Question

. A Simple random sample of 18 recent birth records at the local hospital was taken. In the sample, the average birth. weight was 3390 grams with a standand. deviation of 179.8 grams Asume that in the population of all babus in this hashital the birth weighto follow a Normal distribut? a) Find 95% confidenu interval birth weight of all bables the hospital. for the mean barn in How could we reduce the while beeting the same interval level ? margin of inar 99% confideny

Answer 1

Answer:

sample size (n)=18,

sample mean ($\bar x$) = 3390

sample standard deviation (s) = 179.8

a)

c=95%

formula for confidence interval is

$\bar x\pm t_{\alpha /2} * \frac{s}{\sqrt{n}}$

Where $t_{\alpha /2}$ is the t critical value for c=95% and

df = n-1 = 18-1 = 17

using t table we get critical value as

$t_{\alpha /2}$ = 2.110

$3390\pm 2.110 * \frac{179.8}{\sqrt{18}}$

3390 − 89.412 < $\mu$ < 3390 + 89.412

3300.588 < $\mu$ < 3479.412

Thus we get 95% confidence interval as (3300.588 , 3479.412)

b)

By increasing the sample size (n) we can reduce the margin of error,

because increasing the sample size reduces the standard error. which results into the reduced margin of error