Justin is interested in buying a digital phone. He visited 16 stores at random and recorded the price of the particular phone he wants. The sample of prices had a mean of 262.49 and a standard deviation of 27.57.

A) What t-score should be used for a 95% confidence interval for the mean, μ, of the distribution?
t* = ()

B) Calculate a 95% confidence interval for the mean price of this model of digital phone:
(Enter the smaller value in the left answer box.)

() to ()

We have to construct a confidence interval for the mean. Since, value of population standard deviation is unknown, and value of sample standard deviation is known, we will use t-distribution to find the confidence interval.

Degrees of freedom = df = n - 1 = 16 - 1 = 15

The critical t-value which we have to use should be checked again 15 degrees of freedom and 95% confidence level. From the t-table this value comes out to be:

t = 2.131

Part b) Confidence Interval

The formula to calculate the confidence interval is:

[tex](x- t_{\frac{\alpha}{2}} \times \frac{s}{\sqrt{n}}, x- t_{\frac{\alpha}{2}} \times \frac{s}{\sqrt{n}})[/tex]

Here, [tex]t_{\frac{\alpha}{2} }[/tex] is the critical t-score we found in the previous part. Using the values in the formula, we get:

[tex](262.49-2.131 \times \frac{27.57}{\sqrt{16} }, 262.49-2.131 \times \frac{27.57}{\sqrt{16} })\\\\ (247.802,277.178)[/tex]

Therefore, the 95% confidence interval for the mean price of this model of digital phone is (247.80, 277.18)

codiepienagoya codiepienagoya · Answer 2 · 2021-12-17T16:06:13+01:00

Following are the solution to the given points:

For point a:

Determine the t score that is [tex]95\%[/tex] used to calculate a confidence interval for the distribution's mean "[tex]\mu[/tex]".

The t-score formula is [tex]t* = t_{\alpha, n-1}\\\\[/tex]

Here,

[tex]\to \alpha = 0.05 \\\\ \to n =16\\\\[/tex]

Therefore,

[tex]\to t*= t_{\alpha, n-1} = t_{0.05, 16-1} = t_{0.05,15}[/tex]

According to the t distribution table, [tex]t_{0.05,15} = 2.131.[/tex] As a result, the t score for a [tex]95\%[/tex] confidence interval again for distribution's mean [tex]\mu[/tex] =[tex]2.131[/tex].

For point b:

Determine a confidence interval [tex]95\%[/tex] for the mean price of this model of the digital phone.

Formula for [tex]95\%[/tex] confidence interval:

[tex]\to CI = \bar{x} \pm t\times \frac{s}{\sqrt{n}}[/tex]

Here,

[tex]\to \bar{x} = 262.49,\\\\ \to s = 27.57 \\\\\to n=16\\\\[/tex]

Therefore,

[tex]\to CI = \bar{x} \pm t \times \frac{s}{\sqrt{n}}[/tex]

[tex]= 262.49 \pm (2.131 \times \frac{27.57}{\sqrt{16}})\\\\= 262.49 \pm (2.131 \times \frac{27.57}{ 4})\\\\= 262.49 \pm (2.131 \times 6.8925)\\\\= 262.49 \pm 14.6879 \\\\= (247.8021,277.1779) \\\\[/tex]

As a result, [tex]95\%[/tex] the confidence interval again for the mean price of this model of digital phone is [tex](247.8021\ \ to\ \ 277.1779)[/tex].

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