Question

z-test a sample for the null hypothesis using α = 0.05 as our threshold

a) Given a population with mean of 550 and variance of 100, what would be minimum sample mean to reject the null hypothesis if our sample size is 20?

b) Given a population with mean of 550 and variance of 100, what would be maximum sample mean to reject the null hypothesis if our sample size is 20?

c) Let’s assume the same population mean and use the same sample size of 20. If we have a sample with a standard deviation of 11 and use that to estimate the sample standard deviation, what is the new minimum sample mean for which we would reject the null hypothesis?

d) New Maximum

Answer 1

som (niven that does @ Pop" mean, M: 550 Sample sire no 20 Popn variance, 6²: 100 Sample mean 7 We kerous that; XH NHCO,1) trio z imeä-H) FM1011] P[1214.7.112.17 1-6 tahun So Rejec Hokithin 777 or I L-70/2 al M=0.05value of Zain = 1096 For min, s: 27242 ) - 550 7 1996 2 6 / 10 TIMU 242 ITTTTIIIIII! L> 550 +1.96%.5 =SSO+ 4.382 bomo minimum, [2 min = 554-38 2 ans: for max ã, for which Ho is rejected. 24-2012 → 2-SSD $1.96 =) $550-196885 ī ļ 550 -4382 - 545.618 maximum, max = 545618] Ang – for which Ho rejected. Juo 20 Given that pop mean = U=550, sample standard devialion Sall sample size, n = 20 sample mean = ł - pop" variance is not lanown.

them since popn warlance is not lanown then vee the it's estimate ie sample Stendard deviation at distribution s lie. t= X-M. Sn (8-4) ontnt sin Then we know that S Pľ H7 tn. ca] =4 l i e. we reject null hypothesis, when t> Bagabond or t <- toru Ceva) Unir ..tap for min t ty tn. 1/2) te: 150 (-SS0) > 2.00 critical value at ig def. and 220.05 tig =2.09 5 So 22.99 4.472 550 + 1.x2009 SSD + 22. 9 120. 520 ã> 5.50 +5.140 - 555.140 min value of lã - sss.140 lane: for max. a: t <-tnt1012) Soo (7-850) < -2.09 gã 4550 4 550 – 5.140 - 344.260 max value of 7 = 544-860% and 2009 .