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Conduct an approrimate Hypothesis Test (using & 0.05 to determine if there is difference between students at the university and the population ofthe USA. Show a...

Question

Conduct an approrimate Hypothesis Test (using & 0.05 to determine if there is difference between students at the university and the population ofthe USA. Show all of your calculations and steps. Your answer needs to (the 7 step process meets these requirements)State the hypotheses in terms of the parameter(s, of interest. Calculate sd(estimator)_ Calculate the test statistic, and give its distribution under the null hypothesis_ Give the P-value for the test , using the tables in the Problem

Conduct an approrimate Hypothesis Test (using & 0.05 to determine if there is difference between students at the university and the population ofthe USA. Show all of your calculations and steps. Your answer needs to (the 7 step process meets these requirements) State the hypotheses in terms of the parameter(s, of interest. Calculate sd(estimator)_ Calculate the test statistic, and give its distribution under the null hypothesis_ Give the P-value for the test , using the tables in the Problem Booklet_ State our conclusion in the context of the data. Use Minitab to perform an exact Hypothesis Test to determine if there is a difference between students at the university and the American population: You only need to provide the Minitab output and a conclusion (in context)_ Explain why there is such large difference between the P-values for the tests you performed in (a) and (b). You may include an additional calculation, but this is not required: Based on the evidence available, it is believed that the proportion of students at the university who nominate climate change as the mnost urgent issue is 10 more than 30%. Researchers would like to estimate the proportion using 90% confidence interval based On a normal approximation, with a maximum margin of error of 0.02_ What sample size would be required to achieve this? Show your calculations as well as your answer:



Answers

Please provide the following information. (a) What is the level of significance? State the null and alternate hypotheses. (b) Find the value of the chi-square statistic for the sample. Are all the expected frequencies greater than 5 ? What sampling distribution will you use? What are the degrees of freedom? (c) Find or estimate the $P$ -value of the sample test statistic. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis that the population fits the specified distribution of categories? (e) Interpret your conclusion in the context of the application. Marketing: Compact Discs Snoop Incorporated is a firm that does market surveys. The Rollum Sound Company hired Snoop to study the age distribution of people who stream music. To check the Snoop report, Rollum used a random sample of 519 customers and obtained the following data: Using a $1 \%$ level of significance, test the claim that the distribution of customer ages in the Snoop report agrees with that of the sample report.

Oh we have a sample with size and equal 56 r equals 12 objects in that sample. Satisfying particular conditions we want to observe, we want to test the claim with the population proportion. P does not equal 120.24 Based on this data sample, a confidence level of alpha equals 0.5 Now that we just identify the confidence level, we can proceed to the following procedural steps to conduct this hypothesis test First. We have to answer whether or not the normal distributions appropriate. Yes, it is. Because both M. P and Q. P greater than five. What are the hypotheses being tested? Are no H not? Is P equals 50.24 Are alternative. H A S P does not equal 0.24 meaning we're conducting a two tailed tests for this. Next compute P hat and a test statistic P hat is simply are over end 20.214 are tested to succeed. Is given by the formula on the right. Which equates the negative 0.45 Using the P hat P to an end for this problem. Next compute the P value based on the Z set. So we use a Z. Table to identify that this Z score on a two tailed tests corresponds to P equals 20.65 to 7. We've highlighted that as the area outside the positive negative Z score in yellow on the right next. Using RP value. Do we reject H not? No. We do not. Because he is greeting the alpha. We interpret this to me that we lack evidence that P does not equal 70.24

So we're looking at the reading of professors and we would be assuming that the mean rating for females is equal to the mean rating for males and alternately that they're different. We just want to find out if there's a difference between so we will be doing a two tailed test and I always like to draw a little picture. But remember we're assuming that the difference between these two is zero, and we can also write hypotheses that way, if we so choose so our smallest sample sizes, let's see. Our smallest sample size is 40. Therefore we will use 39 degrees of freedom and we will find our test statistics by taking the difference in our means. And so we're going to take that 3.79 minus the 4.1 And then we will be dividing it by the first standard deviation is 0.51 squared divided by that sample size. And then the second standard deviation is 0.53 square divided by its sample size, which is 53. And when I do that calculation, I find out that that test statistic comes out to be negative 2.25 And now we want to find, because we're doing a two tailed tests, we have this as one of our test statistics and the other one would be up here for our two tailed tests. And these two areas together give us our p values. So we want to find the probability of getting a test statistic less than or equal to negative 2.25 And then we want to double it for the other table, other tail. And we could look up the critical value for this as well. But I'm not going to set that up this way, I'm going to go through and use my T. C. D. F. So my T. C. D. F. I'm gonna have my left bound be like negative uh 10,000 and then my upper will be this negative 2.25 And I will have my degrees freedom being 39. Again that's like this isn't a conservative way. And then we're going to multiply that probability times two. And when I do that I find out that I get 0.4975 By just a tad bit. This is less than 5%. So we do have evidence to reject the not we have sufficient evidence to reject. And now and claim that the ratings are different mm for males and females. Now let's find that confidence interval. And because we did a two tailed test we would be finding them with a 5% significance level. We would be finding a 95% confidence interval and our confidence interval with 39 degrees of freedom. We don't have it in our table. So I'm going to actually use my inverse t to find that. And so I use my inverse T. I can put in the area and I'm gonna put in that lower areas point oh 25 since it's a 2.5% at the bottom tail, 2.5% of the top tail. And then I'll put in the degrees of freedom as 39. Again, this is a conservative approach and then we'll get that with that T value is so our T. Star value will be negative 2.227 So now when we get our confidence interval we need the difference. Let's get what that difference is. So that difference that 4.1 Well, I should say it the other way around the 3.79 minus the 4.1 That comes out to a difference of negative 0.22 it's a negative 0.22 plus or minus. And then we have our value for our T. Star value and then we'll use that big standard deviation and uh let me go back and find what those standard deviations are. So 0.51 point 53 0.51 squared and the 0.53 squared. And let's go back for the sample size is 40 and 53. Yeah. Right. Money. Mhm. Well, I never noticed before how this was 0.53 and that was 53 interesting. Not that makes any difference. So let's find the margin of error, first of all. Yeah, so 2.227 times the square root of 0.51 squared, divided by the 40 plus 400.53 squared divided by 53. And that margin of air comes out to be oh goodness! Isn't this interesting about 0.22? So when we find this interval this bottom while the negative and then minus is going to be minus 0.44 and then when we add them, that's going to be zero. So we're right at that cut off point that uh those values would uh we would only have negative values, we don't have zero. So this would be just at that cut off point to say that we would have evidence to reject the null but notice how close we were to 5% here and that's why this is such a close and we're here to so then it talks about the comparison between another example that did this and had smaller sample sizes. And yes, there is a difference. When we use larger sample sizes, we can make differences. The same differences be significant with larger sample sizes. If these were smaller sample sizes, we would not have been able to say and got the same meaning, same standard deviation, we would not have been able to have sufficient evidence to reject them all.

Right, where will the population has mean new equals 14 From the following sample data, we want to calculate the sample mean X bar and the sample standard deviation S. To do so let's remember the definition of these terms X bar Is the some of the data divided by n or in this case 15.1, I'm a sample standard deviation S is the sum of deviations about the mean square divided by n minus one or 2.51 Next we want to implement a right tail test. That is we want to test whether or not the population mean should actually be greater than the no mean 14. With us using the sample data with a significance level alpha equals 140.1 Where we are noted that X is approximately normally distributed. So to implement this test, we have to answer the following questions in order First, what is the significance of hypotheses? We've alpha equals 0.1 H not is new equals 14 H. A. Is that I mean is greater than 14. What distribution will use computer associated test statistic? Since the population standard deviation sigma is unknown. We have to use a student's T distribution which we know is okay to use because the shape of the distribution is normal, which is both symmetrical amount shaped from this. We calculate the T stat Which is given by this formula and reduces down to 1.386 for this problem? Next compute the p interval and sketch it out so we have degree of freedom and minus 29. We use the one tailed T. Table to identify that this tea interval falls between a p interval of 10.75 point one. That is because our T statistic falls between associated T values for these p values. We can graph this as the area under the student's t distribution. To the right of our T stat 1.386 as is highlighted in yellow on the right. What can we conclude from this? Well, we can conclude that P is greater than alpha, so we have statistically insignificant findings and we fail to reject astronaut, which means that we lack sufficient evidence that suggests our population mean is greater than the No. Mean 14.

We have a sample with N equals 83. And of that 83 members of the sample 64 satisfy a certain condition. Want to meet our equal 64. Now we want to use the sample data to test the claim that p does not equal 2.75 for the population at a confidence level of 5% or alpha equals 0.5 Now that we've identified the confidence level, we can go to the following procedural steps to conduct this hypothesis test first. Is the normal distribution appropriate to use? Yes, it is. Because both N. P and Q. P. R greater than five. B. One of the hypotheses were testing we're testing whether or not H not P equals 10.75 or H a p does not equal 0.75 Ak We're conducting a two tailed test. Next compute P hot. And the test statistic P hot is all over. N equals 20.77 Z is given by the formula on the right, taking in as input P hat P Q n N producing down to Z equals 0.42 Next we compute the P value. We use the table to see the area outside Z equals plus and minus 0.42 as we shaded in yellow and the graph on the right. This P value corresponds to 0.6745 Next we reject H not, no, we do not. P is greater than alpha and we interpret this signing to suggest that we lack evidence that P does not equal .75.


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