What to do if some of the assumptions are not satisfied: Assumption 1. Notice that it is normally distributed with a mean of 10 and a standard deviation of 3.317. Sampling distribution of the difference between mean heights. Assumption 2: Are these large samples or a normal population?

This condition is satisfied; the problem statement says that we used simple random sampling. At 5% level of significance, the data provide sufficient evidence that the new machine packs faster than the old machine on average. Since the above requirements are satisfied, we can use the following four-step approach to construct a confidence interval. Since it does not require computing degrees of freedom, the z score is a little easier.

Compute alpha (α): α = 1 - (confidence level / 100) = 1 - 99/100 = 0.01 Find the critical probability (p*): p* = 1 - α/2 = 1 - 0.01/2 A random sample of 100 current students today yields a sample average of 2.98 with a standard deviation of .45. The formula for the obtained t for a difference between means test (which is also Formula 9.6 on page 274 in the textbook) is: We also need to calculate the degrees Assumption 3.

We present a summary of the situations under which each method is recommended. If the confidence interval includes 0 we can say that there is no significant difference between the means of the two populations, at a given level of confidence. (Definition taken from Find standard error. Click on the 'Minitab Movie' icon to display a walk through of 'Using Minitab to Perform a Separate Variance 2-sample t Procedure'.

We use the sample variances as our indicator. Since responses from one sample did not affect responses from the other sample, the samples are independent. Can this estimate miss by much? From the variance sum law, we know that: which says that the variance of the sampling distribution of the difference between means is equal to the variance of the sampling distribution

Finally, check the box for Assume equal variances. Again, the problem statement satisfies this condition. 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. Although the two-sample statistic does not exactly follow the t distribution (since two standard deviations are estimated in the statistic), conservative P-values may be obtained using the t(k) distribution where k

Find the margin of error. But what exactly is the probability? The difference between the means of two samples, A andB, both randomly drawn from the same normally distributed source population, belongs to a normally distributed sampling distribution whose overall mean is Draw the conclusion using the p-value.

In this case, the test statistic is defined by the two-sample t statistic . The SE of the difference then equals the length of the hypotenuse (SE of difference = ). Using the MINITAB subcommand "POOLED" with the two-sample t test gives the following results: Two Sample T-Test and Confidence Interval Two sample T for C1 C2 N Mean StDev SE Mean Please try the request again.

Resources by Course Topic Review Sessions Central! Remember the Pythagorean Theorem in geometry? The area above 5 is shaded blue. Assumption 2.

Applied Statistical Decision Making Lesson 6 - Confidence Intervals Lesson 7 - Hypothesis Testing Lesson 8 - Comparing Two Population Means, Proportions or Variances8.1 - Comparing Two Population Proportions with Independent Identify a sample statistic. Therefore a 95% z-confidence interval for is or (-.04, .20). Perform the required hypothesis test at the 5% level of significance.

The value is this case, 0.7174, represents the pooled standard deviation \(s_p\). The range of the confidence interval is defined by the sample statistic + margin of error. This tells us the equal variance method was used. Find the p-value from the output.

Using the formulas above, the mean is The standard error is: The sampling distribution is shown in Figure 1. The distribution of the differences between means is the sampling distribution of the difference between means. The samples must be independent. A typical example is an experiment designed to compare the mean of a control group with the mean of an experimental group.

Similarly, 2.90 is a sample mean and has standard error . To find the critical value, we take these steps. The calculations for these test statistics can get quite involved. Select a confidence level.

The range of the confidence interval is defined by the sample statistic + margin of error. Note: The default for the 2-sample t-test in Minitab is the non-pooled one: Two sample T for sophomores vs juniors N Mean StDev SE Mean sophomor 17 2.840 0.520 0.13 Nonetheless it is not inconceivable that the girls' mean could be higher than the boys' mean. Search Course Materials Faculty login (PSU Access Account) I.

For girls, the mean is 165 and the variance is 64. p-value = 0.36 Step 5. To calculate the standard error of any particular sampling distribution of sample-mean differences, enter the mean and standard deviation (sd) of the source population, along with the values of na andnb, Using this convention, we can write the formula for the variance of the sampling distribution of the difference between means as: Since the standard error of a sampling distribution is the

The confidence level describes the uncertainty of a sampling method. The following dialog boxes will then be displayed. Interpret the above result: We are 99% confident that \(\mu_1 - \mu_2\) is between -2.01 and -0.17. We are working with a 99% confidence level.

The samples are independent. And the uncertainty is denoted by the confidence level. Search Course Materials Faculty login (PSU Access Account) Lessons Lesson 0: Statistics: The “Big Picture” Lesson 1: Gathering Data Lesson 2: Turning Data Into Information Lesson 3: Probability - 1 Variable Specify the confidence interval.