Question #333676

A random sample of 10 chocolate energy bars of a certain brand has, on average, 230 calories per bar, with a standard deviation of 15 calories. Construct a 99% confidence interval for the true mean calorie content of this brand of energy bar. Assume that the distribution of the calorie content is approximately normal.

1
Expert's answer
2022-04-29T08:55:22-0400

The critical value for α=0.01\alpha = 0.01 and df=n1=9df = n-1 = 9 degrees of freedom is tc=z1α/2;n1=3.249823.t_c = z_{1-\alpha/2; n-1} = 3.249823.

The corresponding confidence interval is computed as shown below:


CI=(Xˉtc×sn,Xˉ+tc×sn)CI=(\bar{X}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{X}+t_c\times\dfrac{s}{\sqrt{n}})

=(2303.249823×1510,=(230-3.249823\times\dfrac{15}{\sqrt{10}},

230+3.249823×1510)230+3.249823\times\dfrac{15}{\sqrt{10}})

=(214.5847,245.4153)=(214.5847,245.4153)

Therefore, based on the data provided, the 99% confidence interval for the population mean is 214.5847<μ<245.4153,214.5847<\mu<245.4153, which indicates that we are 99% confident that the true population mean μμ  is contained by the interval (214.5847,245.4153).(214.5847,245.4153).



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