Answer to Question #109499 in Statistics and Probability for Ibrahim

Question

A sample of 100 apparently normal adult males, 25 years old, had a mean systolic blood pressure of
125. It is believed that the population standard deviation is 15.

Answer 1

$n=100,\bar{X}=25,\sigma=15.$

$$CI=(\bar{X}-z_{1-\alpha/2}\cdot{\sigma\over\sqrt{n}}, \bar{X}+z_{1-\alpha/2}\cdot{\sigma\over\sqrt{n}})$$

(a) Construct a 97% confidence interval.

$\alpha=0.03, z_{1-\alpha/2}=2.17$

$$CI=(125-2.17\cdot{15\over\sqrt{100}}, 125+2.17\cdot{15\over\sqrt{100}})=$$

$$=(121.745, 128.255)$$

Therefore, based on the data provided, the 97% confidence interval for the population mean is $121.745<\mu<128.255$ , which indicates that we are 97%  confident that the true population $\mu$  is contained by the interval $(121.745, 128.255).$

(b) How many adult males must we select so that the margin of error is not greater than 2.5 kg

considering the confidence level of 98%?

$\alpha=0.02, z_{1-\alpha/2}=2.326$

$$margin\ of\ error=z_{1-\alpha/2}\cdot{\sigma\over\sqrt{n}}\leq E=2.5$$

$$N\geq({z_{1-\alpha/2}\cdot\sigma \over E})^2$$

$$N\geq({2.326\cdot15\over 2.5})^2$$

$$N\geq195$$

$195$ adults.