Question

{Exercise 8.21 (Algorithmic)}

Consumption of alcoholic beverages by young women of drinking age has been increasing in the United Kingdom, the United States, and Europe (The Wall Street Journal, February 15, 2006). Data (annual consumption in liters) consistent with the findings reported in The Wall Street Journal article are shown for a sample of 20 European young women.

87	82	199	174	97
170	222	115	133	169
164	113	99	171	0
93	0	93	110	329

Assuming the population is roughly symmetric, calculate the sample mean and sample standard deviation. Then construct a 95% confidence interval for the mean annual consumption of alcoholic beverages by European young women (Round all answers to 1 decimal).

The mean is  and the standard deviation is

The confidence interval is (  ,  )

Answer 1

Solution

$\small \mathbf{X}$	X-X	(X-)
87	-44	1936
82	-49	2401
199	68	4624
174	43	1849
97	-34	1156
170	39	1521
222	91	8281
115	-16	256
133	2	4
169	38	1444
164	33	1089
113	-18	324
99	-32	1024
171	40	1600
0	-131	17161
93	-38	1444
0	-131	17161
93	-38	1444
110	-21	441
329	198	39204
X = 2620		Σ (x - X) = 104364

Mean (X) = ΣΧ2620 – 131.0 Γ 20.

Standard deviation (s) = 1 - (x – X)? _ ./104364 _ 0 -12 = V 20 -1 = 74.1

Degrees of freedom df) = n-1 = 20 -1 = 19

For 95% confidence level with degrees of freedom = 19

** = 2.093 (Refer t-table)

formula : Confidence interval for the population mean,

= X ****

=131.0 +2.093 * 74.1 ;

= 131.0±34.7

= (131.0 -34.7.131.0 +34.7)

= (96.3.165.7)

... The confidence interval is (96.3, 165.7)