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 11 online advertisements of entry-level road bikes. He determines that the mean price for these l l 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 What is the upper limit of this confidence interval? Please give your answer to the nearest cent.

Answer 1

Answer: The lower limit of this confidence interval = $576.41

The upper limit of this confidence interval = $831.09

Step-by-step explanation:

Let $\mu$ be the population mean price of a used road bike.

As per given have,

n=11

Degree of freedom = 10

$\overline{x}=\ 703.75$

s= $189.56

T-critical value for 95% confidence :

$t_{(df, \alpha/2)}=t_{(10,0.025)}=2.228$

Now, 95% confidence interval for μ, the mean price of a used road bike. will be :

$\overline{x}\pm t_{\alpha/2}\dfrac{s}{\sqrt{n}}$

$\ 703.75\pm (2.228)\dfrac{189.56}{\sqrt{11}}$

$\ 703.75\pm \ 127.34$

$(\ 703.75-\ 127.34,\ \ 703.75+\ 127.34)$

$(\ 576.41,\ \ 831.09)$

Thus , the lower limit of this confidence interval = $576.41

The upper limit of this confidence interval = $831.09