In: Statistics and Probability
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.) |
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 /n)
= (26.49 / 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 /n)
= 2.228* (26.49 / 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 )