Question #347218

The lengths (in minutes) of a random selection of popular children’s animated films are

listed below. Estimate the true mean length of all children’s animated films with 95%

confidence.

90 84 83 91 75 88 78 96 78 79 77


1
Expert's answer
2022-06-03T12:38:31-0400

Sample mean 


xˉ=111(90+84+83+91+75+88\bar{x}=\dfrac{1}{11}(90+84+ 83+ 91 +75+ 88




+78+96+78+79+77)=91911=84.515+78 +96+ 78 +79+ 77)=\dfrac{919}{11}=84.515

Sample variance


s2=Σ(xixˉ)2n1=47.072727272727s^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n-1}=47.072727272727s=47.0727272727276.861s=\sqrt{47.072727272727}\approx6.861


The critical value for α=0.05\alpha = 0.05 and df=n1=10df = n-1 = 10 degrees of freedom is tc=z1α/2;n1=2.2281t_c = z_{1-\alpha/2; n-1} = 2.2281

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}})

=(84.5152.2281×6.86111,=(84.515-2.2281\times\dfrac{6.861}{\sqrt{11}},

84.515+2.2281×6.86111)84.515+2.2281\times\dfrac{6.861}{\sqrt{11}})

=(79.906,89.124)=(79.906,89.124)

Therefore, based on the data provided, the 95 confidence interval for the population mean is 79.906<μ<89.124,79.906 < \mu < 89.124, which indicates that we are 95% confident that the true population mean μ\mu is contained by the interval (79.906,89.124).(79.906, 89.124).



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