Question

4 Consider a large-sample level 0.01 test for testing Ho: P = 0.2 against Hp > 0.2. (a) For the alternative value p = a) 0.21, compute B(0.21) for sample sizes n = 64, 1600, 14,400, 40,000, and 90,000. (Round your answers to four decimal places.) л B 64 1600 14,400 40,000 90,000 b) (b) For p = xin = 0.21, compute the P-value when n = 64, 1600, 14,400, and 40,000. (Round your answers to four decimal places.) P-value 64 1600 14,400 40,000 C) (c) In most situations, would it be reasonable to use a level 0.01 test in conjunction with a sample size of 40,000? Why or why not? Yes, even when the departure from Ho is significant from a practical point of view, a statistically significant result is not likely to appear; it is difficult for the test to detect departures from Ho Yes, it is always advantageous to have a very large sample size, because it will detect very small departures from Ho No, even when the departure from Ho is insignificant from a practical point of view, a statistically significant result is highly likely to appear; the test is too likely to detect small departures from Ho. No, it is never advantageous to have a very large sample size, because it cannot detect very small departures from Ho.

Answer 1

value AS 70211-0-2) n n So we 2 33 we know & Po don't reject Hol Ho is false) As ox=0.01, so we So we reject Ho, if Computed of test static 276 2.33 - P ß-0.2 POL reject. Ho, if z7 - 2012 > 2.33 70.16 te Î 70.2 +2:33 So, we reject to if 0.2+. Now p-Pol Don't reject to 1 P=0.21) ho (502 +0.937 1 1 220-21) (na pri holz < 0.2+0.93% in -0.21 No.al <D (20.011770.932 No. 210.79) 0.4 nn 0,932 m 0.21 (1-0-21)

The table with different value so n are as

n	$\beta$
64
0091	0.9042
14400	02553
40000	0.0044
90000	70000000

(b) Value of test static is

0.21 – 0.20 0.01vn 0.4 0.025n 0.21-0.2) n

As test is right tailed we have the P-value as

Pr(2 > 0.025 Vn

Thus we have the table as

n	P-value
64	0.4207
0091	0.1587
14400	0.0013
40000	00000003

c)   First option is correct

AS for 40000 sample we get P-value 0.0000003 and beta value 0.0044