Hypothesis testing examples on z test

1. Hypothesis Testing Examples

2. Test of Hypothesis Concerning normal population, infinite, Large Samples, with known population variance. ( Z Test) If the size of sample exceeds 30 it should be regarded as a large sample. Testing Hypothesis about population mean μ . Z= x - μ S.E. of x Z= x– μ σ p /√n x = sample mean σ p = population s.d . n= sample size Population size is infinite .

3. Some Hypothesis Testing Examples Illustration1 (two-tailed Test) The mean lifetime of a sample of 100 light tubes produced by a company is found to be 1,580 hours Test the hypothesis at 5% level of significance that the mean lifetime of the tubes produced by the company is 1,600 hours with standard deviation of 90 hours. Solution: z= -2.22. Critical value of Z = ± 1.96

4. Illustration2 One Tailed (Upper Tailed ) An insurance company is reviewing its current policy rates. When originally setting the rates they believed that the average claim amount will be maximum Rs180000 . They are concerned that the true mean is actually higher than this, because they could potentially lose a lot of money. They randomly select 40 claims, and calculate a sample mean of Rs195000. Assuming that the standard deviation of claims is Rs50000 and set α = .05, test to see if the insurance company should be concerned or not.

5. Solution Step 1 : Set the null and alternative hypotheses H : μ≤ 180000 H 1 : μ > 180000 Step 2 : Calculate the test statistic z= = x – μ σ/√n = 1 .897 Step 3 : Set Rejection Region

6. 1.65

7. Step 4 : Conclude We can see that 1 .897 > 1.65, thus our test statistic is in the rejection region. Therefore we fail to accept the null hypothesis. The insurance company should be concerned about their current policies.

8. Illustration3 One Tailed (Lower Tailed) Trying to encourage people to stop driving to campus, the university claims that on average it takes at least 30 minutes to find a parking space on campus. I don’t think it takes so long to find a spot. In fact I have a sample of the last five times I drove to campus, and I calculated x = 20. Assuming that the time it takes to find a parking spot is normal, and that σ = 6 minutes, then perform a hypothesis test with level α = 0.10 to see if my claim is correct.

9. Solution Step 1: Set the null and alternative hypotheses H : μ ≥ 30 H 1 : μ < 30 Step 2 : Calculate the test statistic Z= x– μ σ /√n = - 3.727

10. Step 3: Set Rejection Region

11. Step 4: Conclude We can see that - 3.727 <-1.28 ( or absolute value is higher than the critical value) , thus our test statistic is in the rejection region. Therefore we Reject the null hypothesis in favor of the alternative. We can conclude that the mean is significantly less than 30, thus I have proven that the mean time to find a parking space is less than 30.

12. Exercise

13. Rejection region and conclusion

14. Test of Hypothesis Concerning normal population, finite, Large Samples, with known population variance. ( Z Test) Z= x– μ ( σ /√n) √(N-n )/(N-1) N= population size

15. Illustration 4 Suppose we are interested in a population of 20 industrial units of the same size, all of which are experiencing excessive labour turnover problems. The past record show that the mean annual turnover is 320 employees, with a standard deviation of 75 employees. A sample of 5 of these industrial units is taken at random which gives a mean of annual turnover as 300 employees. Is the sample mean consistent with the population mean? Test at 5% level.

16. Testing Hypothesis about difference Between two population means We assume that the populations are normally distributed. The null hypothesis is H : μ 1 = μ 2 i.e. H : μ 1 - μ 2 = Z = x 1 - x 2 √σ 1 2 / n 1 + σ 2 2 / n 2 In case σ 1 2 and σ 2 2 are not known then s 1 2 and s 2 2 can be used.

17. Illustration 1 A test given to two groups of boys and girls gave the following information: Gender Mean score S.D. Sample Size Girls 75 10 50 Boys 70 12 100 Is the difference in the mean scores of boys and girls statistically significant? Test at 1% level. Z=2.695, table value Z =2.58.

18. Illustration 2 Suppose you are working as a purchase manager for a company. The following information has been supplied to you by two manufacturers of electric bulbs: Company A Company B Mean life ( in hours) 1,300 1,288 Standard deviation( in hours) 82 93 Sample size 100 100 Which brand of bulbs are you going to purchase if you desire to take a risk of 5%

19. Solution Hint Take the null hypothesis that there is no significant difference In the quality of two brands of bulbs i . e. H : μ 1 = μ 2 Z= 0.968

20. Population normal, population infinite, sample size small (30 or less) and variance of population unknown t statistic . t = x – μ ( σ s /√n) with degree of freedom =(n-1) σ s = ∑ (X-X) 2 (n-1)

21. Illustration A tea stall near the New Delhi Railway Station is making a sales of 500 tea cups per day. Because of the development of bus stand nearby , it expects to increase its sales. During the first 12 days after the start of the bus stand, the daily sales were recorded which are as under: 550, 570,490, 615,505,580,570,460,600,580,530,526 . On the basis of this sample information, can one conclude that the tea stall’s sales have increased? Use 5% level of significance. t = 3.558

22. Testing Hypothesis about difference Between two population means We assume that the populations are normally distributed. Variance of population is unknown. The null hypothesis is H : μ 1 = μ 2 Small Sample t test t = x 1 – x 2 s√ (1/n 1 + 1/n 2 ) s = ∑(x 1 - x 1 ) 2 + ∑( x 2 – x 2 ) 2 ( n 1 + n 2 -2 ) with degree of freedom = ( n 1 + n 2 -2 )