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

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

Given the data :

The confidence interval is defined as :

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

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

Using a calculator :

Lower boundary = [tex] 217.7 - 2.26(\frac{17.49}{\sqrt{10}}) = 205.20[/tex]

Lower boundary = [tex] 217.7 + 2.26(\frac{17.49}{\sqrt{10}}) = 230.20[/tex]

Therefore, the confidence interval is (205.20 ; 230.20)

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

The 95% confidence interval is [tex]\mathbf{(205,19, 230.20)}[/tex]

The given parameter is:

[tex]\mathbf{n = 10}[/tex]

Start by calculating the degree of freedom

[tex]\mathbf{df = n - 1}[/tex]

[tex]\mathbf{df = 10 - 1}[/tex]

[tex]\mathbf{df = 9}[/tex]

Next, calculate the mean

[tex]\mathbf{\bar x = \frac{\sum x}{n}}[/tex]

[tex]\mathbf{\bar x = \frac{225 + 240 + 215 + 206 + 211 + 210 + 193 +250 + 225 + 202}{10}}[/tex]

[tex]\mathbf{\bar x = \frac{2177}{10}}[/tex]

[tex]\mathbf{\bar x = 217.7}[/tex]

Calculate the standard deviation

[tex]\mathbf{\sigma = \sqrt{\frac{\sum (x - \bar x)^2}{n-1}}}[/tex]

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

[tex]\mathbf{\sigma = \sqrt{\frac{2752.1}{9}}}[/tex]

[tex]\mathbf{\sigma = 17.49}[/tex]

Calculate the standard error

[tex]\mathbf{SE = \frac{\sigma}{\sqrt n}}[/tex]

[tex]\mathbf{SE = \frac{17.49}{\sqrt {10}}}[/tex]

[tex]\mathbf{SE = 5.53}[/tex]

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

[tex]\mathbf{t= 2.262}[/tex]

So, the confidence interval is:

[tex]\mathbf{CI = \bar x \pm t \times SE}[/tex]

So, we have:

[tex]\mathbf{CI = 217.7 \pm 2.262 \times 5.53}[/tex]

Split

[tex]\mathbf{CI = (217.7 - 2.262 \times 5.53, 217.7+ 2.262 \times 5.53)}[/tex]

[tex]\mathbf{CI = (205,19, 230.20)}[/tex]

Hence, the 95% confidence interval is [tex]\mathbf{(205,19, 230.20)}[/tex]

Read more about confidence intervals at:

https://brainly.com/question/2396419

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