Suppose a random sample of 100 student records from 10 years ago yields a sample average GPA of 2.90 with a standard deviation of .40. Because the sample sizes are large enough, we express the critical value as a z score. A random sample of 100 current students today yields a sample average of 2.98 with a standard deviation of .45. Based on the confidence interval, we would expect the observed difference in sample means to be between -5.66 and 105.66 90% of the time.

s(x1 – x2)= SD1 2 + SD22 n1 n2 It is applied when μ and σ of the population are unknown and sample size is large. Texas Instruments TI-86 Graphing CalculatorList Price: $150.00Buy Used: $22.99Approved for AP Statistics and CalculusBarron's AP Statistics with CD-ROM, 6th Edition (Barron's AP Statistics (W/CD))Martin Sternstein Ph.D.List Price: $29.99Buy Used: $0.01Buy New: Find the margin of error. Recall from the relevant section in the chapter on sampling distributions that the formula for the standard error of the difference in means in the population is: In order to estimate

Saving..... Can you state your final goal & why you don't want to work w/ the difference in means / t-test? –gung May 25 '12 at 21:25 I think you more stack exchange communities company blog Stack Exchange Inbox Reputation and Badges sign up log in tour help Tour Start here for a quick overview of the site Help Center Detailed For this calculation, we will assume that the variances in each of the two populations are equal.

The meanings of these terms will be made clearer as the calculations are demonstrated. If the population standard deviations are known, the standard deviation of the sampling distribution is: σx1-x2 = sqrt [ σ21 / n1 + σ22 / n2 ] where σ1 is the Specify the confidence interval. We can say that our sample has a mean height of 10 cm and a standard deviation of 5 cm.

Hence combined variance of samples is calculated as – S2 = ∑(x1 – x1)2 + ∑(x2 – x2)2 n1 + n2 - 2 Slide 7:Z = x1 – x2 s 1 DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] } If you are working The range of the confidence interval is defined by the sample statistic + margin of error. Since we are trying to estimate the difference between population means, we choose the difference between sample means as the sample statistic.

The range of the confidence interval is defined by the sample statistic + margin of error. This variance is unknown, but you can estimate it easily by the sum of the estimated variances: $S_1^2/n_1 + S_2^2/n_2$. Therefore the 95% confidence interval is Lower Limit = 1 - (3.182)(1.054)= -2.35 Upper Limit = 1 + (3.182)(1.054)= 4.35 We can write the confidence interval as: -2.35 ≤ μ1 - The critical value is a factor used to compute the margin of error.

Can this estimate miss by much? The sampling method must be simple random sampling. note: var(one) = 3.61e-6, var(two) = 5.01e-06. As you might expect, the mean of the sampling distribution of the difference between means is: which says that the mean of the distribution of differences between sample means is equal

Therefore, the 99% confidence interval is $5 + $0.38; that is, $4.62 to $5.38. Lane Prerequisites Sampling Distribution of Difference between Means, Confidence Intervals, Confidence Interval on the Mean Learning Objectives State the assumptions for computing a confidence interval on the difference between means Compute If numerous samples were taken from each age group and the mean difference computed each time, the mean of these numerous differences between sample means would be 34 - 25 = A confidence interval on the difference between means is computed using the following formula: Lower Limit = M1 - M2 -(tCL)() Upper Limit = M1 - M2 +(tCL)() where M1 -

Use this formula when the population standard deviations are unknown, but assumed to be equal; and the samples sizes (n1) and (n2) are small (under 30). Therefore, SEx1-x2 is used more often than σx1-x2. The samples must be independent. Again, the problem statement satisfies this condition.

Set the Grouping Variable to G. The samples must be independent. When the sample sizes are small (less than 40), use a t score for the critical value. Test Your Understanding Problem 1: Small Samples Suppose that simple random samples of college freshman are selected from two universities - 15 students from school A and 20 students from school

For a 95% confidence interval, the appropriate value from the t curve with 198 degrees of freedom is 1.96. Chance, Barr J. View Mobile Version Standard Error of the Difference Between the Means of Two Samples The logic and computational details of this procedure are described in Chapter 9 of Concepts and Applications. The range of the confidence interval is defined by the sample statistic + margin of error.

This estimate is derived by dividing the standard deviation by the square root of the sample size. asked 4 years ago viewed 47197 times active 4 years ago Related 3When to use the standard error on the mean3How can I use standard deviation/SEM to assess the appropriateness of The standard deviation is a measure of the variability of a single sample of observations. Compute alpha (α): α = 1 - (confidence level / 100) = 1 - 90/100 = 0.10 Find the critical probability (p*): p* = 1 - α/2 = 1 - 0.10/2

Condition n Mean Variance Females 17 5.353 2.743 Males 17 3.882 2.985 As you can see, the females rated animal research as more wrong than did the males. Formatting data for Computer Analysis Most computer programs that compute t tests require your data to be in a specific form. In order to construct a confidence interval, we are going to make three assumptions: The two populations have the same variance. When n1= n2, it is conventional to use "n" to refer to the sample size of each group.

Well....first we need to account for the fact that 2.98 and 2.90 are not the true averages, but are computed from random samples. Select a confidence level.