Question

In a sample of 167 children selected randomly from one town, it is found that 37 of them suffer from asthma. At the 0.05 significance leveltest the claim that the proportion of all children in the town who suffer from asthma is 11%.

A public bus company official claims that the mean waiting time for bus number 14 during peak hours is less than 10 minutes. Karen took bus number 14 during peak hours on 18 different occasions. Her mean waiting time was 76 minutes with a standard deviation of 2.3 minutes . At the 0.01 significance level, test the claim that the mean waiting time is less than 10 minutes. Use the P-value method of testing hypotheses .

Answer 1

• One sample z test for proportion (sample size=167, it is sufficiently large)

P 11 =0.11 100 Let, be proportion of all children in a town, who suffer from asthma we want to test the claim : " The proporation of all children in the town, who suffer from asthma is 11%"sie, p = .: The problem of testing is Ho: p=0.11 v/s H, :p #0:11 Test-statistic : Following test -statistie is appropriate for this problem- 2= þ-Po Po (1-) Undero Ho: 7 ~ Normal 0,1) · where, Þ= sample proporation of children in the town, suffesoing from asthma. 37 167 Ž0.221557 Po = hypothe sized value of Þ, specified under n = size of the random sample selected. =164 The observed value of the test statistic,day Lobs. id- +) - 0.11 ñ 4.6075 0.11(1-0. 11) 167 - Rule of rejection: At given, a = level of significance = 0.05, we have the following rule- • We reject Ho, at <=0.05, if | Zobs.! Zq- upper 100 % point foro a standard mommal dis Faribution, ie, N(0,1). 21.96 (using 2-table) 12068.1 = 14.60451–4.6045 We observe, 1 Zobo.lx. 1.96 , thus, we reject Ho. = 0.11 37 167 Zoba. ४/N where, For, 2-0.05, 20.05 - 20.025 2

. Conclusion: As we reject Ho at, 2=0.05 we can conclude that there is sufficient that the proportion of all children in the town who suffer from asthma is 11% $

• One sample t test for mean (population standard deviation is unknown and sample size =18, it us small)

Ho : M = 10 S · Let, H be mean waiting time for bus number 14. (umit: minutes We want to test the daim of a public bus company official that - "Meam waiting time for bus number 14 'is less than 10 minutes", ºien < 10. : The testing problem 18- V/8 Hi: <10 Test - Statistic. Assuming the data is coming from a normal distribution, the appropriate fest-stabiotic, is - say, Tamm (ā-lo such that under Ho: Tnt, where, Hoz hypothesized value of Me = 10, M= size of the sample = 18 (different occasions) X = sample mean waiting time = 46 minutes. S = Sample standard deviation = 2.3 minutes, :: the observed value of the test - statistic , 8ay Tobs. N18 ( 46-10) Toba. : Under - Prvalue of the dest is probability of getting a sample, at least as extremecin direction of Hi) as the one we observe, assuming Ho is troue. : P-value = P. (T<121.15) where the boy) (using t-distribution table) At, x=0.01 as p-value>, we fail to reject Ho. · Conclusion: As we fail to peject Ho, at a=0.01, the claim of the bus company official is rejected. is Ż 121175 2.3 Ho: Tat (*نام ñ 1.00

(P.S: I felt that,76 minutes is too high to consider it as sample mean waiting time, if you have meant 7.6 minutes, calculations are shown. Please ignore this part, if it is not the case!)

Toba. N18 (7.6-10) 2-4.4271. 9.3 P-value = Pho (T3.-4.4241) where THO +0+) ÑO : At =0.01, p-value <<, we reject Ho. Hence, the claim of the Jous company official can be accepted.