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Question 7
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Texts: 2. (30 points) We want to compare the average airspeed velocity of an unladen European swallow to the average airspeed velocity of an unladen African swallow. To test this, a sample of n = 16 unladen European swallows was found to have an average airspeed velocity of μ = 8.4 m/s with a standard deviation of s = 0.25 m/s. The airspeed velocity of unladen swallows is known to be normally distributed. The average airspeed velocity of unladen African swallows is known to be 8.5 m/s. We want to test at 95% confidence if the average airspeed velocity of unladen European swallows is less than that of unladen African swallows. a) (10 points) Clearly write out the hypotheses. Ho: The average airspeed velocity of unladen European swallows is greater than or equal to the average airspeed velocity of unladen African swallows. Ha: The average airspeed velocity of unladen European swallows is less than the average airspeed velocity of unladen African swallows. b) (5 points) Decide on your test statistic and write out the formula. Support your answer. t-test The formula for the t-test is: t = (x̄ - μ) / (s / √n) where x̄ is the sample mean, μ is the population mean, s is the sample standard deviation, and n is the sample size. c) (10 points) Calculate your test statistic and P-value. Round to two decimal places. Test statistic: t = (8.4 - 8.5) / (0.25 / √16) = -0.40 / 0.0625 = -6.40 P-value = 0.0001 d) (5 points) Draw a conclusion. Since the P-value (0.0001) is less than the significance level of 0.05, we reject the null hypothesis. Therefore, there is sufficient evidence to conclude that the average airspeed velocity of unladen European swallows is less than the average airspeed velocity of unladen African swallows at a 95% confidence level.
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Question 14
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Views: 5,115

a) Hypotheses: Ho: There is no clear "favourite" among John, Paul, George, and Ringo. HA: There is a clear "favourite" among John, Paul, George, and Ringo. b) Test statistic: I used the chi-square test statistic (χ2) because we are comparing categorical data (the number of fans for each Beatle) to determine if there is a clear "favourite" among them. c) Calculation of test statistic: To calculate the chi-square test statistic, we need to compare the observed frequencies (O) with the expected frequencies (E) under the assumption of no preference. The formula for the chi-square test statistic is: χ2 = Σ((O - E)^2 / E) First, we need to calculate the expected frequencies. Assuming no preference, each Beatle would have an equal chance of being chosen, so the expected frequency for each Beatle would be 200/4 = 50. For John: (O - E)^2 / E = (57 - 50)^2 / 50 = 0.98 For Paul: (O - E)^2 / E = (55 - 50)^2 / 50 = 0.50 For George: (O - E)^2 / E = (48 - 50)^2 / 50 = 0.08 For Ringo: (O - E)^2 / E = (40 - 50)^2 / 50 = 2.00 Now, we sum up these values: χ2 = 0.98 + 0.50 + 0.08 + 2.00 = 3.56 Rounded to two decimal places, the test statistic is 3.56. d) Conclusion at 95% confidence (a = 0.05): To determine if there is a clear "favourite" among John, Paul, George, and Ringo, we compare the calculated test statistic (χ2 = 3.56) with the critical value from the chi-square distribution table. If the calculated test statistic is greater than the critical value, we reject the null hypothesis (Ho) and conclude that there is a clear "favourite." If the calculated test statistic is less than or equal to the critical value, we fail to reject the null hypothesis and conclude that there is no clear "favourite." Since the degrees of freedom for this test is 3 (number of categories - 1), the critical value for a chi-square test at 95% confidence level is 7.815. Since 3.56 < 7.815, we fail to reject the null hypothesis. Therefore, we conclude that there is no clear "favourite" among John, Paul, George, and Ringo based on the survey results.
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