Question

Consider a sample of 30 college students from Victoria. Based on
this sample, it was found that the average amount spent on
beverages during the Halloween weeekend was $40. The population
standard deviation is $6.50. Assume the following: the population
of college students is very large relative to the sample size, and
the population is normally distributed.

What is the upper confidence limit for an 88% confidence interval
around the population mean?

Express your solution rounded to two decimal places.

Answer #1

Solution :

Given that,

= 40

= 6.50

n = 30

At 88% confidence level the z is ,

= 1 - 88% = 1 - 0.88 = 0.12

=0.12

Z_{}= Z_{0.12} = 1.175

Margin of error = E = Z_{/2}* ( /n)

= 1.175* (6.50 / 30)

= 1.39

At 88% upper confidence interval of the population mean is,

- E < < + E

40-1.39 < < 40+1.39

38.61< < 41.39

(38.61,41.39 )

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