Question

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 (____)

Answer 1

Answer:

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)