Question

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).

Answer 1

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

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

For the normal distribution it is known that confidence intervals can be expressed like $\mu \pm (z*\sigma)$, where z multiplier comes from values of standard normal distribution $N(0,1)$ 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: $1782.7 \pm (2.57583*150.7)=1782.7 \pm 388.2 =\{1395; 2171\}$.

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