The prices (in dollars) for a particular model of digital camera with 6.0 megapixels and an optical 3X zoom lens are shown below for 10 online retailers. Estimate the true mean price for this particular model with 93% confidence. Assume the variable is normally distributed. Use a graphing calculator and round the answers to at least one decimal place.
225 236 217 202 231 185 193 185 245 217

Answers

Answer 1

Solution :

Let [tex]$x_i$[/tex] represents the prices for the digital camera.

              [tex]$x_i$[/tex]                   [tex]$x_i^2$[/tex]

           225              50625

           236              55696

            217               47089

           202              40804

           231               53361

           185               35225

           193               37249

           185               34225

           217               47089

[tex]$\sum x_i = 2136$[/tex]       [tex]$\sum x_i^2 = 460388$[/tex]

Mean, [tex]$(\overline X) = \frac{\sum x_i}{n}$[/tex]

                 [tex]$=\frac{2136}{10}$[/tex]

                 = 213.6

Variance, [tex]$\sigma^2 = \frac{\sum x_i^2}{n}-(\overline X)^2$[/tex]

                     [tex]$ = \frac{460388}{10}-(213.6)^2$[/tex]

                    = 46038.8 - 45624.96

                    = 413.84

Therefore, standard deviation,

[tex]$\sigma = \sqrt{Var}$[/tex]

   [tex]$=\sqrt{413.84}$[/tex]

   = 20.343

Now to find the 93% confidence interval, the formula us

[tex]$(\overline X)\pm (z) \times \frac{\sigma}{\sqrt{n}}$[/tex]

For 93%, the value of z is 1.81

∴ Confidence interval = [tex]$(\overline X)\pm (z) \times \frac{\sigma}{\sqrt{n}}$[/tex]

C.I. = [tex]$213.6\pm 1.81 \times \frac{20.343}{\sqrt{10}}$[/tex]

      [tex]$=213.6 \pm 11.6438$[/tex]

Therefore, lower bound = 213.6 - 11.6438

                                        = 201.9562

                  Upper bound = 213.6 + 11.6438

                                         = 225.2438

∴ [tex]$201.9562 < \mu < 225.2438$[/tex]

or [tex]$202< \mu < 225.2$[/tex]

       


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