Test the claim about the difference between two population mea...

Question

Test the claim about the difference between two population means and at the level of significance α. Assume the samples are random and independent, and the populations are normally distributed.

Claim: μ1=μ2, α=0.01, Assume (σ1)^2=(σ2)^2

Accepted Answer

Hello there. Today we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A school administrator wants to test whether the average test scores of students from two different districts are the same. Independent random samples are taken, and the scores are assumed to be normally distributed with equal variances. District 1, X 1 is equal to 78.5, S1 squared is equal to 16, and N1 is equal to 30. District 2 X 2 is equal to 74.3, S2 squared is equal to 14.44, and N2 is equal to 28. Significance level alpha equals 0.01. Test the claim that the population means are equal. Awesome. So it appears for this particular problem we're asked to take all the information that is provided to us by the prom itself, and we're asked to test the claim that the population means are equal. So now that we know what we're ultimately trying to solve for, let's take a moment to read off our multiple choice answers to see what our final answer might be. A is there is sufficient evidence to conclude that the mean test scores are same. There and then B will be, there is insufficient evidence to conclude that the mean test scores differ between the two school districts. C is there is sufficient evidence to conclude that the mean test scores differ between the two school districts, and D is insufficient data. Awesome. So our first step that we need to take in order to solve this particular problem is we need to state our hypotheses, starting with our null hypothesis first, which will denote as H0. And the null hypothesis for this particular problem, as we should recall, states that mu 1 is equal to mu 2, where mu denotes the mean. And our alternative hypothesis, which will denote as HAA or alternative hypothesis, as we should recall, will state that mu1 cannot equal m2. And we can say this because, as we should be able to recognize, this is a two-tailed test. So now our second step now is we need to use the pooled variances formula. So we need to recall and use the pool variances formula, and we can use this equation or rather formula because or due to that since equal variances are assumed for this particular prompt. So you go ahead and state that SP. Squared, so SP 2 is equal to parentheses, N1 minus 1 multiplied by S1 squad. Plus parentheses N2 minus 1 multiplied by S2 squad. All divided by N1 plus N2 minus 2. So when we plug in all of our known data. Values into our expression to solve for SP 2, we will find that SP 2 is equal to once again plugging your known variables, parentheses 30 minus 1 multiplied by 16, and then we need to take, so 30 minus 1 in parenthes multiplied by 16 plus 28 minus 1 multiplied by 14.44. Divided by, so everything divided by. 30 + 28 minus 2. So when we plug this into our calculator, this expression, and we round to two decimal places, we will find that it's approximately equal to 15.25. Fantastic. So now, Moving right along here, our third step is we need to calculate the test statistic. And as we should recall, the test statistic T is equal to X 1 minus X2 all divided by. The square root of SP2 multiplied by parentheses 1 divided by N1 plus 1 divided by N2. Fantastic. So we plug in our known variables. We will find that T is equal to 78.5 minus 74.3. Divided by the square root of 15.25 multiplied by parentheses. 1 divided by 30 plus 1 divided by 28, and when we plug in this expression into our calculator and we round to two decimal places, we will find that it is approximately equal to 4.09. Hooray, we're on a roll. So now, Our fourth step is we need to find the critical value, but in order to find the critical value, we must first determine the degrees of freedom, which we'll call DF for short. And as we should recall, the degrees of freedom DF is equal to N1 plus N2 minus 2, which in this case will be equal to 30 + 28 minus 2, which will be equal to 56. And we need to note that for a level of significance, alpha equals 0.01, also noting we're dealing with a two-tail test which I'll call TTT for short. We can go ahead and state that our critical value T subscript alpha divided by 2. DF is equal to 56, so we could say that our critical value is approximately equal to plus or minus 2.660. Fantastic. So, moving right along here. Now, our 5th step is we need to make our final decision. So we need to recall that our test statistic T is approximately equal to 4.09. And we're also told that our critical values are plus or minus 2.660, and we need to note that since 4.09 is greater than 2.660. We reject the null hypothesis, so I could put an X next to H0, which 80, H0 denotes the null hypothesis. So once again, because 4.09 is bigger, we can reject the null hypothesis, and because we can reject the null hypothesis, we can make the following conclusions. At the 0.01 level of significance, there is sufficient evidence to conclude that the mean test scores differ between the two school districts, which will mean When we look at our multiple choice answers that our final answer, without a doubt, will have to be multiple choice answer letter C, which once again is, there is sufficient evidence to conclude that the mean test scores differ between the two school districts. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Test the claim about the difference between two population means and at the level of significance α. Assume the samples are random and independent, and the populations are normally distributed.
Claim: μ1=μ2, α=0.01, Assume (σ1)^2=(σ2)^2

Key Concepts

Hypothesis Testing

Level of Significance (α)

Assumption of Equal Variances

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