Below is a sample of 25 sample means. Each sample is based on a sample of two scores. These were drawn from the population of scores (composed of values 2.3.4.5. and 6] and it has a p = 4 and a o = 1.41. lhave taken all possible combinations of the scores. so the table below reects all possible values contained in the Sampling Distribution. The Mean of this Sampling Distribution is: Table: A Sampling Distribution of the Means (all possible combinations) Sample Scores Sample Sample Scores Sample Number Sampled Mean Value Number Sampled Mean Value 1 2.2 2 14 4.5 4.5 2 2.3 2.5 15 4.6 5 3 2.4 3 16 5.2 3.5 4 2.5 3.5 1? 5.3 4 5 2.6 4 18 5.4 4.5 6 3.2 2.5 19 5.5 5 7 3.3 3 20 5.6 5.5 8 3,4 3.5 21 6,2 4 9 3.5 4 22 6.3 4.5 10 3,6 4.5 23 6,4 5 11 4.2 3 24 6.5 5.5 12 4,3 3.5 25 6,6 6 13 4.4 4 Using the formula 2(Sample Mean title\")2 mm.ans , the variance of this Sampling Distribution is: O a. 1.41 O b. 1.99 O c. 1 Q d. 0.28 Clear my choice Using the Table depicting the Sampling Distribution (Q#'l), if you used the formula o/v'n for calculating the Standard Error, it would equal 0 a. 1.41 O b. 0.28 @ c. 1 0 d. 2 Clear my choice Using the Sampling Distribution above, what is the probability of drawing a sample mean = 6? O a. p = .0228 O b. .025 O c. .05 0 d. .078 Using the Sampling Distribution from Q#l, having drawn at random a sample mean of 6, if you drew a 95% Confidence Interval around it, 0 a. it would include u = 4 0 b' it would include u but a u that would not be equal to 4 @ c. it would not include u but would include a sample mean equal to 4 0 d. we can 't know because we can't draw a .95 Cl because we don't actually know the true standard error Clear my choice Having drawn a random sample (see Q#5), the 95% confidence interval stretches from O a. 3.24 to 8.79 O b. 4.04 to 196 O c. 5.95 to 6.05 0 d. As noted in Q#5, we cannot calculate the 95% Confidence Intervals