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1 of 3 ID: MST.FET.HT.CMHT.01.0020A [2 marks] Standard normal cumulative distribution function A hypothesis test is to be performed in order to test the proportion

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1 of 3 ID: MST.FET.HT.CMHT.01.0020A [2 marks] Standard normal cumulative distribution function A hypothesis test is to be performed in order to test the proportion of people in a population that have some characteristic of interest. Select all of the pieces of This table gives values of the standard normal cumulative information that are needed in order to calculate the test statistic for the hypothesis test: distribution function, F(z), for certain values of z. That is, the table gives the area under the standard normal probability the actual population proportion density function from negative infinity to z. Note that the identity F(-z) = 1 - F(z) can be used for negative values of z. the proposed population proportion O the level of significance used If you would like help in using this table, some instructions have been provided (show instructions) the characteristic of interest F(z) - Standard normal cumulative distribution function the size of the sample selected 2 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 the sample proportion calculated 0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359 0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5753 2 of 3 ID: MST.FET.HT.CMHT.03.0030A [1 mark] 0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0 8 0.5987 0.6026 0.6064 0.6103 0.6141 0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0. 81 0.6368 0.6406 0.6443 0.6480 0.6517 You may find the standard normal table useful in answering this question. Consider the hypothesis test with the following null and alternative hypotheses: 0.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.6879 Ho: It = 0.5 0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.7224 HA: TT = 0.5 0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.7549 If the level of significance used is a = 0.025, the region of rejection is the set of numbers: 0.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.7852 0.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.8133 O whose absolute value is less than 1.96 0.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.8389 greater than 2.24 1.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.8621 O whose absolute value is greater than 1.96 1.1 0.8643 0.8665 0.8686 0.8708 0. 29 0.8749 0.8770 0.8790 0.8810 0.8830 greater than 1.96 1.2 0.8849 0.8869 0.8888 0.8907 0.8925 0. 44 0.8962 0.8980 0.8997 0.9015 O whose absolute value is greater than 2.24 whose absolute value is less than 2.24 1.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.9177 1.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319 O less than -2.24 O less than -1.96 1.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.9441 1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.9545 1.7 0.9554 0.9573 2 0.9591 0.9599 0.9608 0.9616 0.9625 0.9633 -3 of 3 ID: MST.FET.HT.CMHT.02.0030A [2 marks] 1.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.9706 Painadol is a well trusted, popular pain-killer used by millions throughout the country. Recently an epidemic of 1.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.9767 headaches has hit the nation and the manufacturers of Painadol have created a new, stronger painkiller called Painadene. The manufacturers are interested in testing whether the speed of pain relief is different with Painadene. 2.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.9817 It is known that the mean time taken by Painadol tablets to relieve pain is 27 minutes and the standard deviation is Painado 2.1 0.9821 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9854 0.9857 5 minutes. 2.2 0.9861 0.9864 0.9868 0.9871 0.9875 0. 0.9878 0.9884 0.9887 0.9890 The manufacturers would like to construct a hypothesis test for the mean time (p) taken for Painadene tablets to Painadene 0.9893 0.9896 relieve pain assuming the same population standard deviation () as the Painadol tablets. A random sample of 32 people with headaches tried the new Painadene tablets and the mean time taken to relieve the pain was calculated 0.9918 0.9920 0.9922 as 25.08 minutes. The hypotheses that will be used by the manufacturers are Ho: U = 27 and HA: U # 27. 2.5 0.9938 0.9940 0.9941 You may find this standard normal table useful throughout the following questions. 2.6 0.9953 0.9955 0.9956 0.9962 0.9963 0.9964 a) Calculate the test statistic (z) that corresponds to the sample and hypotheses. Give your answer as a decimal to 0.9965 0.9966 0.9967 0.9972 0.9973 0.9974 3 decimal places. 2.8 0.9974 0.9975 0.9976 0.9977 0.9977 0 accept 0.9978 0.9979 0.9979 0.9980 0.9981 2 = reject 2.9 0.9981 0.9982 0.9982 0.9983 0.9984 0.9984 0.9985 0.9985 0.9986 0.9986 not reject 3.0 0.9987 0.9987 0.9987 0.9988 0.9988 0.9989 0.9989 0.9989 0.9990 0.9990 b) Using the test statistic for Painadol's hypothesis test and a level of significance of a = 0.05, Painadol should the null hypothesis

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