To estimate the mean height of female high school juniors, you take a random sample of 30 female students and get these results (in inches):

72,51, 67, 68, 61, 69, 58, 56, 60, 56

66, 61, 60, 59, 59, 54, 58, 53, 68, 63

57, 62, 63, 64, 56, 62, 58, 67, 57, 70

If the o is 5.3, based on past research, and you want to construct a 97%

confidence interval using (estimate) + (margin of error)= (estimate)

what's your point estimate of ?

A. 30

B. 5.4

C. 65

D. 61.17

61.16667 ; (58.958, 63.375)

Step-by-step explanation:

Given the data :

72,51, 67, 68, 61, 69, 58, 56, 60, 56, 66, 61, 60, 59, 59, 54, 58, 53, 68, 63, 57, 62, 63, 64, 56, 62, 58, 67, 57, 70

Standard deviation, σ = 5.3

Sample size, n = 30

The point estimate of the population mean:

Mean, m = Σx/n

Σx = 1835 ; n = 30

m = 1835 / 30

m = 61.16667

Confidence interval :

m ± Tn-1,α/2 * σ/sqrt(n)

61.16667 ± T30-1,0.03/2 * 5.3/sqrt(30)

61.16667 ± T29,0.015 * (5.3/sqrt(30))

61.16667 ± 2.282175 * (0.9676431)

Lower bound = 61.16667 - 2.20833 = 58.95834

Upper bound = 61.16667 + 2.20833 = 63.375

(58.958, 63.375)

the final price would be 29.6925

step-by-step explanation:

39.59 minus 25% equals 29.6925

so then what are you asking?

step-by-step explanation: