Answer to Question #269024 in Statistics and Probability for zaqwan

Question

Start-up costs (in thousand RM) for a random sample of candy stores are given 95 173 129 95 75 94 116 100 85 a) Verify that the mean sample of RM106.9 thousand and the sample standard deviation is RM29.4 thousand. b) Find a 90% confidence interval for the population average start-up cost for candy store franchises

Answer 1

a)

sample mean:

$\overline{x}=\sum x_i/n=\frac{95+173+129+95+75+94+116+100+85}{9}=106.9$

sample standard deviation:

$\sigma=\sqrt{\frac{(x_i-\overline{x})^2}{n-1}}=\sqrt{\frac{(95-106.9)^2+(173-106.9)^2+(129-106.9)^2+(95-106.9)^2+(75-106.9)^2+(94-106.9)^2+}{8}}$

$\sqrt{\frac{+(116-106.9)^2+(100-106.9)^2+(85-106.9)^2}{8}}=29.4$

b)

$t=\frac{\overline{x}-\mu}{\sigma/\sqrt n}=\frac{106.9-\mu}{29.4/\sqrt 9}$

$df=9-1=8$

critical values for 90% confidence interval:

$t=\pm 1.860$

then:

$-1.86<\frac{106.9-\mu}{29.4/\sqrt 9}<1.86$

$-18.23<106.9-\mu<18.23$

$88.67<\mu<125.13$