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(Type your name here) 1 Name: Name of Tutor: Subject: Date: Hypothesis Testing 2 A hypothesis refers to a claim or an opinion given about an item or issue .This claim has to be tested statistically to establish whether it is correct or not. The two hypothesis to be tested are the null hypothesis (H0 ) and the alternative hypothesis (H1 ). Both the null and alternative hypotheses are mutually exclusive and exhaustive. They cannot share the same outcome in common. Together they account for all possible outcomes. The null hypothesis refers to the hypothesis being tested.eg. a bureau of standard may walk into a sugar making company with the intention of confirming that 2kg bags of sugar produced are actually 2kgs and no less. Hence the hypothesis test conducted has null hypothesis being Ho = each bag weigh 2kgs. The test results will either confirm or disprove this. The alternative hypothesis is a contradiction of the null hypothesis formulated on the fact that the belief might be untrue. Therefore, we will reject it. When the null hypothesis is rejected, we accept the alternative hypothesis. The procedure to be followed in hypothesis testing include; formulating the hypothesis H0 and H1; determining the analysis plan in form of the significance level; identifying the test statistic of the sample data and finally formulating the decision. For small sample size where (n<30) the arithmetic mean are not normally distributed. Hence, students t distribution must be used to estimate the population mean. In our case, we are going to carry out three research question and test their hypothesis Question 1 A tyres manufacturing company by the name Super Trends Tyres Ltd introduced new brands of tyres, which in their advertisement claim to be superior to their only competitor brand, ‘Roadsmaster.’ The Roadmaster brand manager disputed this for quite some time and termed it as an advertisement scheme. To settle this, the brand managers for both companies agreed to run a road test for each brand using ten Nissan vehicles of uniform weight and identical specifications with each car fitted with both brands tyres; one brand in front and the other at the rear. The cars were then driven to cover a distance of 5000 kilometers each and the trend wear was recorded. Results were as follows: 3 Super Trendy tyres Roadmaster Tyres Centimeters Centimeters 1 1.08 1.12 Paired Difference (x) -0.04 2 1.06 1.09 -0.03 3 1.24 1.16 0.08 4 1.20 1.24 -0.04 5 1.17 1.23 -0.06 6 1.21 1.25 -0.04 7 1.18 1.20 -0.02 8 1.10 1.15 -0.05 9 1.22 1.19 0.03 10 1.09 1.13 -0.04 Sum -0.21 The hypothesis to test is: H0=Super trendy tyres are not superior to Roadmasters H1=Super Trendy tyres are superior to Roadmasters We compute the sum paired differences of the samples. N =10 Sum =-0.21 Compute the Mean -0.21/10 x =-0.021 Standard Deviation 4 S= X X 2 n 1 for n < 30 =0.0431 Calculating the test statistics: Test statistic assuming the sampled data is normally distributed with mean of 0 t. calculated = X SX where = - 0.21/ (0.431/√10) =-1.541 From the t tables t value=t9. 0.995 =3.25 Conclusion: Accept H0 :Super Trendy tyres are not superior to Roadmaster tyres From this we note that the Super trend advertising may be just a marketing gimmick that is trying to outdo the Roadmaster tyre Question 2 A research was carried out in a city hospital .A doctor claims that boys at the age if 17 year have an average body temperature that is higher than the commonly accepted average human temperature of 98.60 Fahrenheit. To test this, a sample of 25 boys, each of 17 years of age is selected. The temperature of the 17 year olds boys is found to be an average of 98.9 0, and with a standard deviation of 0.6 0. Hypothesis to be tested 5 The claim being investigated is that; -The average body temperature of 17 year olds is > 98.60 That is x ≥ 98.6. The negation of this is that: -The population average is not greater than 98.6 degrees. That this is x < 98.6. Null hypothesis H0 : x = 98.6. The alternative hypothesis, or H1: x >98.6. let alpha be a 5% level of significance and alpha will be equal to 0.05. Choice of Test Statistic and Distribution We will use the standard normal distribution. A table of z-scores will hence be necessary. Here n=25, which has square root of 5, standard error is 0.6/5 = 0.12. The test statistic is therefore z = (98.9-98.6)/.12 = 2.5 At a 5% significance level, the critical value for a one tailed test is found from the z table of zscores to be 1.645. Since the test statistic does fall within the critical region, we reject the null hypothesis. Using the P Value The z-score of 2.5 has a p-value of 0.0062. Since this is less than the significance level of 0.05, we reject the null hypothesis. Conclusion In conclusion, we see from the statistical evidence above that either a rare event has occurred, or that the average temperature of 17 year olds is in fact greater than 98.60. Question 3 6 I carried out a research on advertising. The research was on a company by the name Star beam Advertising Company has prepared two different television Commercials advert for Sean’s designer women’s jeans. One of the commercial is humorous, and the other is Serious. A test screening involves eight consumers who are asked to view and rate the commercials by using a standard scale of measurement with the higher scores indicating more favorable and a lower score indicating unfavorable responses. The results are listed below. A 0.05 level of significance is used. The sample data used is as follows: User Commercial with humor Serious Commercial A 26.2 B 20.3 C 25.4 D 19.6 E 28.3 F 21.5 G 23.7 H 24.0 24.1 21.3 23.7 18.0 20.1 25.8 21.4 22.4 Hypothesis tested: Ho = Mean difference is zero H1 ≠mean difference of Zero Assumption The assumption made is that the differences between the ratings for each and every consumer are from a population that has a normal distribution To get the we compute the sum paired differences of the samples. Compute the Mean Mean of paired difference x =9.6 Standard Deviation Sd=18.98654 Using p value: To compute the test statistics or p value P(x_Bar < -9.6) =P(Z< (-9.6-0) / (10/√8) 7 =3.822 δ Critical T value = x̄ + Zα/2√𝑛 = x̄ ±2.365 P =0.0065 Conclusion Reject Ho: Mean difference is zero There is sufficient evidence to warrant rejection of the claim that the differences between the ratings of commercial have a mean of zero. There does not appear to be a difference and because humorous commercials have a generally higher rating, this implies the humorous commercials are better.