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The Student's t equation can be rearranged to estimate the number of samples that must be analyzed by a specific technique to determine the mean with a desired...

Question

The Student's t equation can be rearranged to estimate the number of samples that must be analyzed by a specific technique to determine the mean with a desired confidence,t283where n is the number of samples needed, 8s is the sampling standard deviation, t is the Students t value at the desired confidence interval, and e is the sought for uncertainty: This equation assumes that the analytical uncertainty is negligible compared to the sampling uncertainty: Soil samples from a local dog park

The Student's t equation can be rearranged to estimate the number of samples that must be analyzed by a specific technique to determine the mean with a desired confidence, t283 where n is the number of samples needed, 8s is the sampling standard deviation, t is the Students t value at the desired confidence interval, and e is the sought for uncertainty: This equation assumes that the analytical uncertainty is negligible compared to the sampling uncertainty: Soil samples from a local dog park are being analyzed for nitrate content: The sampling standard deviation is known to be +8.0%. Estimate the number of samples that need to be analyzed to give a 98% confidence that the mean is known t0 within #5.0%. Assume the analytical uncertainty is much smaller than the sampling uncertainty Use the Student's t values in this table



Answers

A random sample has 49 values. The sample mean is 8.5 and the sample standard deviation is $1.5 .$ Use a level of significance of 0.01 to conduct a left-tailed test of the claim that the population mean is $9.2 .$ (a) Check Requirements Is it appropriate to use a Student's $t$ distribution? Explain. How many degrees of freedom do we use? (b) What are the hypotheses? (c) Compute the $t$ value of the sample test statistic. (d) Estimate the $P$ -value for the test. (e) Do we reject or fail to reject $H_{0} ?$ (f) Interpret the results.

Right, we have a sample with N is equal to 49 values. That sample has mean 8.5 and standard deviation 1.5. We want to use a significant level of 0.1 or 1% to conduct the left field test for the population. Mean new equal 9.2 1st. We want to answer whether or not a this is appropriate to use the student's T distribution to solve. Yes, it is because we have any greater than equal to 30 samples. Note that we're gonna have a degree of freedom equal to n minus one or 48. Since students T tables do not use uh 48 or values typically between intervals of five for very large numbers, we're going to approximately this down to 45 so as not overestimate the p value. Next let's take the hypotheses are null hypothesis. Station on his musical 9.2 are alternative hypothesis. Is that new is less than 9.2 next computers. T statistic. Remember the T stat is defined as x minus mu, divided by estimate of the standard deviation. Then in this case that's T equals 5.6. Next use the one tail tea table to derive RP interval. We see that this test statistic T is extreme, so we have the interval P is less than 10.5 incredibly small. P value. Next we reject H not, Yes, we do, because our P value is less than equal to alpha. And we interpret this to mean that we have evidence suggesting new is actually less than 9.2.

We have n equals 25 values, transformers, symmetric, mound shaped distribution sample mean is equal to 10 and the sample standard deviation is equal to two. Using a significant level of alpha equals 20.5 conducted to tail test of new equals 9.5 on the stand for the population. Me first, you have to answer A. Is it appropriate to use a distribution for hypothesis test for this? Yes. It's the distribution of symmetric and mount shape so we can use a student's T distribution. Note that the degrees of freedom is equal to end minus one is equal to 24 in this case, next one of the hypotheses, the hypotheses are h not mu equals 9.5. An alternative hypothesis. You does not equal 9.5. Next let's compute t remember the formula for tea, which is given here. This computes out the T value 1.25 Next let's compute the p p interval using a T interval. We see that P falls between 0.2 and 0.25 for this T statistic. Next we deserve and whether or not we reject asian art. No, the entire P interval is greater than our significant level alpha. So we do not reject a chart, and we interpret this to mean that we lack evidence that suggest new. It's not equal to 9.5.

We have n equals 25 values, transformers, symmetric, mound shaped distribution sample mean is equal to 10 and the sample standard deviation is equal to two. Using a significant level of alpha equals 20.5 conducted to tail test of new equals 9.5 on the stand for the population. Me first, you have to answer A. Is it appropriate to use a distribution for hypothesis test for this? Yes. It's the distribution of symmetric and mount shape so we can use a student's T distribution. Note that the degrees of freedom is equal to end minus one is equal to 24 in this case, next one of the hypotheses, the hypotheses are h not mu equals 9.5. An alternative hypothesis. You does not equal 9.5. Next let's compute t remember the formula for tea, which is given here. This computes out the T value 1.25 Next let's compute the p p interval using a T interval. We see that P falls between 0.2 and 0.25 for this T statistic. Next we deserve and whether or not we reject asian art. No, the entire P interval is greater than our significant level alpha. So we do not reject a chart, and we interpret this to mean that we lack evidence that suggest new. It's not equal to 9.5.

One tale test and two tailed tests when pretty much everything else is the same. So recall. It's a horrible line. We'll try that again. And a one tailed test, whether it's the left or the right. I have my test statistic which I'll just call see. So my P value will be this area And a two tailed test actually twice as big if everything else is the same because I'm counting both sides. So a two tailed test is twice the area of a one tailed. So the p value will be bigger for a two tailed tests than a one tailed test. Then your ass well bigger than alpha and you're asked or no bigger than each other. I'm sorry. Then let's see. Is it possible to reject the null from the one tailed? So we reject and then failed to reject. So remember we reject the null when the P value is less than the level of significance. And we failed to reject when it's greater than the level of significance. So we could reject it here because the P value is a smaller P value, and it would be indeed smaller than alpha, but we would fail to reject it here because in the two tailed tests or P value would actually be greater, which could cause it to be greater than alpha. So for letter B, the answer is yes. And then for letters C were asked if we reject the null and a two tailed test, will we reject it in the one tail test? And the answer is yes again, because this one's bigger, the P value from here will be bigger than the P value and the one tail tests. So if we reject it in the bigger the value, we will reject it also in the lower P value, because it's already smaller than alpha here. And then finally, is it possible when you're told that you're rejecting the hypothesis is a good idea to know which type of test, and the answer again is yes. It is important to know because of this situation where you might reject it with a one tailed and not with a two tailed, and fail to reject it with a two tailed. So you need to make sure you find out which Which one it is.


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