Question

The prices (in dollars) for a particular model of digital camera with 6.0
megapixels and an optical 3X zoom lens are shown here for 10 randomly selected online
retailers. Find 95% confidence interval of the true mean price for this particular
model.
225 240 215 206 211 210 193 250
225 202

Answer 1

Using the data given, the 95% confidence interval of the true mean is (205.20 ; 230.20)

Given the data :

• 225 240 215 206 211 210 193 250 225 202

The confidence interval is defined as :

Since, sample size ls less than 30 we use a t-distribution table :

$C.I = \bar{x} ± t_{α/2, n-1} \frac{s}{\sqrt{n}}$

• Zcritical = $t_{0.05/2, 9} = 2.26$
• n = sample size = 10

Using a calculator :

• Mean, $\bar{x} = 217.7$
• Standard deviation, σ = 17.49

Lower boundary = $217.7 - 2.26(\frac{17.49}{\sqrt{10}}) = 205.20$

Lower boundary = $217.7 + 2.26(\frac{17.49}{\sqrt{10}}) = 230.20$

Therefore, the confidence interval is (205.20 ; 230.20)

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Answer 2

The confidence interval is simply a range that contains the true mean of the population

The 95% confidence interval is $\mathbf{(205,19, 230.20)}$

The given parameter is:

$\mathbf{n = 10}$

Start by calculating the degree of freedom

$\mathbf{df = n - 1}$

$\mathbf{df = 10 - 1}$

$\mathbf{df = 9}$

Next, calculate the mean

$\mathbf{\bar x = \frac{\sum x}{n}}$

$\mathbf{\bar x = \frac{225 + 240 + 215 + 206 + 211 + 210 + 193 +250 + 225 + 202}{10}}$

$\mathbf{\bar x = \frac{2177}{10}}$

$\mathbf{\bar x = 217.7}$

Calculate the standard deviation

$\mathbf{\sigma = \sqrt{\frac{\sum (x - \bar x)^2}{n-1}}}$

$\mathbf{\sigma = \sqrt{\frac{(225 -217.7)^2+ (240 -217.7)^2+..........+ (202 -217.7)^2}{10-1}}}$

$\mathbf{\sigma = \sqrt{\frac{2752.1}{9}}}$

$\mathbf{\sigma = 17.49}$

Calculate the standard error

$\mathbf{SE = \frac{\sigma}{\sqrt n}}$

$\mathbf{SE = \frac{17.49}{\sqrt {10}}}$

$\mathbf{SE = 5.53}$

At 95%, df = 9. the t-value is:

$\mathbf{t= 2.262}$

So, the confidence interval is:

$\mathbf{CI = \bar x \pm t \times SE}$

So, we have:

$\mathbf{CI = 217.7 \pm 2.262 \times 5.53}$

Split

$\mathbf{CI = (217.7 - 2.262 \times 5.53, 217.7+ 2.262 \times 5.53)}$

$\mathbf{CI = (205,19, 230.20)}$

Hence, the 95% confidence interval is $\mathbf{(205,19, 230.20)}$

Read more about confidence intervals at:

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