Question

Suppose that Randy is an analyst for the bicyling industry and wants to estimate the asking price of used entry-level road bikes advertised online in the southeastern part of the United States. He obtains a random sample of n = 14 online advertisements of entry-level road bikes. He determines that the mean price for these 14 bikes is ¯¯¯ x = $ 714.19 and that the sample standard deviation is s = $ 184.56 . He uses this information to construct a 99% confidence interval for μ , the mean price of a used road bike. What is the lower limit of this confidence interval? Please give your answer to the nearest cent.

Answer 1

Here, we have given that,

X: price of the bike

n= Number of bikes=14

= sample mean of the price of bikes= $ 714.19

S= sample standard deviation of price of bikes= $ 184.56

The sample is a simple random sample and the population standard deviation is not known. here we are using the one-sample t-test approach to find the 99% confidence interval.

Now, we want to find the 99% confidence interval for the population mean

The formula is as follows,

Where

E=Margin of error =

Now,

Degrees of freedom = n-1 = 14-1=13

c=confidence level =0.99

=level of significance=1-c=1-0.99=0.01

and we know that confidence interval is always two-tailed

t-critical = 3.012  ( using students t table )

Now,

E=

=

=148.57

The Margin of error is 148.57

We get the 99% confidence interval for the population mean

The lower limit of this confidence interval is 565.62

Interpretation:

we conclude that we are 99% confident that the population mean will fall within this confidence interval