Question

In: Statistics and Probability

Consumer Reports tested 11 brands of vanilla yogurt and found these numbers of calories per serving:...

Consumer Reports tested 11 brands of vanilla yogurt and found these numbers of calories per serving:

130, 160, 150, 120, 90, 110, 170, 140, 110, 130, 90

If we want to create a 95% confidence interval for the average calorie content of vanilla yogurt, what would be the standard error of the sample average? (Round to two decimal places.)

Solutions

Expert Solution

Solution:

x	x2
130	16900
160	25600
150	22500
120	14400
90	8100
110	12100
170	28900
140	19600
110	12100
130	16900
90	8100
∑x=1400	∑x2=185200

Mean ˉx=∑xn

=130+160+150+120+90+110+170+140+110+130+90/11

=1400/11

=127.2727
Sample Standard deviation S=√∑x2-(∑x)2nn-1

=√185200-(1400)211/10

=√185200-178181.8182/10

=√7018.1818/10

=√701.8182

=26.4919

The standard error = (s / $\sqrt$ n)

= (26.49 / $\sqrt$ 11)

=7.987

The standard error = 7.99

Degrees of freedom = df = n - 1 = 11 - 1 = 10

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t /2,df = t0.025,10 =2.228

Margin of error = E = t/2,df * (s / $\sqrt$ n)

= 2.228* (26.49 / $\sqrt$ 11)

= 26.49

Margin of error = 26.49

The 95% confidence interval estimate of the population mean is,

- E < < + E

127.27 - 26.49 < < 127.27 + 26.49

109.47 < < 145.07

(109.47 , 145.07 )

orchestra answered 1 year ago

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