36 28.52 33.14 32.74 34.34 34.33 29.95

Brand B: 41.08 38.22 39.59 38.82 36.24 37.73 35.03 39.22 34.13 34.33 34.98 29.64 40.60

Let μX represent the population mean for Brand B and let μY represent the population mean for Brand A.

Find a 98% confidence interval for the difference μX−μY. Round down the degrees of freedom to the nearest integer and round the answers to three decimal places.

Respuesta :

Answer:

98% confidence interval for the difference μX−μY = [ 0.697 , 7.303 ] .

Step-by-step explanation:

We are give the data of Measurements of the sodium content in samples of two brands of chocolate bar (in grams) below;

Brand A : 34.36, 31.26, 37.36, 28.52, 33.14, 32.74, 34.34, 34.33, 29.95

Brand B : 41.08, 38.22, 39.59, 38.82, 36.24, 37.73, 35.03, 39.22, 34.13, 34.33, 34.98, 29.64, 40.60

Also, [tex]\mu_X[/tex] represent the population mean for Brand B and let [tex]\mu_Y[/tex] represent the population mean for Brand A.

Since, we know nothing about the population standard deviation so the pivotal quantity used here for finding confidence interval is;

        P.Q. = [tex]\frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }[/tex] ~ [tex]t_n__1+n_2-2[/tex]

where, [tex]Xbar[/tex] = Sample mean for Brand B data = 36.9

            [tex]Ybar[/tex] = Sample mean for Brand A data = 32.9

              [tex]n_1[/tex]  = Sample size for Brand B data = 13

              [tex]n_2[/tex] = Sample size for Brand A data = 9

              [tex]s_p = \sqrt{\frac{(n_1-1)s_X^{2}+(n_2-1)s_Y^{2} }{n_1+n_2-2} }[/tex] = [tex]\sqrt{\frac{(13-1)*10.4+(9-1)*7.1 }{13+9-2} }[/tex] = 3.013

Here, [tex]s^{2}_X[/tex] and [tex]s^{2} _Y[/tex] are sample variance of Brand B and Brand A data respectively.

So, 98% confidence interval for the difference μX−μY is given by;

P(-2.528 < [tex]t_2_0[/tex] < 2.528) = 0.98

P(-2.528 < [tex]\frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }[/tex] < 2.528) = 0.98

P(-2.528 * [tex]s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}[/tex] < [tex](Xbar -Ybar) -(\mu_X-\mu_Y)[/tex] < 2.528 * [tex]s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}[/tex] ) = 0.98

P( (Xbar - Ybar) - 2.528 * [tex]s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}[/tex] < [tex](\mu_X-\mu_Y)[/tex] < (Xbar - Ybar) + 2.528 * [tex]s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}[/tex] ) = 0.98

98% Confidence interval for μX−μY =

[ (Xbar - Ybar) - 2.528 * [tex]s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}[/tex] , (Xbar - Ybar) + 2.528 * [tex]s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}[/tex] ]

[ [tex](36.9 - 32.9)-2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9}[/tex] , [tex](36.9 - 32.9)+2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9}[/tex] ]

[ 0.697 , 7.303 ]

Therefore, 98% confidence interval for the difference μX−μY is [ 0.697 , 7.303 ] .

                     

Otras preguntas