Question

1 TUTORIAL 9 Inferential Statistics - Hypothesis Testing I & II Question 1 Cro According to a local report, 23% of the physicians in Malaysia were women in 2010. A recent sample of 300 physicians in Malaysia found that 51 of them were women. Can you conclude that the current percentage of women in the population of all physicians in Malaysia is less than 23%? Use a 0.05. 51/3000? <P < 25 Question 2 Has the recent drop in airplane passengers resulted in better on-time performance? Before the recent economic downturn, one airline bragged that 92% of its flights were on time. A random sample of 165 Nights completed this year reveals that 153 were on time. Can we conclude at the 5% significance level that the airline's on-time performance has improved? Proportion 0.93 Question 3 The Labor Department Peninsular Malaysia collects data on unemployment insurance payments. Suppose that during 2009 a random sample of 80 unemployed people in Ampang received an average weekly benefit of RM 201.85, whereas a random sample of 65 unemployed people in Kajang received an average weekly benefit of RM 185.31. Assume that the population standard deviations of all weekly unemployment benefits in Ampang and Kajang are RM 32.48 and RM 26.15, respectively. Using the 4% significance level, can you conclude that the means of all weekly unemployment benefits in Ampang and Kajang paid in 2009 are different? Use both approaches to make this test. Question 4 2 souple unknown The budget director for Nexus Media would like to compare the daily travel expenses for the sales staff and the audit staff. The collected sample information are shown below. Sales (S) 131 135 146 165 143 136 149 142 120 Audit (S) 130 102 129 139 (1) At the 1% significance level, can the director conclude that the mean daily expenses are greater for the sales staff than the audit staft? What assumptions did you make for the test you used above? (ii) Question 5 Surveys have been widely used by politicians around the world as a way of monitoring the opinions of the electorate. Six months ago, a survey was undertaken to determine the degree of support for a national party leader. Of a sample of 1100, 56% indicated that they would vote for this politician. This month, another survey of 800 voters revealed that 46% now support the leader. At 5% significance level, can we infer that the national leader's popularity has decreased? n 1100 P, SA a = 800 P first data CH, A T greater M. Mi-M0 Hois second dah

i need answer for question 3,4 and 5

Answer 1

(3) This is a test for two population means. Since we have the population SD we will use the normal distribution. The probabilties and the critical values are found using the normal probability and percentage tables.

(3) (3) = 4% level of significance Ampang 650 Gitical value Ampang W Kajang (2) So 201.85 185.31 32.48 26:13 We want to test whether the mean benefits are different . Two-tailed eest. at x = 40% Z 9 X disiding by Tinte it is two tailed) Test - Ho: Mi = M₂ I M - p2 =0 Lo - value ap (2 > IT.S.D. 22 P ( 7 > 3-397) = 2 r - P(243.397) :20 - 0.9997) p-value = 0.00681 Test Statistic :- (x - X2-146-M20) Since Test Stat > Critical Value 20185 - 185.31 - 32.482 26.15² 80 65 1 The null mean difference to Test Slatinic : 3.397 Also p-value < 0.04 (level of significance We reject the null hypotheris at 0.04 level and conclude the mean benefits are significantly different

(4)

	Sales(1)	Audit(2)
1	131	130
2	135	102
3	146	129
4	165	143
5	136	149
6	142	120
7		139
Total	855	912
Mean	142.500	130.286
SD	12.243	15.787
Var	149.900	249.238

Mean =

ur

SD =

Σ(1 – 1)2 η – 1 V

This is a test for two population means. Since we have unknown population SD we will use the t-distribution. The probabilties and the critical values are found using the t-dist tables.

pooled variance =

si(nı - 1) + si (n2 - 1) ni + n2 - 2

Assumptions:

The samples are independent

The samples are randomly selected

The samples come from a normal population.

The samples have same population SD

(5) This is a test for two population proportions so we are going to use the normal distribution.

pooled proportion $\hat{p}=\frac{\hat{p_{1}}(n_{1})+\hat{p_{2}}(n_{2})}{n_{1}+n_{2}}$

Level of siagnificance - 5% Critical value = 1.6449 p Six months agolt) a month ago (2) 1100 800 56% 46% I We are test it popularity has decreant Cone tailed test) There to test clain proportion a month ago is less than six pronths ago. Test Ho : R p piepe swine- Hi P PIZP2 Test Statistic ß. - P - U-PC Yon love pooled proportion i Pinit Pena Test Stat - 56% – 46% (0.518)01-0.513) (41018001 Io lo Test Stat 4.307 ualue = P(Z > TS.) = P(Z >4.307) 1- P(Z<4,307) = 0.0000 Since Test Start > Critical D -walue < 0.05 hialue L ni+n2 We reject the mill hypothesis and Conclude that the popularity has significantly decreased.