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A medical research team studied the Intracranial Pressure (ICP) from sample of 50 patients who suffered ffom head injury The data below has been rounded off for sim...

Question

A medical research team studied the Intracranial Pressure (ICP) from sample of 50 patients who suffered ffom head injury The data below has been rounded off for simplicity11 12 13 14 18 33 16 17 18 18 20 20 21 22 23 23 23 23 23 24 26 26 27 27 27 28 29 30 31 33 35 36 37 38 39 40 40 43 46 47 48 50 52 54Column n Mean Std. dev Min QlMedian Q3Max ICP 50 26.2 13.110208 18 23.5 36 54Construct frequency distribution for the ICP measurements_ Use classes beginning With lower class limit of and class widt

A medical research team studied the Intracranial Pressure (ICP) from sample of 50 patients who suffered ffom head injury The data below has been rounded off for simplicity 11 12 13 14 18 33 16 17 18 18 20 20 21 22 23 23 23 23 23 24 26 26 27 27 27 28 29 30 31 33 35 36 37 38 39 40 40 43 46 47 48 50 52 54 Column n Mean Std. dev Min QlMedian Q3Max ICP 50 26.2 13.110208 18 23.5 36 54 Construct frequency distribution for the ICP measurements_ Use classes beginning With lower class limit of and class width of 10_ Construct histogram of ICP. Is the distribution symmetric Or skewed? If skewed, is it skewed to the left or to the right? What is the appropriate measure of center. What is the measure of variation that corresponds to the chosen measure of center. Construct the interval of usual values Are there any outliers? Describe the data using the characteristics above The table below" shows the results of 3025 students from university . Students Tere asked how many times they saw tutor for help in TWo week period_ Erent Times tutoring_Frequency 505 1026 942 430 Find the probability of each event in the table Are the events mutually exclusive? Find the probability of seeing tutor more than twice in tio wreek period. Find the probability of seeing tutor exactly three times given that the tutor was seen at least once.



Answers

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. (Note: Answers in Appendix D include technology answers based on Formula 9 - 1 along with "Table" answers based on Table $A$ - 3 with df equal to the smaller of $\boldsymbol{n}_{I}-\boldsymbol{I}$ and $\boldsymbol{n}_{2}-\boldsymbol{I} .$ ) Magnet Treatment of Pain People spend around $\$ 5$ billion annually for the purchase of magnets used to treat a wide variety of pains. Researchers conducted a study to determine whether magnets are effective in treating back pain. Pain was measured using the visual analog scale, and the results given below are among the results obtained in the study (based on data from "Bipolar Permanent Magnets for the Treatment of Chronic Lower Back Pain: A Pilot Study", by Collacott, Zimmerman, White, and Rindone, Journal of the American Medical Association, Vol. $283,$ No. 10). Higher scores correspond to greater pain levels. a. Use a 0.05 significance level to test the claim that those treated with magnets have a greater mean reduction in pain than those given a sham treatment (similar to a placebo). b. Construct the confidence interval appropriate for the hypothesis test in part (a). c. Does it appear that magnets are effective in treating back pain? Is it valid to argue that magnets might appear to be effective if the sample sizes are larger? Reduction in Pain Level After Magnet Treatment: $\quad n=20, \bar{x}=0.49, s=0.96$ Reduction in Pain Level After Sham Treatment: $n=20, \bar{x}=0.44, s=1.4$

It's a kind of an interesting question if we we would be assuming that the anxiety level if we are going from easy to difficult is equal to the anxiety level. If the questions were difficult to easy and alternately, we just want to know if there's a difference and easy to difficult and not equal to difficult to easy. And so we're going to assume that that difference is actually zero so that the, I'll just write E G minus D. E. That that difference is zero. And when I put this into my calculator, um we have the lists have different sample sizes. We have that first group of the E. T. D. I have, there are 25 numbers versus the D two E. There are 16 numbers and we could use the degrees of freedom of 15, but I'm going to use the degrees of freedom from the formula. And that degrees of freedom ends up being uh 30 almost 39. So 38 point well, I'm just going to call it 39 it's approximately 39 degrees of freedom. And that test statistic that we're going to get, we need to take the mean, which was 27.1152 minus the mean of the other group, which was 31.7 to 8125 And then divided by the square root of. And the first standard deviation all around that a bit is 6.857 one square, divided by the sample size, which was 25. And then the second standard deviation was 4.26 square divided by the sample size of 16. And when I got that test statistic, the test statistic came out to be, I'll just read it up here negative 2.6 566 And so it came out down here. And since we're doing a two tailed test, we also use this one. And so what's the likelihood of getting a test statistic in this distribution that is less than or equal to that negative 2.6566 That gives us this tale and then we want the other tales to double it. And that p value comes out to be a 0.114 So at a 5% significant level, this is smaller than 5%. So we would have sufficient evidence to reject now. Yeah. And say that the mean anxieties are different so that they're different. The means are different. However, at a 1% significance level we would fail to reject the novel. Mhm. And we'd have to say here that they're actually not. They appeared to be the same. So it does depend on our significance level. How did pick you want to be. But it is an interesting idea in all my years of teaching, I never thought about the arrangement of having them all go easy, too difficult or difficult to easy. So what interesting concept.

What we want to conduct A p D. T. A pair differences tests at the alpha equals 1% confidence level. Testing the claim that the population means A bar next pr are not equal. We have the data for A and B. Given below assuming amount shapes, not your distribution on the right. I've already calculated D. Bar noted that an equal seven and calculated SD as 70.47 So we proceeded to do the five steps listed below to solve this first. We evaluate the requirements and hypotheses. So the requirements to use the students distribution have been met because the distribution shape we have degree of freedom and minus one equals six. Are null hypothesis is mute equals zero. Or alternative is beauty does not equal zero. And we're testing at alpha equals 00.1 confidence Next will compute the test statistic and P value. So the statistic is T equals D. Bar divided by SD over. Uh huh. 2.083 from a tea table. This puts p between .1.05. So we can conclude that P is greater than alpha, which means we fail to reject the null hypothesis, which means that we lack evidence that beauty does not equal to about.

We want to conduct a pair differences test at the alpha equals 5% level testing the claim that population mean X bar A is greater than population X barbie. We have data a be given here, we assume amounts to mr distribution as you can see on the right. I've already calculated the mean difference D bar 6.125 The sample size and eight and the sample standard deviation of differences SD and 8.7 We complete the five steps us to blow to solve this problem first, let's evaluate the requirements to use a student's T distribution of the hypotheses because of the distribution shape it is appropriate to use a student distribution your degree of freedom and minus 27. No hypothesis mute equals zero. Alternative media is greater than zero and alpha equals 00.5 for confidence nexus, complete the test at and the P value our test that is T equals D. Bar over SDR. Again this gives 2.14 U. T. Table. This gives us a P value between 0.5 point 025 That means we can include that P is less than equal to alpha. So we reject the null hypothesis H not which means that we have evidence and you D. Is greater than zero.

We want to conduct a pair differences test at the output equals 1% confidence level testing and claim that sample or rather population means that our A. Is greater than expire. Be given the data for A and B. Below here assuming the mound shaped in symmetric distribution. As you can see, I've already computed D. Bar the mean the difference is 12.6 and equals five. And a standard deviation differences 22.66 on the right. Using the appropriate formulas, we proceeded the five steps below to compute this PDT first we check the requirements, evaluate hypotheses because the distribution shape it's appropriate to the students distribution degree of freedom is n minus one equals four. Null hypothesis is not equal zero. Alternative nuclear than zero and alpha equals 00.1 significance. Next our test statistic is T equals D. Bar over SD over route and or 1.24 from the tea table. This gives p value between 0.25 point 125 Thus we can conclude That thing is greater than α.. So we fail to additional hypothesis, meaning we lack evidence, moody is greater than zero


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