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In: Statistics and Probability

A random sample of 28 shoppers spent an average of $21.53 per visit at Barnes &...

A random sample of 28 shoppers spent an average of $21.53 per visit at Barnes & Noble with a standard deviation of $2.31. Find the 95% confidence interval of the true mean cost per visit.

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Expert Solution

Solution :

Given that,

= 21.53

s =2.31

n = Degrees of freedom = df = n - 1 = 28- 1 = 27

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2= 0.05 / 2 = 0.025

t /2,df = t0.025,27 = 2.052 ( using student t table)

Margin of error = E = t/2,df * (s /n)

= 2.052* (2.31 / 28)

= 0.896

The 95% confidence interval mean is,

- E < < + E

21.53 -0.896 < < 21.53+ 0.896

20.634 < < 22.426

orchestra answered 2 years ago
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