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Council of Student Organizations De La Salle Universityβ Manila Rm. 402 Bro. Connon Hall, De La Salle University 2401Taft Avenue, Manila ENGSTAT Quiz 3 Reviewer Prepared by: Ma. Elizabeth Ann L. Uy Sampling Distribution of the Sample Mean EXAMPLE 1: Heights of Staring Players Player Height A 76 B 78 C 79 D 81 E 86 π = 80 Obtain the sampling distribution of the sample mean for samples of size 2 and for samples of size 4. Compare. Solution: For size 2: Sample A, B A, C A, D A, E B, C B, D B, E C, D C, E D, E Height mean 76 78 77 76 79 77.5 76 81 78.5 76 86 81 78 79 78.5 78 81 79.5 78 86 82 79 81 80 79 86 82.5 81 86 83.5 (x-ΞΌ) -3 -2.5 -1.5 1 -1.5 -0.5 2 0 2.5 3.5 (x-ΞΌ) Ξ£ (x-ΞΌ)2 β(π₯ β π)2 43.5 π=β =β = π. ππ π 10 2 9 6.25 2.25 1 2.25 0.25 4 0 6.25 12.25 43.5 For size 4: Sample A, B, C, D A, B, C, E A, B, D, E A, C, D, E B, C, D, E Heights 76 76 76 76 78 78 78 78 79 79 π=β 79 79 81 81 81 81 86 86 86 86 mean 78.5 79.75 80.25 80.5 81 (x-ΞΌ) -1.5 -0.25 0.25 0.5 1 Ξ£ (x-ΞΌ)2 2 (x-ΞΌ) 2.25 0.0625 0.0625 0.25 1 3.625 β(π₯ β π)2 3.625 =β = π. ππ π 5 The higher the sample size, the higher the accuracy there would be because Ο will be smaller. EXAMPLE 2: Prices of New Mobile Homes 53.8 53.1 35.9 42.0 54.4 59.4 42.5 62.7 Prices of new mobile homes (in thousand pesos) 45.2 42.9 49.9 48.2 41.6 49.7 43.7 52.7 47.7 41.5 57.2 45.1 50.3 50.0 41.9 62.8 46.6 60.3 43.9 56.4 58.9 35.3 37.3 49.8 48.6 58.9 39.7 63.9 a) Determine Point Estimate; mean = 49.27 b) Identify the Distribution of the variable; ππ₯ = 7.2 β36 = π. π c) Show that 95.44% of all samples have the property that the interval from x β 2.4 to z + 2.4 contains ΞΌ. ππ₯ β 2ππ₯ π‘π ππ₯ + 2ππ₯ 49.27 β 2(1.2) π‘π 49.27 + 2(1.2) π·ππ ππ, πππ ππ π·ππππ, πππ The β95.44β part of the 68.26-95.44-99.74 rule states that, for a normally distributed variable, 95.44% of all possible observations lie within two standard deviations to either side of the mean. Applying the rule to the variable x, we see that 95.44% of all samples of 36 new mobile homes have mean prices within 2.1 to 2.4 of ΞΌ or equivalently 95.44% of all sample of 36 new mobile homes have the property that the interval from x-2.4 to x+2.4 contains ΞΌ. EXAMPLE 3: Age of the civilian labor force Find a 95% confidence interval for the mean age, ΞΌ, of all people in the civilian labor force. Assume Ο=12.1 years, n = 50. 22 32 33 43 60 51 27 28 42 35 58 34 16 37 41 37 31 39 40 29 40 45 49 19 28 65 33 43 31 33 42 38 29 21 35 37 24 26 34 32 43 19 30 62 37 26 34 38 38 33 Step 1: Confidence level = 95% 1 β πΌ = 1 β 0.95 = 0.05 ππ₯/2 = π0.025 = 1.96 Refer to the table for this part* Step 2: Confidence Interval π₯ β (ππΌ π₯ 2 35.98 β (1.96) ( π βπ ) π‘π π₯ + (ππΌ π₯ 2 π βπ ) 12.1 12.1 ) π‘π 35.98 + (1.96) ( ) β50 β50 ππ. πππ ππ ππ. πππ HYPOTHESIS TESTING EXAMPLE 4: Quality Assurance (taken from the book) A company that produces snack foods uses a machine to package a 454 g bags of pretzels. We assume that the net weights are normally distributed and that the population standard deviation of all such weights is 7.8 g. A simple random sample of 25 bags of pretzels has the net weights, in grams, displayed in the Table 1. Do the data provide a sufficient evidence to conclude that the packaging machine is not working properly? We use the following steps to answer the question. Table 1. Weights, in grams, of 25 randomly selected bags of pretzels 465 449 468 446 447 456 442 433 447 456 438 449 454 456 456 454 446 463 452 435 447 447 450 444 450 a) State the null and alternative hypothesis for the hypothesis test. b) Discuss the logic behind carrying out the hypothesis test. c) Identify the distribution of the variable x, that is, the sampling distribution of the sample mean for sample size 25. d) Obtain a precise criterion for deciding whether to reject the null hypothesis in favor of the alternative hypothesis. e) Apply the criterion in part d to the sample data and state the conclusion. Solution: a) The null and alternative hypothesis for the hypothesis test are: a. π»π : π = 454 π (π‘βπ πππππππππ πππβπππ ππ π€ππππππ ππππππππ¦) b. π»π : π πππ‘ = 454 π (π‘βπ πππππππππ πππβπππ ππ πππ‘ π€ππππππ ππππππππ¦) b) Basically, the logic behind carrying out the hypothesis test is this: If the null hypothesis is true, that is, if ΞΌ=454 g, the mean weight, x, of the sample of 25 bags of pretzels should approximately equal 454g. We say βapproximately equalβ because we cannot expect a sample mean to equal exactly the population mean; some sampling error is to be anticipated. However, if the sample mean weights differ βtoo muchβ from 454 g, we would be inclined to reject the null hypothesis and conclude that the alternative hypothesis is true. As we show in part d, we can use our knowledge of the sampling distribution of the sample mean to decide how much difference is βtoo muchβ. c) Because n=25, Ο=7.8, and the weights are normally distributed a. ππ₯ = π (π€βππβ π€π πππβ² π‘ππππ€) π 7.8 b. ππ₯ = π = = 1.56 β β25 c. X is normally distributed In other words, for samples of size 25, the variable x is normally distributed with mean ΞΌ and standard deviation 1.56 g. d) The β95.44β part of the 68.26-95.44-99.74 rule states that, for a normally distributed variable, 95.44% of all possible observations lie within two standard deviations to either side of the mean. Applying this part of the rule to the variable x and referring to part c, we see that 95.44% of all samples of 25 bags of pretzels have mean weights within 2(1.56)=3.12 g of ΞΌ. Or equivalently, only 4.56% of all samples of 25 bags of pretzels have mean weights that are not within 3.12 g of ΞΌ as illustrated in Figure 1. Figure 1. 95.44% of all samples of 25 bags of pretzels have mean weights within two standard deviations (3.12 g) of ΞΌ Thus if the mean weight, x, of the 25 bags of pretzels sampled is not within the two standard deviations (3.12g) of 454g, we have evidence against the null hypothesis. Why? Because observing such a sample mean would occur by chance only 4.56% of the time if the null hypothesis, ΞΌ=454g, is true. In summary, we have obtained the following precise criterion for deciding whether to reject the null hypothesis. The criterion is portrayed graphically in Figure 2a. Figure 2. a) criterion for deciding whether to reject the null hypothesis; b) normal curve associated with x if the null hypothesis is true, superimposed on the decision criterion If the mean weight, x, of the 25 bags of pretzels sampled is more than two standard deviations (3.12g) from 454g, reject the null hypothesis, ΞΌ=454g, and conclude that the alternative hypothesis, ΞΌ not = 454g, is true. Otherwise, do not reject the null hypothesis. If the null hypothesis is true, the normal curve associated with x is the one with parameters 454 and 1.56; that normal curve is superimposed on Figure 2a in Figure 2b. e) The mean weight, x, of the sample of 25 bags of pretzels whose weights are given in Table 1 is 450 g. Therefore, π₯ β 454 450 β 454 π§= = = β2.56 1.56 1.56 That is, the sample mean of 450 g is 1.56 standard deviations below the null hypothesis population mean of 454 g, as shown in Figure 3. Because the mean weight of the 25 bags of pretzels sampled is more than two standard deviations from 454 g, we reject the null hypothesis, ΞΌ=454g, and conclude that the alternative hypothesis, ΞΌ not = 454g, is true. Figure 3. Graph showing the number of standard deviations that the sample mean of 450 g is from the null hypothesis population mean of 454g. Interpretation: The data provided sufficient evidence to conclude that the packaging machine is not working properly. Reference: 5th Edition Scheaffer, Mulekar, and McClave (2010). Probability and Statistics for Engineers,