Assume that the two samples are independent simple random...

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. Do the following: a. Test the given claim using the Pvalue method or critical value method. b. Construct a confidence interval suitable for testing the given claim. In a study of proctored and nonproctored tests in an online Intermediate Algebra course, researchers obtained the data for test results given below (based on “Analysis of Proctored versus Nonproctored Tests in Online Algebra Courses,” by Flesch and Ostler, MathAMATYC Educator, Vol. 2, No. 1). Use a 0.01 significance level to test the claim that students taking nonproctored tests get a higher mean than those taking proctored tests. Group 1 ( Proctored ) : n = 30 , x ? = 74.30 , s = 12.87 Group 2 ( Nonproctored ) : n = 32 , x ? = 88.62 , s = 22.09

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. Do the following:
a. Test the given claim using the Pvalue method or critical value method.
b. Construct a confidence interval suitable for testing the given claim.
In a study of proctored and nonproctored tests in an online Intermediate Algebra course, researchers obtained the data for test results given below (based on “Analysis of Proctored versus Nonproctored Tests in Online Algebra Courses,” by Flesch and Ostler, MathAMATYC Educator, Vol. 2, No. 1). Use a 0.01 significance level to test the claim that students taking nonproctored tests get a higher mean than those taking proctored tests. Group 1 ( Proctored ) : n = 30 , x ? = 74.30 , s = 12.87 Group 2 ( Nonproctored ) : n = 32 , x ? = 88.62 , s = 22.09

Key Concepts

Assumptions of Independence and Normality

Most parametric tests, including the independent samples t-test, rely on the assumptions that the data are drawn from independent random samples and that the underlying populations follow a normal distribution. These conditions ensure that the test statistics have the assumed distribution under the null hypothesis, which is critical for the validity of the inference.

Hypotheses Formulation

In hypothesis testing, clear formulation of the null hypothesis (H0) and the alternative hypothesis (H1) is crucial. The null hypothesis typically states that there is no difference or no effect, while the alternative hypothesis reflects the research claim, such as one group having a higher mean than the other. Well-defined hypotheses help in selecting the appropriate test and interpreting its results.

Significance Level (Alpha)

The significance level, often denoted by alpha, is the threshold for making a decision in hypothesis testing. It represents the maximum probability of committing a Type I error (rejecting a true null hypothesis). A smaller alpha, such as 0.01, implies stricter criteria for rejecting the null hypothesis in favor of the alternative hypothesis.

Pvalue Method and Critical Value Method

These are two standard approaches to hypothesis testing. The pvalue method quantifies the probability of observing test results at least as extreme as the current results under the assumption that the null hypothesis is true, and it is compared to a predetermined significance level (alpha). The critical value method involves determining cutoff values from the probability distribution of the test statistic, beyond which the null hypothesis is rejected.

Two?Sample t-Test for Independent Samples

This statistical test compares the means of two independent groups to determine if there is a statistically significant difference between them. When population variances are unknown and potentially unequal, a variation of the t-test (commonly referred to as Welch's t-test) is used, which adjusts the degrees of freedom appropriately based on the sample sizes and variances.

Confidence Interval for the Difference of Means

A confidence interval for the difference between two means provides a range of plausible values for the true difference. If the interval does not contain the value specified under the null hypothesis (often zero), it suggests that there is a significant difference between the groups. This interval estimation complements the hypothesis testing procedure by quantifying the uncertainty associated with the point estimate.

Transcript

00:01 Looking at people getting proctored tests versus unproctored tests.

00:04 And so the mean for proctors, we're going to assume is equal to the mean of non -proctored tests.

00:10 And alternately, that the mean of the non -proctored test is actually higher.

00:18 And so we're doing a, i think the significance level was 1%.

00:24 I'm going to quick peek back at the question.

00:27 Yes, 1 % significance level.

00:30 And so our test statistic will be a t value, and we're going to calculate that by taking that 74 .3 minus the 88 .62.

00:42 Looks like that's pretty high difference.

00:45 And we have the standard deviation in the first one is 12 .87 squared divided by the sample size, and then the 22 .09.

00:55 Oh, that has quite a bit of variability, divided by the sample size of 32.

01:01 And conservatively, we would use 29 degrees of freedom.

01:04 So let's find out what that test statistic is.

01:08 And let me quick change one thing that i just hit in wrong.

01:14 And so we get that test statistic is a 3 .142.

01:19 And if we want to find that p value, we want to find what the probability is for 29 degrees of freedom of being less than or equal to negative 3 .142.

01:29 And again if we use software the software would tell us that we would use degrees of freedom of 50 .44 and we'll just use the conservative estimate and so i'll use my tcdf to find this i can look it up in the table too but again we don't we can't find accurate accurate results with the table so negative 3 .147 oops that's a two make it look like a two and our degrees of freedom conservatively would be 29.

02:02 And so when i do that, i get that value, that p value is 0 .0019.

02:09 So this is definitely less than the 1 % significance level.

02:15 So we would have sufficient evidence to reject the null.

02:23 So it does appear as though the non -proctor does seem to have a higher test result mean.

02:29 Now if we're going to calculate a competence interval, for this setting...

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