Question

5. Given below are descriptive statistics, as well as boxplots, of the moisture content (by percent) for random samples of different fruits and vegetables, probability plots for the two variables, and output from the test of equality of population means. Descriptive Statistics: Fruits, Vegetables Statistics Variable N N Mean SE Mean StDer Minimum Q1 Median 03 Maximum Fruits 3 O 82.88 244 690 7200 76.00 85.00 87.75 92.00 Vegetables 9 0 9167 125 374 85.00 88.50 92.00 95.00 96.00 Boxplot of Fruits Vegetables Probability Plot of Fruits Vegetables Normal-95% a is.' BG.

Answer 1

(a) Conclusion: The conclusion is based on the boxplots is that the null hypothesis (Ho) that the population means are equal. The mean of the box plot of fruits is 82.88 less than that of the mean of the boxplot of vegetables is 91.67. Hence, we can justify that the null hypothesis is rejected and can conclude that the two population means are not equal. ----- (6) The probability plot delineates the data points that are concetrated around the centred line. Hence, we can say that the distribution of data points are approximately normally distributed. We can use the t - distribution to test the difference between the two population means. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

(c) Let 4 and Hy be the population means and let Xi and x, be the means of the respective samples. The summary statistics of the Two-sample t - test output of the Minitab18: Samplel: n = 8, X = 82.88, 5ı = 6.90 Sample2: , = 9, X3 = 91.67, s2 = 3.74 The level of significance is a =0.05. Step1: State the null and alternative hypotheses: The null and alternative hypotheses are, respectively We need to use two-tailed test. Ho:-=0 (The two population means are equal) versus H:44-470 (The population mean 44 is not equal to 4) ------------------------------------------------ Step2: Select the distribution to use: Here, the two samples are independent, q, and o, are unknown but equal, and the sample sizes are small but both populations are normally distributed Hence, all conditions are satisfied. Consequently, we will use the t distribution Step3: Determine the rejection and non-rejection regions. The + sign in the alternative hypothesis indicates that the test is two-tailed. The significance level is 0.05. From the t distribution table, the critical value of t for df = 15 and 0.05 area in the right-tail of the t distribution is 2.13145. From the t distribution table, the critical value of t for df = 15 and 0.05 area in the left-tail of the t distribution is - 2.13145.

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Step 4: Calculate the value of the t-test statistic. The value of the test statistic t for x1 - xz is computed as follows. The pooled standard deviation: (n-1) s +(23-1): V n+1-2 (8-1)47.61+(9-1)13.9876 V 8+9-2 (7)(47.61)+8(13.9876) 17-2 333.27+111.9008 V 15 445.1708 15 = 29.67805 = 5.447757 s, = 5.447757 --- ------ Step 5: The standard error of the difference between the means: - - - - - - - - - - - - - - - - - - - - - - - - - VMM = 5.447757.** = 5.4477570.125+0.111111 = 5.44775770.236111 = 5.447757(0.485913) = 2.647134 S-t = 12.647134 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Step6: The formula for calculating the t-statistic is , (7- x?)-(4–163) _(82.88-91.67)-(0) 2.647134 -8.79 2.647134 = -3.32057 t = -3.321 Therefore, the t-test statistic is t = 1-3.321 ----------------------------------------- The P-value approach: 2P(t>to)=2(1-P(t<t,)) =2(1-P(t <3.321)) =2(1-0.997671) =2(0.002329) = 0.004658 P-value = 0.004658<a=0.05. We reject the null hypothesis H. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Step7: The conclusion and the decision is based on the critical value method: Since, the test statistic is 41=3.321 > 0.05.15 = 2.13145. We reject the null hypothesis H, There is sufficient evidence to conclude that the difference between the two population means are not equal at the 5% level of significance.

(d) The null and alternative hypotheses are given below. HO:6 = 42 versus 4:47 Step1: Let 41 and 4 be the population means. And let Xand xz be the means of the respective samples. The summary statistics of the Two-sample t-test output of the Minitab18: Samplel: n1 = 8, X = 82.88, S = 6.90 Sample2: ng = 9, x3 = 91.67, sy = 3.74 The level of significance is a =0.05. ----- - - - - - - - - - - - - - - Step 2: The pooled standard deviation: (n-1) s +(12-1) V 4+1, -2 (8-1)47.61+ (9-1)13.9876 L 8+9-2 (7)(47.61)+8(13.9876) 17-2 333.27 +111.9008 V 15 445.1708 V 15 = 29.67805 = 5.447757 5,= 5.447757 The standard error of the difference between the means: =5.447757 = 5.447757/0.125+0.111111 = 5.447757 10.236111 = 5.447757(0.485913) = 2.647134 Sep - = 12.647134 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Step3: The 95% confidence interval for the mean difference between the two population means is (1, - 2) = 42.**2-287-7 =(X-X) +605128+9–267-76 =(x-)+60025:17-287- =(X-X) +60125.1997 =(82.88-91.67)+2.13145(2.647134) =>-8.795.642232 =(-14.4322,-3. 14777) (3,-2)6212--14:4354-43315 - - - -

Interpretation: We can be 95% confident that the true population mean difference is between |- 14.43 and -3.15. Since, the digit 'O'is not included in the confidence interval caclulated. We reject the null hypothesis H. Hence, we can conclude that the difference between the two population means are not equal at the 5% level of significance.