This document provides examples of hypothesis testing in statistics. It covers different scenarios and explains the steps involved in conducting hypothesis tests. Topics include null and alternative hypotheses, significance level, test statistics, and conclusion.
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STATISTICS [Document subtitle] [DATE]
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Question 1 The requisite hypotheses are as stated below. Null Hypothesis: μ = 22oC Alternative Hypothesis: μ ≠ 22oC Level of significance = 0.05 As the population standard deviation is known and also the sample size is greater than 30, hence in accordance with the Central Limit Theorem, the relevant test statistics would be Z and not T. The formula for computation of t stat is shown below. Z=(/ (σ/n0.5) Putting the respective input values, we get Z = (20-22)/(1.5/400.5) = -8.43 The requisite two tail p value for the above z value comes out to be zero. Since the computed p value (0.00) is lower than the assumed level of significance (0.05), hence the available evidence would lead to rejection of null hypothesis and acceptance of alternative hypothesis. Hence, it can be concluded that the population mean temperature is different from 22oC. As a result, the original claim is false. Question 2 The requisite hypotheses are as stated below. Null Hypothesis: sf2≤ sM2i.e. the variation in blood pressure of females is lower than or equal to variation in blood pressure of males Alternative Hypothesis: sf2> sM2i.e. the variation in blood pressure of females is greater than variation in blood pressure of males
Level of significance = 0.05 The relevant test statistic for the given case would be F value Test statistic i.e. F value = sf2/ sM2= 22.72/20.12= 1.275 The critical value for F with df1= 15, df2= 16 and level of significance = 0.05 comes out as 2.35 Since computed test statistic is lower than F critical value, hence the available evidence would not lead to rejection of null hypothesis and acceptance of alternative hypothesis. Thus, it can be concluded that the variation in blood pressure of females is lower than or equal to variation in blood pressure of males. Question 3 Step 1 Null hypothesisProduction process is not out of control when defect does not exceed 3%. Alternative hypothesisProduction process is out of control when defect does exceed 3%. Step 2 Significance level = 0.01 Step 3 The z stat would be computed as shown below. Sample proportion P= 5.9% Sample size = 85 items
Step 4 The p value for z = 1.57 and right tailed hypothesis testing Step 5 The null hypothesis would only be rejected when the p value is lower than the significance level. In present case, the p value comes out to be higher than the significance level and hence, null hypothesis would not be rejected. Therefore, alternative hypothesis would not be accepted. Therefore, it can be concluded that the production process is not out of control when defect does not exceed 3%. Question 4 Sample size = 50 Response time for two companies A and B Step 1 Null hypothesisMean response time of company A is same as company B. Alternative hypothesisMean response time of company A is not same as company B. Step 2 Significance level = 0.01 Step 3
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The z stat would be computed as shown below. Step 4 The p value for z = 2.25 and two tailed hypothesis testing Step 5 The null hypothesis would only be rejected when the p value is lower than the significance level. In present case, the p value comes out to be higher than the significance level and hence, null hypothesis would not be rejected. Therefore, alternative hypothesis would not be accepted. Therefore, it can be concluded that the mean response time of company A is same as company B. Question 5 Competitors scores before and after training Sample size = 7 gymnasts Step 1 Null hypothesisMean scores of gymnasts before and after training is same. Alternative hypothesisMean scores of gymnasts before is lower than after training.
Step 2 Significance level = 0.01 Step 3 The t stat would be computed as shown below. Step 4 Critical value approach Degree of freedom = 7-1 = 6 Significance level = 0.01 The given critical value = -3.143 (left tailed hypothesis test) Step 5 The null hypothesis would only be rejected when the t calculated is lower than the t critical value.
In present case, the t calculated comes out to be higher than the t critical value and hence, null hypothesis would not be rejected. Therefore, alternative hypothesis would not be accepted. Therefore, it can be concluded that mean scores of gymnasts before and after training is same. Thus, it is essential to improve the training program so as to make the gymnasts competition score effective. Question 6 The requisite hypotheses are as stated below. Null Hypothesis: μ = 2,20,000 miles Alternative Hypothesis: μ > 2,20,000 miles Significance level = 0.01 Since the population standard deviation is unknown and also the sample size is less than 30, hence the appropriate test statistic is T. The formula for computation of t stat is shown below. T =(/ (s/n0.5) Putting the respective input values, we get Test statistics (T) = (226450-220000)/(11500/230.5) = 2.690 The t critical value for one tail with df = 23-1 = 22 and 0.01 significance level comes out as 2.508. Since computed test statistic exceeds the t critical value, hence the available evidence would lead to rejection of null hypothesis and acceptance of alternative hypothesis. Hence, it can be concluded that the mean lifetime of car engines is greater than 2,20,000 miles.