One-tailed Test PDF

# One-tailed Test ## Cape Peninsula University of Technology **Department of Clothing and Textile Technology** **Faculty of Engineering and the Built Environment** **Cape Peninsula University of Technology** **Bellville Campus** **Cape Town, 7535** **Subject code: MTS150X and MTS150S** ## One-tai...

# One-tailed Test ## Cape Peninsula University of Technology **Department of Clothing and Textile Technology** **Faculty of Engineering and the Built Environment** **Cape Peninsula University of Technology** **Bellville Campus** **Cape Town, 7535** **Subject code: MTS150X and MTS150S** ## One-tailed Test of Ho * α = 0.05 * Region of Acceptance (No significant difference) * **Chapters - 13 part 2 (Hypothesis Testing: One-tailed test)** ## One-Tail Hypothesis Test: Left tail * **Ho: u ≥** * **Ha: u <** * α the left tail * Z (critical) * Fail to Ho ## One-Tail Hypothesis Test: Right tail * **Ho: u ≤** * **Ha: u >** * α the right tail * Z (critical) * Fail to Ho ## Test Statistics ## Example 1: Testing a Claim About Average Fabric Strength * A textile company claims that the average strength of their fabric is at least 22.7 kg. *A sample of 40 fabric pieces has an average strength of 21.8 kg with a standard deviation of 2.3 kg. * Test this claim at a 5% significance level. #### Null * **Hypothesis (Ho): μ ≥ 22.7 kg** #### Decision Rule * Reject H0 if z < -1.645 (for a one-tailed test at α = 0.05) * **Conclusion:** Since -2.48 < -1.645, we reject the null hypothesis. There is sufficient evidence to conclude that the average strength of the fabric is less than 22.7 kg. #### Alternative Hypothesis * **(Ha): μ < 22.7 kg** #### Significance Level * 0.05 #### Test Statistic * = -2.48 ## The END **Sampling** | **Parameters** ------- | -------- Mean | Standard deviation | σ Population (sample) size | N Population (sample) proportion | ## Standard error: ## Interval Estimates * The probability associated with an interval estimate is called the **confidence level**. * 90% confidence level means 1.65 (10% level of significance) * 95% confidence level means 1.96 (5% level of significance) * 99% confidence level means 2.58 (1% level of significance) ## Areas Under the One-Tailed Standard Normal Distribution This table provides the area between the mean and some Z score. For example, when Z score = 1.45, the area = 0.4265. | Z | 0.00 | 0.01 | 0.02 | 0.03 | 0.04 | 0.05 | |---|---|---|---|---|---| |0.0| 0.0000 | 0.0040 | 0.0080 | 0.0120 | 0.0160 | 0.0195 | |0.1| 0.0398 | 0.0438 | 0.0478 | 0.0517 | 0.0557 | 0.0596 | |0.2| 0.0793 | 0.0832 | 0.0871 | 0.0910 | 0.0948 | 0.0987 | |0.3| 0.1179 | 0.1217 | 0.1255 | 0.1293 | 0.1331 | 0.1368 | |0.4| 0.1554 | 0.1591 | 0.1628 | 0.1664 | 0.1700 | 0.1736 | |0.5| 0.1915 | 0.1950 | 0.1985 | 0.2019 | 0.2054 | 0.2088 | |0.6| 0.2257 | 0.2291 | 0.2324 | 0.2357 | 0.2389 | 0.2422 | |0.7| 0.2580 | 0.2611 | 0.2642 | 0.2673 | 0.2704 | 0.2734 | |0.8| 0.2881 | 0.2910 | 0.2939 | 0.2967 | 0.2995 | 0.3023 | |0.9| 0.3159 | 0.3186 | 0.3212 | 0.3238 | 0.3264 | 0.3289 | |1.0| 0.3413 | 0.3438 | 0.3461 | 0.3485 | 0.3508 | 0.3531 | |1.1| 0.3643 | 0.3665 | 0.3686 | 0.3708 | 0.3729 | 0.3749 | |1.2| 0.3849 | 0.3869 | 0.3888 | 0.3907 | 0.3925 | 0.3944 | |1.3| 0.4032 | 0.4049 | 0.4066 | 0.4082 | 0.4099 | 0.4115 | |1.4| 0.4192 | 0.4207 | 0.4222 | 0.4236 | 0.4251 | 0.4265 | |1.5| 0.4332 | 0.4345 | 0.4357 | 0.4370 | 0.4382 | 0.4394 | |1.6| 0.4452 | 0.4463 | 0.4474 | 0.4484 | 0.4495 | 0.4505 | |1.7| 0.4554 | 0.4564 | 0.4573 | 0.4582 | 0.4591 | 0.4599 | |1.8| 0.4641 | 0.4649 | 0.4656 | 0.4664 | 0.4671 | 0.4678 | |1.9| 0.4713 | 0.4719 | 0.4726 | 0.4732 | 0.4738 | 0.4744 | |2.0| 0.4772 | 0.4778 | 0.4783 | 0.4788 | 0.4793 | 0.4798 | |2.1| 0.4821 | 0.4826 | 0.4830 | 0.4834 | 0.4838 | 0.4842 | |2.2| 0.4861 | 0.4864 | 0.4868 | 0.4871 | 0.4875 | 0.4878 | |2.3| 0.4893 | 0.4896 | 0.4898 | 0.4901 | 0.4904 | 0.4907 | |2.4| 0.4918 | 0.4920 | 0.4922 | 0.4925 | 0.4927 | 0.4929 | |2.5| 0.4938 | 0.4940 | 0.4941 | 0.4943 | 0.4945 | 0.4946 | |2.6| 0.4953 | 0.4955 | 0.4956 | 0.4957 | 0.4959 | 0.4960 | |2.7| 0.4965 | 0.4966 | 0.4967 | 0.4968 | 0.4969 | 0.4970 | |2.8| 0.4974 | 0.4975 | 0.4976 | 0.4977 | 0.4978 | 0.4979 | |2.9| 0.4981 | 0.4982 | 0.4982 | 0.4983 | 0.4984 | 0.4984 | |3.0| 0.4987 | 0.4987 | 0.4987 | 0.4988 | 0.4988 | 0.4989 | |3.1| 0.4990 | 0.4991 | 0.4991 | 0.4991 | 0.4992 | 0.4992 | |3.2| 0.4993 | 0.4993 | 0.4994 | 0.4994 | 0.4994 | 0.4995 | |3.3| 0.4995 | 0.4995 | 0.4995 | 0.4996 | 0.4996 | 0.4996 | |3.4| 0.4997 | 0.4997 | 0.4997 | 0.4997 | 0.4997 | 0.4997 | |3.5| 0.4998 | 0.4998 | 0.4998 | 0.4998 | 0.4998 | 0.4998 | |3.6| 0.4998 | 0.4998 | 0.4999 | 0.4999 | 0.4999 | 0.4999 | |3.7| 0.4999 | 0.4999 | 0.4999 | 0.4999 | 0.4999 | 0.4999 | |3.8| 0.4999 | 0.4999 | 0.4999 | 0.4999 | 0.4999 | 0.4999 | |3.9| 0.5000 | 0.5000 | 0.5000 | 0.5000 | 0.5000 | 0.5000 | ## Standard Normal Table | Z | 0.00 | 0.01 | 0.02 | |---|---|---|---| | 0.0 | 0.0000 | 0.0040 | 0.0080 | | 0.1 | 0.0398 | 0.0438 | 0.0478 | | 1.6 | 0.4452 | 0.4463 | 0.4474 | | 1.9 | 0.4713 | 0.4719 | 0.4726 | | 2.5 | 0.49379 | 0.49396 | 0.4940 | * 90% confidence level means 1.65 (0.90/2 = 0.45, If we look up for above value in the Z table, the value is 1.65) * 95% confidence level means 1.96 (0.95/2 = 0.475, If we look up for above value in the Z table, the value is 1.96) * 99% confidence level means 2.58 (0.99/2 = 0.495, If we look up for above value in the Z table, the value is 2.58) ## Estimation ### Estimating when σ is known: * Finite population, (σ is known) * Infinite population, (σ is known) ### Estimating when σ is unknown, n > 30: * Finite population, (σ is unknown) * Infinite population, (σ is unknown) ## Example 3 (σ is unknown) * The smart packaging company wants to estimate the average spending per customer. * A sample of 100 customers spent an average of R3.50 each with a sample standard deviation of R0.75. * Estimate the true average expenditure with a 90% confidence level. ## Example 4 (σ is unknown) * The A1 company has received a shipment of 100 computers and the manager wants to estimate the average mass of the computers. * A sample of 36 computers has shown that the average mass was 3.6 kg with a standard deviation of 0.6 kg. * Construct an interval estimate of the true average mass of the computers with a 99% confidence level. #### Finite case: * N = 100 * n = 36 * Z = 2.58 * s = 0.6 #### Finite population, * = 3.6 2.58 = 3.6 0.207 * ≈ 3.393 < < 3.807 #### Sample Size * n=( ) #### Where: * n = sample size * s = standard deviation * Z = confidence interval ## Hypothesis Testing * It is an assumption to be tested with the objectives of making statistical decision based on a scientific procedure. #### Null hypothesis (Ho): * No difference between the values of the parameters #### Alternative hypothesis (H1): * Null hypothesis is not true. ## Two-tailed Test ## Cape Peninsula University of Technology **Department of Clothing and Textile Technology** **Faculty of Engineering and the Built Environment** **Cape Peninsula University of Technology** **Bellville Campus** **Cape Town, 7535** **Subject code: MTS150X and MTS150S** ## Two-tailed Test of Ho * α = 0.05 * Region of Acceptance (No significant difference) * Region of rejection significant difference 2.5% * **Ho: μ = ** * **H1: μ ≠ ** * α = 0.05 * ☐ = rejection region * Reject Ho if: Z ≤ -1.96 or Z ≥ 1.96 * If Ho: =; then Ha: ≠ ; two- tail or * ≤ : one- tail to the left * ≥ : one- tail to the right * If Ho: ≥; then H1: < ; (one-tail) * If Ho: ≤; then Ha: > ; (one-tail) ## Example * The owner of the Mr. Cake company stated that the average number of breads sold daily was 1500. * An employee wants to test the accuracy of the owner’s statement. * A random sample of 36 days showed that the average daily sales were 1450 breads. * Using a 1% level of significance and assuming that o = 120 breads, what should the employee conclude? * **Ho: =1500** * **Ha: ≠ 1500** * α = 0.01,, σ = 120, n = 36 #### Decision rule: * Reject Ho if Z ≤ -2.58 or Z ≥ 2.58 #### Test statistics: * = -2.5 #### Conclusion: * The Null hypothesis should not be rejected (accept), the average daily sales were 1500 breads. ## Two-tail Test * Mr. Soda thinks that his business sells an average of 17 cans of Soda Delight daily. * His partner, Mr. Water thinks this average is wrong and went to investigate. * A random sample of 36 days showed a mean of 15 cans and a sample standard deviation of 4 cans. * Test the accuracy of Mr. Water’s statement at 10% level of significance. * **Ho: = 17 Ha: ≠ 17** * α = 0.10,, s = 4, n = 36 #### Decision rule: * Reject Ho if Z ≤ -1.65 or Z ≥ 1.65 #### Test statistics: * = -3 #### Conclusion: * Since Z < -1.65, the Null hypothesis should be rejected, the average daily sales were not 17 cans. ## Test for Proportions * A local newspaper has stated that only 25% of all college students read newspapers daily. * A random sample of 200 students showed that 45 of them were daily readers of newspapers. * Test the accuracy of the newspapers statement using a significance level of 5%. * **Ho: = 0.25** * **Ha: ≠ 0.25 α = 0.05_p = 45/200 =0.225, n = 200_q = 1-0.225 = 0.775** #### Decision rule: * Reject Ho if Z ≤ -1.96 or Z ≥ 1.96 #### Test statistics: * = -0.85 #### Conclusion: * Since Z falls between±1,96, null hypothesis cannot be rejected, only 25% student read news papers daily. ## Thank you.