The average prices for a product in ten stores in a city are

The average prices for a product in ten stores in a city are shown below.

$1.99, $1.85, $1.25, $2.00, $1.99, $1.76, $2.50, $1.85, $2.75, $2.85

Test the hypothesis that the average price is not equal to $1.85. Use level of significance a = 0.05.

Find the 95% confidence interval estimate for the average price for this product. Assume normality

sample size =n =10

sample mean =2.079

sample standard deviation =0.4873842

Let mu be the population mean

The test hypothesis:

Ho: mu =1.85 (i.e null hypothesis)

Ha: mu not equal to 1.85 (i.e. alternative hypothesis)

The test statistic is

t=(xbar-mu)/(s/vn)

=(2.079-1.85)/(0.4873842/sqrt(10))

=1.49

The degree of freedom =n-1=10-1=9

It is a two-tailed test.

Given a=0.05, the critical values are t(0.025, df=9) =-2.26 or 2.26 (from student t table)

The rejection regions are if t<-2.26 or t>2.26, we reject the null hypothesis.

Since t=1.49 is between -2.26 and 2.26, we do not reject Ho.

So we can not conclude that the average price is not equal to $1.85

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So the lower bound is

xbar-t*s/vn= 2.079-2.26*(0.4873842/sqrt(10)) =1.730679

So the upper bound is

xbar +t*s/vn=2.079+2.26*(0.4873842/sqrt(10))=2.427321