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Let Us Do Activity 2: Express Me Directions: Formulate the null and alternative hypothesis, identify the level of significance, illustrate if CLT can be used and illustrate the rejection region. Teacher Sara reported that out of her 300 students, 84% prefer to be contacted through online messaging. Test the hypothesis with 95% level of confidence. GSC-CID-LRMS-ESSLM, v.r. 02.00, Effective April 21, 2021Let Us Discover Population proportion (p) is a part of the population with a particular attribute or trait expressed as a fraction, decimal, or percentage of the whole population. For example, a principal would like to know the proportion of the students who prefer to have a printed module. The null hypothesis states that the population proportion, p, is equal to a specific value or the hypothesized proportion, po. On the other hand, the alternative hypothesis is the competing claim that the population proportion is less than, greater than, or not equal to po. Null Hypothesis Alternative Hypothesis Ho: P = Po Ha: P # Po (two-tailed) Ho: p 2 Po Ha: P Po (one-tailed) Example: A school principal claims that more than 67% of the 700 students of their school prefer printed modules. In words, In symbols, Ha: 67% of the students prefer printed modules. Ho: p 5 0.67 Ha: More than 67% of the students prefer printed modules. H.: p > 0.67 GSC-CID-LRMS-ESSLM, v.r. 02.00, Effective April 21, 2021 Using the Central Limit Theorem in Testing Population Proportion 1. The conditions for binomial experiment are met. That is, there is a fixed number of independent trials with constant probabilities and each trial has two outcomes that we usually classify as "success" (p) and "failure" (q). The sum of p and q is 1. Hence, we can write p + q = 1 or q = 1- p. 2. The conditions np 2 5 and nq 2 5 are both satisfied so that the binomial distribution of sample proportion can be approximated by a normal distribution with u=np and a= Inpq. Now, let us check the assumptions from the previous example: 1. It is evident that the responses have only two outcomes: "who prefer printed module" (success) or "who do not prefer printed module" (failure). Therefore, the first assumption is met. 2. To be able to satisfy the second condition, we find the hypothesized value of the population proportion po = 0.67 while n = 700. To get q, q = 1-po which makes q = 1 -0.67 = 0.33. Through substitution, it shows that the second assumption is also met, since no = 700 (0.67) = 469 2 5 and nq = 700 (0.33) = 231 2 5. Since we have shown that np, 2 5 and nq 2 5, all conditions are met where the sample size is truly large enough to use CLT. In this Ze = - P - Po Po (1 - Po) condition, the test statistic to be used is the z-test statistic for n proportions denoted by Z. or the computed z-value. The significance level, also denoted as alpha or a, is the probability of rejecting the null hypothesis when it is true, or we aim to reject the null if it is false. For example, a significance level of 0.05 indicates that the difference is said to be significant at 0.05 or 5% even when there is no actual difference. Critical Values of Z Type of Test for Statistical Hypothesis Directional/ Non-directional/ Confidence One-tailed Test Two-tailed Test Significance Level Reject Ho Reject Ho Reject Ho Level (a) 1% or 0.01 99% or 0.99 Z= -2.326 Z = 2.326 Z = + 2.567 5% or 0.05 95% or 0.95 Z = -1.645 Z = 1.645 2 = $1.960 10% or 0.10 |90% or 0.90 Z = -1.282 Z= 1.282 Z = 1 1.645 Example: A school principal claims that more than 67% of the 700 students of their school prefer printed modules. Test the hypothesis using 95% level of confidence. Formulate the null and alternative hypothesis Ho: p $ 0.67, H.: p > 0.67 Identify the level of significance. a = 0.05, One-tailed (right) Can CLT be used? YES, since np = 700 (0.67) = 469 2 5 and nq = 700 (0.33) = 231 2 5 GSC-CID-LRMS-ESSLM, v.r. 02.00, Effective April 21, 2021 Determine the critical value. Z = 1.645 Illustrate the rejection region in the normal curve a =0.05 95% Rejecti Regle 1.645