Question

Adam tabulated the values for the average speeds on each day of his road trip as 60.5, 63.2, 54.7, 51.6, 72.3, 70.7, 67.2, and 65.4 mph. The sample standard deviation is 7.309.Select the 98% confidence interval for Adam’s set of data.a. 46.94 to 71.33b. 46.94 to 79.46c. 55.45 to 79.46d. 55.45 to 70.95

Answer 1

Answer: 55.45 to 70.95

Explanation:

Answer 2

Option D, 55.45 to 70.95

Explanation:

Alpha = 1 – confidence interval

Alpha = 1 – 0.98 = 0.02

Sample size = n = 8

t (alpha/2) ; (n-1) = t (0.02/2) ; (8-1) = t 0.01, 7 = 2.998

Mean = sum of all frequencies /total number of frequency = 505.6/8 = 63.2

s = 7.309

E = t (0.01;7) * s/sqrt n

Substituting the given values, we get –

E = 2.998 * 7.309 /sqrt (8)

E = 7.75

98% confidence interval

Mean – E and Mean + E

63.2 – 7.75 and 63.2 + 7.75

(55.45, 70.95)