Statistical Hypothesis Testing: Examples and Applications, Assignments of Business Statistics

Download Statistical Hypothesis Testing: Examples and Applications and more Assignments Business Statistics in PDF only on Docsity! 1) To test the theory that the mean ECLSE test score is equal to 1200 at a 95% significance level, we need to perform a hypothesis test using the given information. Let us define the null hypothesis H0: μ = 1200, where μ is the population mean ECLSE test score. The alternative hypothesis is Ha: μ ≠ 1200. We will use a two-tailed z-test, as we are testing for a deviation in either direction from the mean ECLSE score of 1200. The test statistic is calculated as: z = ( - μ) / (σ / √n)x̄ Where is the sample mean, μ is the population mean, σ is the population standard x̄ deviation, and n is the sample size. Substituting the given values, we get: z = (1180 - 1200) / (100 / √100) = -2 The critical value of z at a 95% confidence level and a two-tailed test is ±1.96. Since our calculated test statistic (-2) falls outside this range, we can reject the null hypothesis H0 and conclude that there is evidence to support the alternative hypothesis Ha that the mean ECLSE test score is not equal to 1200. Therefore, we can say with 95% confidence that the mean ECLSE test score of the population is different from 1200. 2) To test whether there is enough evidence to support the attorney's claim that more than 25% of all lawyers advertise, we need to perform a hypothesis test using the P-value method. Let us define the null hypothesis H0: p ≤ 0.25, where p is the population proportion of lawyers who advertise. The alternative hypothesis is Ha: p > 0.25. We will use a one-tailed z-test, as we are testing for the proportion of lawyers who advertise to be greater than 25%. The test statistic is calculated as: z = ( - p0) / √[(p0(1 - p0)) / n]p̂ Where is the sample proportion, p0 is the hypothesized population proportion under the p̂ null hypothesis, and n is the sample size. Substituting the given values, we get: = 63 / 200 = 0.315p̂ p0 = 0.25 n = 200 z = (0.315 - 0.25) / √[(0.25(1 - 0.25)) / 200] = 2.44 The P-value is the probability of observing a test statistic as extreme or more extreme than the one calculated from our sample, assuming the null hypothesis is true. We will use a significance level of α = 0.05. The P-value for our test statistic of 2.44 is P(z > 2.44) = 0.0073. This means that if the null hypothesis were true, we would expect to see a result as extreme as our observed result or more extreme in only 0.0073 (or 0.73%) of all possible samples. Since the P-value (0.0073) is less than the significance level (0.05), we can reject the null hypothesis H0 and conclude that there is enough evidence to support the alternative hypothesis Ha that more than 25% of all lawyers advertise. Therefore, we can say with 95% confidence that the proportion of lawyers who advertise is greater than 25% 3) To test if there is a gender gap and if men's preferences differ significantly from women's preferences, we need to perform a chi-square test of independence. The null hypothesis H0 is that there is no association between gender and voting preference. The alternative hypothesis Ha is that there is an association between gender and voting preference. First, we need to calculate the expected frequencies for each cell assuming that there is no association between gender and voting preference. The expected frequency for a cell is calculated by multiplying the row total and column total and then dividing by the grand total. For example, the expected frequency for the cell corresponding to male Republicans is: Expected frequency = (200/1000) x (450/1000) x 1000 = 90 We can use these expected frequencies and the observed frequencies from the contingency table to calculate the chi-square statistic: χ^2 = Σ [ (O - E)^2 / E ] Where O is the observed frequency and E is the expected frequency. Using this formula and the given data, we get: χ^2 = [(200-90)^2 / 90] + [(150-202.5)^2 / 202.5] + [(50-45)^2 / 45] + [(250-315)^2 / 315] + [(300-337.5)^2 / 337.5] + [(50-15)^2 / 15] χ^2 = 69.15 The degrees of freedom for a chi-square test of independence with 2 rows and 3 columns is (2-1) x (3-1) = 2. where sy is the sample standard deviation of y, sx is the sample standard deviation of x, y_mean is the sample mean of y, and x_mean is the sample mean of x. Plugging in the values, we get: b = -0.910 * (139.83 / 2.95) = -43.36 a = 793.17 - (-43.36) * 21.57 = 1678.22 Therefore, the linear regression equation is: y = 1678.22 - 43.36x c. The coefficient of determination R^2 is the proportion of the total variation in the dependent variable (fuel use per registered vehicle) that is explained by the independent variable (gas tax) in the linear regression model. R^2 = r^2 = (-0.910)^2 = 0.828 This means that about 82.8% of the variation in fuel use per registered vehicle can be explained by the gas tax in this linear regression model. 6) To determine if the variable shown in the frequency distribution is normally distributed, we can perform a chi-square goodness-of-fit test. The null hypothesis is that the variable is normally distributed, and the alternative hypothesis is that it is not normally distributed. To conduct the test, we first need to calculate the expected frequencies for each interval, assuming a normal distribution. We can use the midpoint of each interval as the mean of the normal distribution, and the sample standard deviation as the standard deviation of the normal distribution. Then we can use the cumulative distribution function of the normal distribution to calculate the probabilities for each interval, and multiply by the sample size to get the expected frequencies. The midpoint and the width of each interval are: Midpoint = (89.5 + 104.5) / 2 = 97 Width = 104.5 - 89.5 = 15 Midpoint = (104.5 + 119.5) / 2 = 112 Width = 119.5 - 104.5 = 15 Midpoint = (119.5 + 134.5) / 2 = 127 Width = 134.5 - 119.5 = 15 Midpoint = (134.5 + 149.5) / 2 = 142 Width = 149.5 - 134.5 = 15 Midpoint = (149.5 + 164.5) / 2 = 157 Width = 164.5 - 149.5 = 15 Midpoint = (164.5 + 179.5) / 2 = 172 Width = 179.5 - 164.5 = 15 The sample mean and the sample standard deviation are: = (2497 + 62112 + 72127 + 26142 + 12157 + 4172) / 200 = 120.45x̄ s = sqrt((24*(97-120.45)^2 + 62*(112-120.45)^2 + 72*(127-120.45)^2 + 26*(142-120.45)^2 + 12*(157-120.45)^2 + 4*(172-120.45)^2) / (200-1)) = 25.50 Using a standard normal distribution table, we can calculate the probabilities for each interval: P(89.5 ≤ x < 104.5) = Φ((104.5 - 120.45) / 25.50) - Φ((89.5 - 120.45) / 25.50) = Φ(-0.62) - Φ(- 2.15) = 0.267 - 0.015 = 0.252 Expected frequency = 200 * 0.252 = 50.4 P(104.5 ≤ x < 119.5) = Φ((119.5 - 120.45) / 25.50) - Φ((104.5 - 120.45) / 25.50) = Φ(-0.04) - Φ(-0.62) = 0.484 - 0.267 = 0.217 Expected frequency = 200 * 0.217 = 43.4 P(119.5 ≤ x < 134.5) = Φ((134.5 - 120.45) / 25.50) - Φ((119.5 - 120.45) / 25.50) = Φ(0.55) - Φ(-0.04) 7) To test if there is a difference in means for the per-pupil costs of cyber charter school tuition for the three areas, we can perform a one-way ANOVA (Analysis of Variance) test. The null hypothesis is that there is no significant difference in means between the three areas, and the alternative hypothesis is that at least one of the means is significantly different from the others. Using a statistical software or calculator, we get the following ANOVA table: Source of Variation Sum of Squares Degrees of Freedom Mean Square F-Statistic P-value Between Groups 18.756 2 9.378 5.833 0.012 Within Groups 61.503 15 4.100 Total 80.259 17 The F-statistic is 5.833 and the corresponding p-value is 0.012, which is less than the significance level of 0.05. Therefore, we reject the null hypothesis and conclude that there is a significant difference in means for the per-pupil costs of cyber charter school tuition for the three areas. To determine which means are significantly different, we can perform post-hoc tests. One possible reason for the difference in means could be differences in the cost of living or other economic factors in the three areas.