# One Tests of Hypothesis.pdf

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1 Universitas Indonesia 1.   The Liquid Chlorine: No 18 Hal 342-343 The liquid chlorine added to swimming pools to combat algae has a relatively short shelf life before it loses its effectiveness. Records indicate that the mean shelf life of a 5-gallon jug of chlorine is 2,160 hours  (90 days). As an experiment, Holdlonger was added to the chlorine to find whether it would increase  the shelf life. A sample of nine jugs of chlorine had these shelf lives (in hours): At the .025  level, has Holdlonger increased the shelf life of the chlorine? Estimate the p-value. Jawab: Step 1:  State the null hypothesis and the alternate hypothesis.   H0 : µ   2.160 H1 : µ > 2.160 (Kata kunci: increase ) Step 2:  Select the level of significance.   = 0,025 seperti yang tertera di soal df = n-1 = 9-1 = 8 Critical t value = 2,306 Step 3:  Select the test statistic. Menggunakan tabel t. Hal ini diketahui karena n < 30 dan   tidak diketahui. Step 4:  Formulate the decision rule. Ho ditolak jika t > 2,306   2.159 2.170 2.180 2.179 2.160 2.167 2.171 2.181 2.185   2 Universitas Indonesia Step 5: Make a decision and interpret the result. X bar   =   =    = 2.172 S =       =      = 9,382311963 t =    √   =     √    = 3,97912   NoX(X - Xbar)^2 12.159 18122.170 632.180 5742.179 4352.160 15562.167 3072.171 282.181 7392.185 158 Total19.552 704  3 Universitas Indonesia Keputusan = Ho ditolak  . Hal ini disebabkan t value sebesar 3,97912 berada di wilayah Ho ditolak. Kesimpulan = Holdlonger meningkatkan  umur penyimpanan klorin. Step 6: Estimate the p-value. P value dari 3,97912 berada pada line df 8, diantara significant level 0,005 dan 0,0005. Dapat disimpulkan bahwa P value kurang dari 0,005. 2.   Hugger Polls: No 20 Hal 343 Hugger Polls contends that an agent conducts a mean of 53  in-depth home surveys every week. A streamlined survey form has been introduced, and Hugger wants to evaluate its effectiveness. The number of in-depth surveys conducted during a week by a random sample of 15 agents are: At the .05  level of significance, can we conclude that the mean number of interviews conducted by the agents is more than 53 per week? Estimate the  p-value. Jawab: Step 1:  State the null hypothesis and the alternate hypothesis.   H0 : µ   53 H1 : µ > 53 (Kata kunci: more than ) Step 2:  Select the level of significance.   = 0,05 seperti yang tertera di soal df = n-1 = 15-1 = 14 Critical t value = 1,761   535750555854605259626060515956  4 Universitas Indonesia Step 3:  Select the test statistic. Menggunakan tabel t. Hal ini diketahui karena n < 30 dan   tidak diketahui. Step 4:  Formulate the decision rule. Ho ditolak jika t > 1,761 Step 5: Make a decision and interpret the result. X bar   =   =    = 56 NoX(X - Xbar)^2 1 53 122 57 03 50 414 55 25 58 36 54 67 60 138 52 199 59 710 62 3111 60 1312 60 1313 51 2914 59 715 56 0 Total846196

Oct 7, 2019

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Oct 7, 2019
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