Question

Ustin is interested in buying a digital phone. he visited 20 stores at random and recorded the price of the particular phone he wants. the sample of prices had a mean of 195.69 and a standard deviation of 21.02. (a) what t-score should be used for a 95% confidence interval for the mean, μμ, of the distribution? t* = 2.09302 (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.)

Answer 1

(a) 2.09302406 (b) Lower limit of 95% confidence interval = 185.85 (a) To get the t-score, first determine the number of degrees of freedom you have. That's simply the number of samples minus 1, so in this case we have 19 degrees of freedom. Then calculate (1-0.95)/2 = 0.05/2 = 0.025 which is the size of the tail for the confidence interval you want. Finally, lookup in a T-Distribution Table for the 19 degrees of freedom and tail size. The value looked up will be 2.09302406. (b) Once you have the t-score, to calculate your desired interval, calculate the following: V = T*SD/sqrt(n) where V = Variance T = T-score SD = Standard deviation n = number of samples So let's plug in the values and calculate V V = T*SD/sqrt(n) V = 2.09302406*21.02/sqrt(20) V = 2.09302406*21.02/4.472135955 V = 43.99536574/4.472135955 V = 9.837662849 Your 95% confidence interval is now the mean +/- V. So lower limit = 195.69 - 9.84 = 185.85 upper limit = 195.69 + 9.84 = 205.53