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Preliminary data analyses indicate that use of a paired t-test is reasonable. Perform each hypothesis test by using either the critical-value approach or the P-value approach. Sleep. In 1908. W. S. Gosset published "The Probable Error of a Mean" (Biometrika, Vol. 6, pp. $1-25$ ). In this pioneering paper, published under the pseudonym "Student," he introduced what later became known as Student's $t$ -distribution. Gossct used the following data set, which gives the additional slecp in hours obtained by 10 paticnts who used lacvohysocyamine hydrobromide. $$\begin{array}{|ccccc|} \hline 1.9 & 0.8 & 1.1 & 0.1 & -0.1 \\ 4.4 & 5.5 & 1.6 & 4.6 & 3.4 \\ \hline \end{array}$$ a. Identify the variable under consideration. b. Identify the two populations. c. Identify the paired-difference variable. d. Are the numbers in the table paired differences? Why or why not? e. At the $5 \%$ significance level, do the data provide sufficicnt evidence to conclude that laevohysocyamine hydrobromide is effective in increasing sleep? (Note: $\bar{d}=2.33$ and $s_{d}=2.002 .$ ) f. Repeat part (c) at the $1 \%$ significance level.

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.

A classic story involves four carpooling students who missed a test and gave an excuse that they had a flat tire on the makeup test. The instructor asked students to identify which tire went flat. If they really didn't have a flat tire, would they be able to identify the same tire? So the author decided he was going to run a an experiment, and the author asked 41 other students toe identify the tire that they would select, and some of them said left front. Others said right front. Some said Left rear and others said Right rear. And the data that was collected was that 11 people said Left front 15 said Right Front eight said left rear, and six said Right rear. Now that totals up to Onley 40. Even though he asked 41 students, one of the students said That spare tire, So we're gonna leave that out of our data. So this author is making a claim and he believes that the results fit a uniformed distribution. So in order for us to test this claim, we're going to have to construct are null hypothesis and our alternative hypothesis, and you're no hypothesis when you're trying to determine whether observed data fits something that is expected, the statement of fit becomes your null hypothesis. So therefore, our claim is going to be our null hypothesis. So our alternative hypothesis is going to be that the results do not fit a uniformed distribution. Mhm. And the hypothesis test that we're going to run is going to be a chi square goodness of fit test, which requires us to generate a chi square test statistic for our data. And we will have to apply the formula some of observed, minus expected quantity squared, divided by expected. So let's go back up to our data, and the information that we've collected would be classified, as are observed data. Yeah, so now we need to calculate are expected values. And if there were 40 people involved and we would expect a uniformed distribution, that means we would expect each response to get 10 people so we would expect tend to say left front, tend to say right front, tend to say left rear and tend to say right rear. So we're now ready to calculate our chi square test statistic, so we're going to add on to our chart An additional column and we're going to call that column oh minus e quantity squared, divided by e. So we'll take the observed value minus the expected value. We'll get a difference of one. We're going to square that So it's still 1/10, which is that expected value. So we'd get 0.1 and then we'll do 15 minus 10. We get a result of five when we square it. It's 25 divided by the expected value of 10. Yields a value of 2.5. Do the same thing for left rear eight. Minus 10 is negative two. But when we square it, it's positive. Four, divided by the expected value of 10, resulting in 0.4 and then six minus 10 would be negative. Four. When we square that we get positive 16/10 or 1.6 now to calculate the Chi Square test statistic, we will have to add up these values. And when we add up those values, our Chi Square test statistic is 4.6 now. Another component of the hypothesis test is to calculate our P value, and R P value is referring to the probability that Chi Square is greater than the test statistic we just found. Now, to get a better sense of what that's about, we're going to draw a chi square Distribution and chi square distributions are skewed to the right, and their shape is dependent on the degrees of freedom, and our degrees of freedom can be found by doing K minus one. And K represents the number of categories which you've separated your data into. If we go back to our chart, you can see we have separated our data into four different categories, corresponding with the four different tires on the vehicle. So if K is four, then our degrees of freedom will be three. Not only does the degrees of freedom indicate the shape of the graph, but the degrees of freedom also is equivalent to the mean of that chi square distribution. So on our picture we could place the mean, which will be found slightly to the right of the peak on the Chi Square axis. Now, for us to find our P value, we're trying to figure out what's the probability that Chi square is greater than 4.6. So we're trying to determine this shaded region, and in order to do so, the most effective way is to utilize your chi squared cumulative density function in your graphing calculator, and any time you use that function, you have to provide the lower boundary of the shaded area, the upper boundary of the shaded area and the degrees of freedom. So for our data, the lower boundary is the test statistic, the upper boundary. If you imagine that curve continuing infinitely to the right, you're going to get to some extremely high values of Chi Square. So we're going to use 10 to the 99th Power to represent our upper limit, and our degrees of freedom was three. So let me show you where you can find that chi squared cumulative density function. So I'm bringing in my calculator and I'm going to hit the second button and the variables button and select number eight. We're going to put the low boundary of the shaded area, the upper boundary of the shaded area, followed by my degrees of freedom, and I end up with a P value off approximately point 2035 Now there's one more component of a hypothesis test that we could find and that is called your chi square critical value. And to determine that chi square critical value, we're going to look in the back of your textbook. You will find a chi square distribution table and down the left side of the table, you'll find degrees of freedom and across the top of the table, you will find levels of significance, and levels of significance are denoted by the Greek letter Alfa. And we want to run this hypothesis test at a level of significance of 0.5 So we will locate 0.5 across the top of the chart and our degrees of freedom down the side of the chart and where the to correspond or meet up is your chi square critical value, and they meet up at 7.815 So let's recap the three components that we have found so far we have found the Chi Square test statistic to be 4.6. We have found the P value to be point 2035 and we have found our chi square critical value to be 7.815 So what do we do with these values to make a decision about that claim. So when it comes time to make your decision, you can either utilize the P value or you can use the chi square critical value. You do not need to do both. I'm going to show you both and then you can make a determination of which method you prefer. In order to use the P value. You're going to compare your level of significance to your P value. And if your level of significance is greater than your P value than your decision is to reject the null hypothesis. So let's run our test. So our level of significance was 05 and we found RPI value to be 0.2035 So we could say that Alfa is not greater than the P value. So therefore our decision will be fail to reject the no hypothesis. Let me show you how you can use the critical value to arrive at that same decision. To use the critical value, I recommend drawing out another chi square distribution and placing your critical value on that curve. And by placing that on the curve, you have broken your graph into two parts. You have the tail which we're going to define as the reject, the null hypothesis region. And then the other part of the graph is going to be determined or called our fail to reject the null hypothesis region and you're then going to look at your calculated Chi Square test statistic and we found our Chi Square test statistic to be a 4.6. So if 7.8 is right here, then 4.6 would fall back here, which is in the fail to reject region. So again, we end up with the decision that we will fail to reject the null hypothesis. So if we go back to our statements, we are not able to reject this statement. So that's saying it could be true. It might be true. It might not be true, but our evidence don't support throwing it away. So therefore, our conclusion there is insufficient evidence to reject the claim that the results fit a uniforms distribution again. We don't have enough to throw it away, but we don't We're not saying we support it, either. It's just the data is inconclusive and that concludes your hypothesis test

In this problem, we're going to be testing the clean that the survival rates for patients who had cardiac arrest during the the D. Waas the same uh huh, for patients who had carried cardiac arrest at night. So we have two samples way. Have patients who got cardiac arrests. That's not during the day and the proportion off those who survived. He's 1 11,000 604 our tooth 58,000 593. So that's the proportion for patients who survived. You survived during the day, and that's night. The proportion of patients who survived waas 4000 139 out off 28,000 155 So we're testing the clean that the survival rates are the same for day and night. So we're going to do that using the hypothesis test method and using the confidence interval method. So the level of significance Alfa it's 0.1 on the non hypothesis he is P one is equal to P two alternative policies. His P one is not equal to be to, and for that reason, the critical value for Zet he is classroom minus 2.5 seven. So now we can walk out the test statistics and substituting the values into the formula. And when you do that, the calculated values that is 18 point 27 and when we compare that to the critical value, you'll notice that the value 18 is within the critical region. That's 18.27 So since it's within the critical region, we make the conclusion to reject the non hypothesis and when rejected an hypothesis, we conclude that there is not sufficient evidence to support the claim that the survival rates are the same for day and night. Now we move on to the second approach, which is the confidence interval method, and here we work out the value of the margin of error by substituting the values into the formula. And when we do that, we get that e is zero point 00 69 on the confidence interval limits are 0.0 441 and 0.0 five 79 So according to the confidence inter form, the confidence interval limits do not include zero. So since they did not include zero, it appears that the two proportions are not equal because the confidence interval limits include Onley positive values. It appears that they read off survival during the day is different from that at night and therefore it is it is in agreement with the fast test name for this is test. So based on the results, were supposed to tell when it appears that for in house or in hospital patients who have a cardiac arrest, the survival rate is the same for day and night. And since we rejected another hypotheses, we see that one of these writs is much greater than the other, and when we give the percentage off survival during the day, the percentage is 19 8% and at night the percentage is 14.7%. So this since there's a significant difference between the two proportions, then we can conclude that the survival rates are much better during the day compared to the night

All right. We have a sample of size and equals 196. And in that sample we see 29 objects satisfying a certain observation we want to observe that is R equals 29 out of 10 equals 196. We want to use this data to test the claim that P is greater than 092 with alpha equals 0.5 Or confidence level 5%. Now that we've identified the confidence level, we can proceed in the following procedural steps in order to conduct this hypothesis test first. Is it appropriate to use the normal distribution? Yes, it is. Because N. P and Q. P. R both greater than five secondly what hypotheses are retesting? We're testing H and R P equals 50.92 H. A. P greater than 0.92 Which means we're conducting a right tailed or one tailed test. Next compute P. A. And the test statistic he had simply are over end or 0.148 plugging that as well as P. Q and N. Into rz stat formula on the right. Give Z equals 0.271 or 2.71 next let's compute the P value based on our Z equals 2.71 We can use this table to identify the P value. The P value is simply the area under the normal curve to the right of the Z score. Since this is the right tool test from the table, we get P equals 0.34 We've illustrated this in the graph on the right next. We use this P value to reject H. Not. Yes, we do because he is the alpha. And we interpret this to mean that we have evidence that P is greater than 0.92