Green.kiwif Gold.kiwifruit Mean 6.89 5.03 Median 6.85 5.1 SD 0.504 0.368 Range 1.8 1.6 Q1 6.5 4.775 Q3 7.225 5.3 Sample size 30 30A random sample of 60 people were selected to taste two types of kiwifruit, the green and gold kiwifruit. 30 were randomly chosen to taste green kiwifruit, while the remaining 30 tasted gold kiwifruit. Each person gave a liking score from 1 (dislike extremely) to 10 (like extremely). Kiwifruit.xisx contains the summary statistics of the liking scores of these two groups of tasters. Download and use Excel to undertake an appropriate hypothesis test to address the research question. Research Question: On average, is there a significant difference in taste, measured by liking score, between the green and gold kiwifruit?? Figure 1: Green kiwifruit Figure 2: Gold kiwifruit 10 Frequency Frequency o 4.5 50 65 60 6.5 70 7.5 4.5 50 65 6.0 6.5 7.0 7.5 Liking scores Liking scores Figure 3: Green-Gold Figure 4: Liking scores Frequency Liking scores 0.0 0.5 10 1.5 20 2.5 3.0 3.5 Groon Kiwifruit Gold kiwifruit Differences in Liking scores Answer the following questions by choosing the most correct option or typing the answer: 1. (1 mark) The most appropriate test for these data is:2. (1 mark) Which graphical display(s) would be the best to check the normality assumption for this hypothesis? A. Figure 1 B. Figure 2 C. Figure 3 D. Figure 1 & 2 E. Figure 1 & 4 F. Figure 2 & 4 G. Figure 3 & 4 H. No need to check normality assumption 1. All four figures are needed 3. (1 mark) How can you check the assumption of equal variances? A. Comparative boxplots could be used B. Descriptive statistics such as variance, standard deviation and range could be used C. Descriptive statistics such as mean and median could be used D. Both A & B could be used E. Both A & C could be used F. No need to check the assumption of equal variances For the remaining questions you may assume that any relevant assumptions have been met. 4. (1 mark) The value of the standard deviation used in calculating the test statistic is (3dp) 5. (2 marks) The absolute value of the test statistic is equal to (3dp) 6. (1 mark) The degrees of freedom is (Integer) 7. (1 mark) The p-value is larger than 0.05 8. (1 mark) What is the most appropriate conclusion for this test? A. The observed absolute difference 1.86 is statistically significant at the 5% level. B. At the 5% significance level, we do not have enough evidence to reject the null hypothesis, and the liking scores may be the same for both types of kiwifruit. C. At the 5% significance level, we accept the null hypothesis, and the liking scores may be the same for both types of kiwifruit. 9. Calculate a 95% confidence interval to estimate the mean liking scores for people who tasted a green kiwifruit a. (2 marks) The Absolute Value of the Critical Value for this 95% confidence interval is (3dp) b. (2 marks) The range of this 95%% confidence interval is 3dp) (Hint: the range is a positive number)