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

Answers

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 :

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

Zcritical = [tex]t_{0.05/2, 9} = 2.26 [/tex] n = sample size = 10

Using a calculator :

Mean, [tex]\bar{x} = 217.7 [/tex] Standard deviation, σ = 17.49

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)

Learn more :https://brainly.com/question/25173276

Answer 2

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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What is 0.0000128 as a fraction?

To write 0.0000128 as a fraction you have to write 0.0000128 as numerator and put 1 as the denominator. Now you multiply numerator and denominator by 10 as long as you get in numerator the whole number.

0.0000128 = 0.0000128/1 = 0.000128/10 = 0.00128/100 = 0.0128/1000 = 0.128/10000 = 1.28/100000 = 12.8/1000000 = 128/10000000

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0.0000128 as a fraction equals 128/10000000

Answer:

128/10000000

Step-by-step explanation:

To write 0.0000128 as a fraction you have to write 0.0000128 as numerator and put 1 as the denominator. Now you multiply numerator and denominator by 10 as long as you get in numerator the whole number.

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0.0000128 as a fraction equals 128/10000000

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