Question

Answers C. By the CLT, X approx N(a = 10,'- )s P(Z <-1.2)-0.115 ). Thus, P(X <9.8)s Plz < a. 6min X, and 62-2x-1 b·6, is biased (because it is always overestimation) but 02 is unbiased 2. (becase X is unbiased for estimating ) 3. a. 66.2, don't know whether it underestimates or overestimates b.?2o/2 ? = 66.2 ± 1.96 ,44=66.2 ±0.40 C. We don't know as the value of ? is unknown. d. Yes, because is the center of the interval. e. 1600 4, 95% confidence interval is ±to/2(n-1) (-66.2 ± 1966. ? 66.2 ± 0.49. The other answersmmmmth.mm 5. -Question 3 setting: At level 5%, the rejection region of the test Since l66.2-66.8] > 0.4, we can rriget H. : ? 66.8 (at level 5%) and correspondingly accept the alternative (H1: 668) that population mean is not 66.8 inches.. The p-value of the observed sample mean 66.2 is 2P(X < 662) = 2P(417668 < 66 2 v 66.8)-2P(Z <-29-20017 = .00 4.1/V400 4.1/V400 It is (strongly) significant at level 5%, and thus we can reject the null hypothesis and correspondingly conclude that popalation mean is not 66.8 inches. -Question 4 setting: At level 5%, the rejection region of the test Since l66.2-668] > 0.49, we can reject Ho : ? = 66.8 (at level 5%) and correspondingly accept the alternative (Hi : ? 668) that population mean is not 66.8 inches. The p-value of the observed sample mean is 2P(8 < 66.2) = 2P(5/__ < 05/- 1-66.8 , 06.2-66.8 /V400 5/400 ) 2P(t(399) <-24) = 2. .0084-.0168 It is also significant at level 5%, and thus we can reject the null hypothesis and correspondingly conclude that population mean is not 66.8 inches. 41

Answer 1

We know that, from central limit theorem the test statistic of $\bar X$ is given by

$\frac{\bar X-\mu}{\sigma/\sqrt{n}}\sim N(0,1)$

But this is applicable only when population standard deviation ($\sigma$) is known. When population standard deviation is unknown and sample size is small (< 30), $\bar X$ follows t-distribution, and we get

$\frac{\bar X-\mu}{s/\sqrt{n}}\sim t_{n-1}$

where s is sample standard deviation. Hence from this we get confidence interval as

$\bar X- \frac{s}{\sqrt{n}}t_{n-1,\alpha/2}< \mu<\bar X+\frac{s}{\sqrt{n}}t_{n-1,\alpha/2}$

$\mu<\bar X\pm \frac{s}{\sqrt{n}}t_{n-1,\alpha/2}$