In: Statistics and Probability
Reserve Problems Chapter 8 Section 2 Problem 2
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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: 1962, 1842, 1588, 1911, 1779, 1603, 1758, 1771, 1874,
1974;
Men: 2097, 2560, 2328, 2399, 2420, 2292, 2263, 2047.
Calculate a 90% 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: | Enter your answer; Women: confidence interval, lower bound ≤μ≤ Enter your answer; Women: confidence interval, upper bound | |
Men: | Enter your answer; Men: confidence interval, lower bound ≤μ≤ Enter your answer; Men: confidence interval, upper bound |
for women":
sample mean x= | 1806.200 |
sample size n= | 10 |
sample std deviation s= | 134.36 |
std error sx=s/√n= | 42.4913 |
for 90% CI; and 9 df, critical t= | 1.8330 | |
margin of error E=t*std error = | 77.8865 | |
lower bound=sample mean-E = | 1728 | |
Upper bound=sample mean+E= | 1884 |
for men:
sample mean x= | 2300.750 |
sample size n= | 8 |
sample std deviation s= | 168.706 |
std error sx=s/√n= | 59.6465 |
for 90% CI; and 7 df, critical t= | 1.8950 | |
margin of error E=t*std error = | 113.0301 | |
lower bound=sample mean-E = | 2188 | |
Upper bound=sample mean+E= | 2414 |