Tests of Significance

Hypothesis Tests With confidence intervals, we don’t know… p, the population proportion, or the true value Hypothesis Testing for Sample Proportions And we are trying to…. estimate the true value But what if we have a value of p? We want to test this claim. Tests of Significance USES/PURPOSES: What are these tests used for? Decide the truth about a claim What do these tests compare? Sample statistics to population parameters p̂ ↔ p What are the 4 components of a Test of Significance? a. Hypotheses b. Test Statistic c. P-Value(probability) d. Conclusion In Ch. 9, we will be looking at testing a population proportion using sample proportions Components: 1. Hypotheses What do the hypotheses always describe? population parameter What types of symbols do the hypotheses always use? parameters; Greek letters: p, μ, β, σ NULL HYPOTHESIS FORM: H0: parameter = claim value “H naught” Assumed… to be true ALTERNATIVE HYPOTHESIS What is it? What we suspect/think is true FORM: Two types of alternative hypotheses: One Sided Two Sided H A: p < % H A: p ≠ % H A: p > % For this chapter… H0: p = % 2. Test Statistic What is this test statistic used for? To compare sample to the claim What type of a variable is this test statistic? z-score Refresher: A z-score tells us… how many standard deviations a value is above/below the mean value Formula: z= pˆ − p Example: p(1− p) n y So z-scores that are small mean that the sample is close to the parameter p, so we believe the claim. What types of z-scores are “normal”? About 95% are within z = ± 2 About 99.7% are within z = ±3 y So z-scores that are large mean that the sample is far from the parameter p, so we reject the claim. 3. P-Value Definition: The probability of getting our sample statistic if the claim is true. The smaller the p-value… the more likely the claim is false What do we use on the calculator? If HA: p < P(z < #) normalcdf(-10^99, #) If HA: p > P(z > #) normalcdf(#, 10^99) If HA: p ≠ 2P(z > |#|) 2xnormalcdf(|#|, 10^99) Example: How do we use the test statistic to find the P-Value? Since we are using z-scores we can find the normal probabilities 4. Conclusion What is a significance level? A number that we compare… P-Value to We use this to help us decide… If the claim is false What symbol do we use to denote this level? alpha α The common significance levels are: α = 0.10; 0.05; 0.01 If a significance level is not given, what level do we use? α = 0.05 So how do we decide whether the claim is false? If the P-value is LESS THAN the ALPHA, our claim is False If the P-value is GREATER THAN the ALPHA, we can’t say that the claim is false Conclusions: The two conclusions that we can have are: (they are 2 sentences) If the p-value is LESS THAN ALPHA: Since the P-Value is less than alpha (____ < ____), we reject the null hypothesis. We have sufficient evidence that the true proportion of _______ is _______ ___%. Conclusions: If the p-value is GREATER THAN ALPHA: Since the P-Value is greater than alpha (____ > ____), we fail to reject the null hypothesis. We do not have sufficient evidence that the true proportion of _______ is _______ ___%. Example 1: We take a sample of 200 M&M’s from a large bag and find that 32 of them are orange. According to M&M/Mars company the true proportion is supposed to be 20%. Is there evidence that the percent is less than the company’s claim? Use α = 0.05. 32 = 0.16 n = 200 α = 0.05 p = 0.20 pˆ = 200 Hypotheses: H0 : p = 0.20 HA : p < 0.20 Test Statistic: z = pˆ − p = 0.16 − 0.20 p(1− p) ( 0.20)( 0.80) n 200 P-Value: P ( z < −1.4142) = 0.0787 = −1.4142 Conclusion: Since the P-Value is greater than alpha(0.0787 > 0.05) we fail to reject the null hypothesis. There is not enough evidence to say the proportion of orange M&M’s is less than 20%. Example 2: It has been claimed that 40% of all shoppers can identify a highly advertised trademark. If 20 of 36 shoppers were able to identify the trademark, is there evidence that the true proportion is now higher? Use the 1% level of significance. p = 0.40 pˆ = Hypotheses: H0 : p = 0.40 HA : p > 0.40 20 = 0.5556 n = 36 36 α = 0.01 Test Statistic: z = 0.5556 − 0.40 ( 0.40)( 0.60) 36 P-Value: P ( z > 1.9057) = 0.0283 = 1.9057 Conclusion: Since the P-Value is greater than alpha(0.0283 > 0.01) we fail to reject the null hypothesis. There is not enough evidence to say the proportion of shoppers that can identify the trademark is more than 40% Test Statistic: z = 0.4828 − 0.35 ( 0.35)( 0.65) = 3.3527 145 P-Value: 2P ( z >| 3.3527|) = 8.0 ×10−4 = 0.0008 Conclusion: Since the P-Value is less than alpha(0.0008 < 0.05) we reject the null hypothesis. There is sufficient evidence to say the proportion of adults afraid of flying is not 35% Example 3: The proportion of people who are afraid of flying is claimed by the FAA to only be 35%. You do not believe this, and take a sample of 145 random adults and find that 70 of them are afraid of flying. Perform a statistical test of significance on the claim at the 0.05 significance level. Use α = 0.05. 70 p = 0.35 pˆ = = 0.4828n = 145 α = 0.05 145 Hypotheses: H0 : p = 0.35 HA : p ≠ 0.35