Answer to Question #347218 in Statistics and Probability for astrid

Question

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

Answer 1

Sample mean 

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

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

Sample variance

$$s^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n-1}=47.072727272727$$ $$s=\sqrt{47.072727272727}\approx6.861$$

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

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{X}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{X}+t_c\times\dfrac{s}{\sqrt{n}})$$

$$=(84.515-2.2281\times\dfrac{6.861}{\sqrt{11}},$$

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

$$=(79.906,89.124)$$

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