Mathematics, 09.04.2020 16:22 lola78830

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 ()

Answers: 3

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Answer from: 34267

Following are the solution to the given points:

For point a:

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

The t-score formula is $t* = t_{\alpha, n-1}\\\\$

Here,

$\to \alpha = 0.05 \\\\ \to n =16\\\\$

Therefore,

$\to t*= t_{\alpha, n-1} = t_{0.05, 16-1} = t_{0.05,15}$

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

For point b:

Determine a confidence interval $95\%$ for the mean price of this model of the digital phone.

Formula for $95\%$ confidence interval:

$\to CI = \bar{x} \pm t\times \frac{s}{\sqrt{n}}$

Here,

$\to \bar{x} = 262.49,\\\\ \to s = 27.57 \\\\\to n=16\\\\$

Therefore,

$\to CI = \bar{x} \pm t \times \frac{s}{\sqrt{n}}$

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

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

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Answer from: garzamatt7

a) t = 2.131

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

Step-by-step explanation:

Sample size = n = 16

Sample mean = x = 262.49

Sample Standard deviation = s = 27.57

Part a) Value of t-score

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:

$(x- t_{\frac{\alpha}{2}} \times \frac{s}{\sqrt{n}}, x- t_{\frac{\alpha}{2}} \times \frac{s}{\sqrt{n}})$

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

$(262.49-2.131 \times \frac{27.57}{\sqrt{16} }, 262.49-2.131 \times \frac{27.57}{\sqrt{16} })\\\\ (247.802,277.178)$

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

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Justin is interested in buying a digital phone. He visited 16 stores at random and recorded the pric...