Question

Question 5 View Policies Current Attempt in Progress During the nutrition research, the amount of consumed kilocalories per day was measured for 18 people - 10 women and 8 men. Results are as follows: Women: 1818, 1776, 1935, 1976, 1593, 2076, 1572, 1750, 1994, 1590; Men: 2422, 2194, 1993, 2449, 1840, 2082, 1841, 2365. Calculate a 95% confidence interval on the mean for women and men separately. Assume distribution to be normal. Round your answers to the nearest integer (e.g. 9876). Women: susi VIVI Men: Susi

Answer 1

Women

Mean = sum of all terms / no of terms = 18080 / 10 = 1808

SD

I-u – Ο (1 – 'n)

data	data-mean	(data - mean)2
1818	10	100
1776	-32	1024
1935	127	16129
1976	168	28224
1593	-215	46225
2076	268	71824
1572	-236	55696
1750	-58	3364
1994	186	34596
1590	-218	47524

Σα; - )2 V η -1 304706 10 -1 - = 184.0006

t value for 95% of confidence interval at 9 df is TINV(0.05,9) = 2.262

t*o 2.262 * 184.0006 V10 = 131.63

CI = mean +/ - E = 1808 +/- 131.63 = (1676.37 , 1939.63)

CI for women = (1676 , 1940)

Men

Mean = 17186 / 8 = 2148.25

data	data-mean	(data - mean)2
2422	273.75	74939.0625
2194	45.75	2093.0625
1993	-155.25	24102.5625
2449	300.75	90450.5625
1840	-308.25	95018.0625
2082	-66.25	4389.0625
1841	-307.25	94402.5625
2365	216.75	46980.5625

Στη - )2 V η -1 432375.5 -= 248,5315 8 -1

t value for 95% of confidence interval at 7 df is TINV(0.05,7) = 2.365

t*o 2.365 * 248.5315 E = = 207.811 V8

CI = mean +/ - E = 2148.25 +/- 207.811 = (1940.44 , 2356.06)

CI for men = (1940 , 2356)