Answer to Question #129842 in Statistics and Probability for ืnub

Question #129842

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: 1646, 1921, 1990, 1802, 1943, 1606, 1706, 1928, 1645, 1640;


Men: 2287, 2057, 2206, 1978, 2215, 2344, 2081, 2088.


Calculate a 99% 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).


1
Expert's answer
2020-08-17T19:29:16-0400

The mean for women is μW=xinW=1782.7\mu_W = \frac{\sum x_i}{n_W} =1782.7 . The mean for men is μM=xinM=2157\mu_M = \frac{\sum x_i}{n_M} =2157 .

Using the formula for standard deviation σ=(xiμ)n1\sigma = \sqrt{\frac{\sum (x_i -\mu)}{n-1}} we can find: for women σW=150.7\sigma_W = 150.7, for men σM=125.5\sigma_M = 125. 5.

For the normal distribution it is known that confidence intervals can be expressed like μ±(zσ)\mu \pm (z*\sigma), where z multiplier comes from values of standard normal distribution N(0,1)N(0,1) that separate the middle N% from the outer (100-N)%. In our case we need 99% confidence interval. z-value for it is 2.57583.

For women: 1782.7±(2.57583150.7)=1782.7±388.2={1395;2171}1782.7 \pm (2.57583*150.7)=1782.7 \pm 388.2 =\{1395; 2171\}.

For men: 2157±(2.57583125.5)=2157±323.3={1834;2480}2157 \pm (2.57583*125.5)=2157 \pm 323.3 =\{1834; 2480\}.



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