Question #269024

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


1
Expert's answer
2021-11-23T14:24:35-0500

a)

sample mean:

x=xi/n=95+173+129+95+75+94+116+100+859=106.9\overline{x}=\sum x_i/n=\frac{95+173+129+95+75+94+116+100+85}{9}=106.9


sample standard deviation:

σ=(xix)2n1=(95106.9)2+(173106.9)2+(129106.9)2+(95106.9)2+(75106.9)2+(94106.9)2+8\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}}


+(116106.9)2+(100106.9)2+(85106.9)28=29.4\sqrt{\frac{+(116-106.9)^2+(100-106.9)^2+(85-106.9)^2}{8}}=29.4


b)

t=xμσ/n=106.9μ29.4/9t=\frac{\overline{x}-\mu}{\sigma/\sqrt n}=\frac{106.9-\mu}{29.4/\sqrt 9}

df=91=8df=9-1=8

critical values for 90% confidence interval:

t=±1.860t=\pm 1.860

then:


1.86<106.9μ29.4/9<1.86-1.86<\frac{106.9-\mu}{29.4/\sqrt 9}<1.86


18.23<106.9μ<18.23-18.23<106.9-\mu<18.23


88.67<μ<125.1388.67<\mu<125.13

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