Question

Question 3 of 26 (3 points) Attempt 1 of Unlimited View question in a popup 8.2 Section EX Baby weight: Following are weights, in pounds, of 10 two-month-old baby girls. It is reasonable to assume that the population is approximately normal. 12.23 12.32 14.13 9.34 11.48 8.63 12.66 10.30 12.34 12.95 Send data to Excel Part: 0/2 Part 1 of 2 Construct an 80% interval for the mean weight of two-month-old baby girls. Round the answers to three decimal places. An 80% confidence interval for the mean weight of two-month-old baby girls is Skip Part Check Answer Save For Later Submit Assignment

Answer 1

( 3 )

mean = 11.64

std dev = 1.7136

The provided sample mean is X = 11.64 and the sample standard deviation is s = 1.7136. The size of the sample is n = 10 and the required confidence level is 80%. The number of degrees of freedom are df = 10 -1 =9, and the significance level is a = 0.2. Based on the provided information, the critical t-value for a = 0.2 and df = 9 degrees of freedom is te = 1.383 The 80% confidence for the population mean u is computed using the following expression CI = x texs vn Therefore, based on the information provided, the 80 % confidence for the population mean u is 1.383 x 1.7136 .,11.64+ 1.383 x 1.7136 V10 10 = (11.64 -0.749, 11.64 +0.749) = (10.891, 12.389) = (11.64

( 4 )

( a )

The provided sample mean is 8 = 11.35 and the sample standard deviation is s = 1.8794. The size of the sample is n = 11 and the required confidence level is 95%. The number of degrees of freedom are df = 11 - 1 = 10, and the significance level is a = 0.05. Based on the provided information, the critical t-value for a = 0.05 and df = 10 degrees of freedom is te = 2.228 The 95% confidence for the population mean u is computed using the following expression CI= ( cX8 Therefore, based on the information provided the 95 % confidence for the population mean u is 2.228 x 1.8794 2.228 x 1.8794 VIT 11 = (11.35 - 1.263, 11.35 + 1.263) (10.087, 12.613) C1 = (11.35 11.35 -

( b )

yes it is reasonable because 11.1 lies between 10.087 and 12.613

( 5 )

(6) x=46, n=98 , C.L= 80% Cal P = 46 = 0.4694 98. at x=0.20 Lois tal = 1.282 0.4694 1-282 -0.4694) 98 0.4694 Cine 0.4694 + 0.0646 (10:4045, 10.5340]) is 80% ch

( 6 )

The sample proportion is computed as follows, based on the sample size N = 71 and the number of favorable cases X = 47 х = 0.662 p=N=71 The critical value for a = 0.02 is ze = 21-a/2 = 2.326. The corresponding confidence interval is computed as shown below: CI(Proportion) = (P Þ(1 - i) p(1-P) P + V n n 0.662 2.326 x 0.662(1 -0.662) ,0.662 +2.326 x 71 0.662(1 -0.662) 71 (0.531,0.793)