1. Determine the upper- and lower-tail critical values of t in each of the following two- tests assuming that the variances of the two populations are equal. a) a = .10, n1 = 16, n2 21 b) a = .05, n1 = 20, n2 = 25 c) a = .02, n1 = 10, n2 = 12 d) a = .05, n1 = 8, e) a = .01, n1 = 12, n2 = 12 %3D n2 = 9 %D
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- A random sample of 16 statistics examinations from a large population was taken. The average score in the sample was 78.6 with a sample variance of 64. We are interested in determining whether the average grade of the population is significantly more than 75. Assume the distribution of the population of grades is normal . The value of the test statistic is A) 3.6 B) 1.8 C) 0.45The results of a state mathematics test for random samples of students taught by two different teachers at the same school are shown below. Can you conclude there is a difference in the mean mathematics test scores for the students of the two teachers? Use α = 0.01. In addition, assume the populations are normally distributed and the population variances/standard deviations are not equal. Teacher 1 Teacher 2 ?̅1 = 473 ?̅2 = 459 S1 = 39.7 S2 = 24.5 n 1 = 8 n 2 = 18 a. State the null and alternate hypotheses (write it mathematically) and write your claim. b. Find the test statistic c. Identify the Rejection region (critical region) and fail to reject region. Show this by drawing a curve and…The aim of a study is to test the ratio of variances in the weight of two groups of rats being under a certain drugs A ( group 1) and B ( group 2) . A random sample of 16 rats was selected from group land 13 rats from group 2, Then their weight were recorded. The results are shown in the following table n Mean Median sd group 1 16 3.5 3.0 1.05 group 2 13 2.20 2.0 1.15 F15,12,0.05 = 0.404 F 19,16,0.025 = 0.386 F16,10,0.025 =0.335| F16,10,0.05 =0.401 F15,12,0.1 =0.496 F19,16,0.05 =0.451 F10,16,0.025 = 0.286 F 12,15,0.05 =0.382 F16,19,0.025 = 0.371| F10,16,0.05 = 0.354 F12,15,0.1 =0.475 F16,19,0.05 = 0.437| Construct a 95% confidence interval for the ratio of the variances. [ Hint: because of the rounding select the closest interval ] O a. None of these O b. [ 2.488, O0.238 ] O c. [ 2.160 , 0.309] O d. [2.063 , 0.318]
- Data on the weights (Ib) of the contents of cans of diet soda versus the contents of cans of the regular version of the soda is summarized to the right. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. Use a 0.01 significance level for both parts. Diet Regular H2 27 27 0.79037 lb 0.80399 lb 0.00449 lb 0.00756 lb a. Test the claim that the contents of cans of diet soda have weights with a mean that is less than the mean for the regular soda. What are the null and alternative hypotheses? O A. Ho: H1 = H2 OB. Ho: H1#H2 Hq: HyA new study has found that, on average, 6- to 12-year-old children are spending less time on household chores today compared to 1981 levels. Suppose two samples representative of the study's results report the following summary statistics for the two periods. 1981 Levels 2008 Levels 24 minutes 19 minutes 51 = 3.5 minutes 52 - 4.3 minutes ni = 30 n2 - 30 Which of the following is the correct value of the test statistic assuming that the unknown population variances are equal? Multiple Cholce Z = -4.94 tss = 4.94 Z = 4.94 t59 = -4.942. A pet association claims that the mean annual costs of food for dogs and cats are the same. The results for samples of the two types of pets are shown in the following table. On wished to test the claim. Assume both populations follow normal distributions Dogs Cats 21 - 16 T1= 239 n2 = 18 Iz= 203 S1 = 32 82 =28 with equal variances. Which of the following formula should be used to calculate the standardized test statistic? (a) Z = (21-32)- Do lu A n2 = (E1-2)-Do (b) Z V n1+2 (c) T (1-2)-Do (21-1)a7+(n2-1)s I1 11+11タ-3 V 1'n2 (d) T = d- Do sp/Vn (e) ZAssume that the examination marks of students follow normal distri- butivn with unknown mean (4) but known variance (a?) = 0.25. Test the hypothesis Ho : u =8 against H: #8 with the level of significance a = 0.01, the sample mean being equal to 7.8 and the sample size being 50.5. We don't want predictor variables that are too highly correlated with each other or they are redundant (say ≥±.70) or multicollinear (2±.90), which means the variables are measuring pretty much the same thing so why have them all in the equation. What was the finding for redundancy between SPI Scientist and SPI Practitioner? a. No issue with redundancy since the shared variance between the two variables was only 4% (r2 = .04). b. There was an issue with redundancy between the two variables shared 62% of the variance in the relationship.You have data drawn from a normal distribution with a known variance of 16. You set up the following NHST: • Ho: data follows a N(2, 4²) • HÃ: data follows a N(µ, 4²) where µ ‡ 2. Test statistic: standardized sample mean z. Significance level set to a = .05. You then collected n = 16 data points with sample mean 1.5. (a) Find the rejection region. Draw a graph indicating the null distribution and the rejection region. (b) Find the z-value and add it to your picture in part (a). (c) Find the p-value for this data and decide whether or not to reject Ho in favor of HA.Is the proportion of wildfires caused by humans in the south lower than the proportion of wildfires caused by humans in the west? 360 of the 501 randomly selected wildfires looked at in the south were caused by humans while 435 of the 588 randomly selected wildfires looked at the west were caused by humans. What can be concluded at the = 0.10 level of significance? For this study, we should use Select an answer t-test for the difference between two dependent population means z-test for the difference between two population proportions t-test for a population mean t-test for the difference between two independent population means z-test for a population proportion The null and alternative hypotheses would be: Select an answer μ1 p1 Select an answer < ≠ > = Select an answer μ2 p2 (please enter a decimal) Select an answer μ1 p1 Select an answer > ≠ < = Select an answer p2 μ2 (Please enter a decimal) The test statistic ? z t = (please show your…The average size of a farm in Pekan I is 185 acres. The average size of a farm in Pekan II is 100 acre. Assume that the data were obtained from two samples with standard deviations of 38 acres and 12 acres, and sample size of 8 and 9 respectively. Can it be concluded at α = 0.05 that the average size of the farms in Pekan I and Pekan II are different? (Assume equal variances)At a 0.05, test to see if the population variances from which the following samples were drawn are equal. The test statistic is Group 1 n₁ = 21 Group 2 n₂ = 19 $1 = 20 S2 = 18 (A) 1.56 (B) 1.23 (C) 1.26 (D) 1.11SEE MORE QUESTIONS