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=11 online advertisements of entry-level road bikes. He determines that the mean price for these 11 bikes is x⎯⎯⎯=$703.75 and that the sample standard deviation is s=$189.56 . He uses this information to construct a 95% 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.

Step-by-step explanation:

Confidence interval is written in the form,

(Sample mean - margin of error, sample mean + margin of error)

The sample mean, x is the point estimate for the population mean.

Margin of error = z × s/√n

Where

s = sample standard deviation = $189.56

n = number of samples = 11

From the information given, the population standard deviation is unknown and the sample size is small, hence, we would use the t distribution to find the z score

In order to use the t distribution, we would determine the degree of freedom, df for the sample.

df = n - 1 = 11 - 1 = 10

Since confidence level = 95% = 0.95, α = 1 - CL = 1 – 0.95 = 0.05

α/2 = 0.05/2 = 0.025

the area to the right of z0.025 is 0.025 and the area to the left of z0.025 is 1 - 0.025 = 0.975

Looking at the t distribution table,

z = 2.228

Margin of error = 2.228 × 189.56/√11

= 127.34

the lower limit of this confidence interval is

703.75 - 127.34 = $576.41

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