Question

Category Observed Counts A gene called ACTN3 encodes a protein that functions in fast- twitch muscles. People have different variants of this gene, classified as RR, RX, or XX. RR 130 Using the sample data provided in the table, you want to test if the proportions in these categories are 0.4 (for RR), 0.4 (for RX), and 0.2 (for XX). RX XX 126 80 1. State the null and alternative hypotheses for this chi-square goodness-of-fit test. Ho: Ha: 2. What is the sample size (for the entire data set provided above)? STAT 250 Lab 11: Ch 7.1-7.2 April 24, 2020 3. What is the observed number of people with the RR gene variant? What is the expected number of people with the RR gene variant, given Ho? 4. By hand, calculate the chi-square (x2) test statistic. 5. Use your test statistic above and Minitab Express to calculate the p-value: Choose Statistics > Distribution Plot > Display Probability For Distribution, select Chi-Square Enter the Degrees of freedom (k-1). Select 'A specified x value' Choose Right tail (for chi-square distributions, we use the right tail) For X value, enter your test statistic (from Q4 above) • 6. Given a=0.05, can we conclude that the assumed proportions (per Ho) are not accurate?

Answer 1

Solution: Part-1: H.: PRR = 0.4, Prx = 0.4, Pxx = 0.2 H: The proportion in these categories are different from above. Part-2: Sample size = 130 + 126 + 80 = 336 Part-3: If the null hypothesis is true, expected frequencies can be calculated as below: Observed Expected Frequency Category Frequency foffe fo - fo² | (f.-fe)/fe fe f. RR 19.36 70.56 RX XX 130 126 80 Ef. = 336 336 x 0.4 = 134.4 336 x 0.4 = 134.4 336 x 0.2 = 67.2 Efe = 336 -4.4 -8.4 12.8 163.84 0.144 0.525 2.438 Ef.-fe)?/fe = 3.107 Hence, Observed number of people with RR gene = 130 Expected number of people with RR gene = 134.4 Part-4: Test Statistic: x2 => 5 (fo - fel" = 3.107 fe

Part-5: Degrees of Freedom = df = (number of categories - 1) Hence, df = (3-1) = 2 P-value: Using Minitab, p-value = 0.2115 Part-6: Level of significance = a = 0.05 Since p - value > a, we fail to reject the null hypothesis There is not sufficient evidence to conclude that the assumed proportions are not accurate.