Yellowfin Tuna
A marine biologist claims that the mean ...

Question

Yellowfin Tuna

A marine biologist claims that the mean fork length (see figure at the left) of yellowfin tuna is different in two zones in the eastern tropical Pacific Ocean. A sample of 26 yellowfin tuna collected in Zone A has a mean fork length of 76.2 centimeters and a standard deviation of 16.5 centimeters. A sample of 31 yellowfin tuna collected in Zone B has a mean fork length of 80.8 centimeters and a standard deviation of 23.4 centimeters. At ,α=0.01 can you support the marine biologist’s claim? Assume the population variances are equal. (Adapted from Fishery Bulletin)

Illustration of a yellowfin tuna with arrows indicating the fork length measurement from head to tail fork.

Accepted Answer

Hello. In this video we are told that a botanist is studying the average height of a particular plant species in two different forests. In Forest X, a sample of 22 plants has a mean height of 54.3 centimeters and a standard deviation of 8.7 centimeters. In Forest Y, a sample of 28 plants has a mean height of 58.9 centimeters and a standard deviation of 10.2 centimeters. At a significance level of 0.05, can we conclude that there is a significant difference in mean plant height between the two forests, and we want to assume equal population variance. OK, so let's go ahead and review what is given to us. First, we are given two samples from two different forests. So what this means is that we are going to want to use a two-tailed tea test. And we are also using a two-tail T test because we are working with large samples that are greater than 30. OK, so what we want to go ahead and do now is we want to create the hypothesis for this test. Now, because we are working with equal population variances, we want to go ahead and set the stan, the original claim to the alternate hypothesis. Now the original claim. Now the claim here is that there is a significant difference in the mean plant heights, so our alternate hypothesis is going to be that the mean of forest X is not equal to the mean of forest Y. And with that being said, the null hypothesis is going to state that the means between forest X is equal to the means of heights in forest Y. So let's go ahead and list out the given information. For Forest X, we are given a sample size of 22 plants. We have a mean height of 54.3 centimeters. And we are given a sample standard deviation. Of 8.7 centimeters For forced Y We have a sample size of 28 plants. We have a mean height. Of 58.9 And we have a standard deviation. Of 10.2 centimeters Also, one more thing, I'm using the wrong variable here, but the mean height is going to be X bar, so we have XX and XY. Also, we want to go ahead and determine our conclusion with a significance level of 0.05. Now, eventually we are going to have to find a critical value. So let's go ahead and first determine the degree of freedom. Now for a 2 sampled for a 2 tail test with equal population variance, the degree of freedom is going to be defined as the first sample plus the second sample minus 2. This is going to give us a degree of freedom of 22 + 28 minus 2, and that is going to give us a total degree of freedom of 48. OK, let's go ahead and find our test statistic. Now in this case here, the T test statistic is going to be defined as the following. The t test statistic is equal to the mean of force of force x minus the mean of force y minus. The average mean of forest X minus the average mean of forest Y. And this will all be divided by our pooled variance. Multiplied by the square root. Of 1 divided by the first sample size of forest X plus 1 divided by the sample size of forest y. Also, one more thing to note here is that for the for the null hypothesis, we are assuming that the two averages are equal to each other. So because we are assuming that they are equal value, we can go ahead and eliminate mu of x minus m of y, and because that is going to be zero. But there's one thing that we need for the T test statistic, and we need to find the pooled variant or the pooled standard deviation. Now, in order to find the pulled standard deviation, the pulled standard deviation is defined as the following. The pole standard deviation squared is equal to the sample of forest x minus 1 multiplied by the standard deviation of forest x 2 minus the sample size of foresty minus 1. Multiply the sample standard deviation of foresty 2, and this will all be divided. By the degree of freedom So let's go ahead and first calculate our polled variants. Now, by plugging in our known values, we get the pool variance squared, or the pool standard deviation squared is equal to 22 minus 1 multiplied by 8.7 2 minus 28 minus 1 multiplied by 10.2 quantity squared. This will all be divided by 48. Now this is going to give us a pooled standard deviation squared. And by plugging this into a calculator, we're going to get approximately 91.6369. And by taking the square root of both sides of this equation, we get a pooled standard deviation of approximately 9.5727. Now that we have our pooled standard deviation, we can go ahead and calculate the T test statistic. So by plugging in our known values to the T test statistic, our T test statistic is going to be defined as the average of x, which is 54.3. Minus the average of y which is 58.9, this will all be divided by a pole of standard deviation, which is 9.5727. Multiplied by the square root of 1 divided by the sample size of forest x which is going to be 22. Plus 1 divided by the sample size of foresty which is going to be 28. And by plugging this into a calculator, we are going to get an approximatet test statistic. Of -1.687 So we are going to set that to the side for now. Now, let's go ahead and determine our T critical value. Now here, what we want to go ahead and do is we want to determine the T critical value of a degree of freedom of 48. Now, earlier we stated that we are using a two tail test, and so what this means is that we have to rewrite the significance level for the area underneath each tail. Now the given significant level is 0.05, and by cutting this in half or dividing this by two, we get a significance level of 0.025. So we want to see critical value. That uses a degree of freedom of 48 and a significance level of 0.025. And so by using a T chart, we get a T critical value of approximately negative or positive negative 2.011. Now, what we want to go ahead and do is you want to compare the T test statistic to the T critical values. Now here, -1.687 lies within the boundary of the T critical value. Or in other words The tea test statistic. Lies within the t critical value. So what does this tell us about the null hypothesis? What this tells us is that we fail to reject. There are null hypothesis. And so what this means is that at the significance level of 0.05, there is insufficient evidence to conclude that there is a significant difference in the mean plant height between forest X and forest Y. And so with that being said, the solution to this problem is going to be B. So I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

Key Concepts

Two-Sample t-Test for Means

Assumption of Equal Population Variances

Significance Level and Hypothesis Testing

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