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T-distribution with k degrees of freedom? Which of the following statements (are) true about the_ Thc t distribution symmetric aroundthe mean Iler variance (spread)...

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T-distribution with k degrees of freedom? Which of the following statements (are) true about the_ Thc t distribution symmetric aroundthe mean Iler variance (spread) than the t-distribution with k+1 degrees of Tne t distribution with degrees of freedom has freedom. The t-distribution has (spread) than the standard norma distribution: larger varianceI onlyb) IlonlyIIl onlyand IIandIIIBenera now does quadrupling the samp size affect the width of the confidence interval? The width of the interval be

t-distribution with k degrees of freedom? Which of the following statements (are) true about the_ Thc t distribution symmetric aroundthe mean Iler variance (spread) than the t-distribution with k+1 degrees of Tne t distribution with degrees of freedom has freedom. The t-distribution has (spread) than the standard norma distribution: larger variance I only b) Ilonly IIl only and II andIII Benera now does quadrupling the samp size affect the width of the confidence interval? The width of the interval becomes four times as large: The width of the interval becomes two times as large: The width of the intervae bccomc half _ arge_ d) The width of the interva becomes one-quarter as Jarge . We need t0 know the sample size be ablc to determine the cffcct: The 90% confidence interval for the population mean length frog jumps cm to 14.4cm, Which of the following statement correct interpretation of the confidence interval? Of thc total number in random sampb frogs, 90% of them can jump between 12.6cm t0 14.4 cm. There" 90% charce that any particular frog catch can jump between 12.6 cm to 14.4 cm: were repeat this sampling many times, 90% of the confidence intervals constructed in the same Manner would contain the true population mean: 90% of the confidence Intervals we construct after repeated sampling would be from 12.6 cm to 14.4 None cm: of the above: Part ? _ Written Questions According to Newsweck magazine, the number of hours per year spent kids front of screen (including television, comnuter smartphcne, etc) can be modeled by narma distribution with mcune 1500 houls 230 hours: standard deviation of What percent of kids spend more than 1850 nours per Year (this roughly foro than haurs screen? per dav) in front of & If 2 motner claims her child spent less than OCO hours front = Explain vour answer briefly. screen last year; would You believe this mother? Page 120i 4



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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 of independence? (e) Interpret your conclusion in the context of the application. Use the expected values $E$ to the hundredths place. Psychology: Myers-Briggs The following table shows the Myers-Briggs personality preferences for a random sample of 519 people in the listed professions (Atlas of Type Tables by Macdaid, McCaulley, and Kainz). T refers to thinking and $\mathrm{F}$ refers to feeling. $$ \begin{array}{l|c|c|c} \hline & \multicolumn{2}{c} {\text { Personality Preference Type }} & \\ \cline { 2 - 3 } \text { Occupation } & \multicolumn{1}{c} {\text { T }} & \text { Row Total } \\ \hline \text { Clergy (all denominations) } & 57 & 91 & 148 \\ \hline \text { M.D. } & 77 & 82 & 159 \\ \hline \text { Lawyer } & 118 & 94 & 212 \\ \hline \text { Column Total } & 252 & 267 & 519 \\ \hline \end{array} $$ Use the chi-square test to determine if the listed occupations and personality preferences are independent at the $0.01$ level of significance.

All right, we have a population with no means 67 we want to calculate for the following sample the sample mean and the sample standard deviation So to do so, let's remember that the sample mean and sample standard deviation are defined by these formulas X. Bar is the some of the data divided by M. In this case 61.8 S is some of the deviations about the mean square all divided by n minus one or 10.6. Now, we want to implement a two tailed test, that is, we want to test whether or not from the sample we can find statistically significant suggesting the population means differs from the no mean of 67 ida direction. We use an alpha value of 670.1 and we note that X is approximately normally distributed. To implement this test, we have to answer the following questions to start. What's the significance of hypotheses after is equal to 0.1? No hypothesis, H and r is that mu is equal to 67? And our alternative hypothesis is that new is not equal to 67. What distribution of do we use? Let's compute the associated test statistic. The distribution will use the student's T. Because a population standard deviation sigma is unknown. We can use this distribution safely because the shape is described as normal, which is both mound shaped and symmetric from this. We have to derive a T stat defined by this formula. It reduces our negative 2.19 And this problem, let's compute the p interval from this. T stat. This is the degree of freedom is and minus one equals 15. We have the p p interval corresponding 150.75 PM. That's natures of 0.1. We find this p interval by using the two tailed T table, which can be found on google or a textbook for this T stat. The graph looks like this on the right where we have our absolute value, both negative and positive, much for artist at negative 2.19 and 2.19 And the P value is the area under the distribution to the left and right of these values. What can we conclude from these findings? Well, we concluded that since P is greater than alpha, we have statistically insignificant findings. Therefore, we can not reject h not and then we can't reject H not. We interpret this finding to mean that we lack any evidence suggesting that the population mean differs from its known mean of 67.

We have muse equal to 1300 as a known mean of the population. We want to calculate the sample mean exploring the sample standard deviation s for the randomly sampled data that follows. So to do so, we simply have to remember the definition of expire to ask for a sample. That's why, for example, in some of the data to buy buy em in this case, 12 68 S is some of the deviations about the moon squared, all divided by n minus one. In this case. 37.29 Next, what we want to do is implement a two tailed test. That is we want to test whether or not this sample suggests the actual meaning this population differs either higher or lower. Either direction from the no mean 1300. We're going to do so with an alcohol level of 0.1 and we're gonna know at this point that X is approximately normally distributed. So to implement this test, we have to follow the following procedures, one by one 1st. What is the significance of hypotheses? Alpha equals 0.1 The null hypothesis H not is musical 1300. And the alternative hypothesis H is that it is not equal 1300. Next distribution We're going to use compute the associated testes, cystic. We're going to use a student's t distribution because the population standard deviation sigma is not known. We only have a sample standard deviation S It's appropriate to use the distribution because X is approximately normally distributed, meaning it's both symmetric and round shape necessary to use students T. Six reasons to distribution. Let's calculate the T statistic. It's defined by this formula, expire minus mu divided by s over root end, which in this case equates the negative 2.714 next let's compute the p interval and sketch the results. So, since we have a degree of freedom of n minus one equals nine, no, from a two tailed T table, which we can find in google and the stats textbook, we find that the T statistic corresponds to a p value range between 20.0 to 1.5 We can think of this also as the area underneath the T distribution outside of negative 2.714 and 2.714 The T statistic as is graft on the right. Finally, given this p interval range, we can make a conclusion about this test. Since P is greater than alpha, we have statistically insignificant bindings and we cannot reject h shot. We fail to a general hypothesis. Therefore, we interpret this to mean that we lack evidence suggesting that the population mean differs from its no mean of 1300.

Following is a solution video to number 24 and this is where we compare to means uh for the soil, water content for field a compared to feel b. And the first part is just to verify that the mean and the standard deviations are in fact these numbers, So the mean for field day is 12.53 and the standard deviation is 2.39 And then the mean for field B was 10.77 with the standard deviation of 2.4, and it says use a calculator something use this T I 84 I went ahead and typed in the means, or sorry, the data values L one and L two, so L one represents fuel day and then L two represents, you'll be and if you go to stat falcon and it's one of our stats and we're going to change that to L one and calculate and that gives us everything we need. So the X bar, is that 12.53 So that's verified. And then we're looking at the s the standard of the sample standard deviation is about 2.39 So for the field A. That is correct. And then let's just go and double check field be. So we're gonna change it to L. Two now because that's where field B. Is, and then that verifies its 10.77 for the mean, then about 2.40 for the standard deviation. So the first part is done, that's verified. And then the second part it says conduct not formally, but we're basically just going to conduct a uh uh two sample T. Test with an alpha value of point oh five. And we're gonna go back to the calculator because doing it by hand can be pretty cumbersome. So if you go to stat and then test now, since we already have, we're gonna go to two sample t. Test since we already have the data in there. I'm just gonna use the data instead of the summary stats. So list one is L. One that's the field A. List to is L. Two. That's field B. And then these frequencies just keep them the same and then we have the the alternative hypothesis and you kinda have to use your context clues here. But this one actually explicitly states is field A. Is the soil content or water content higher or greater than. So I'm going to change this to greater than So μ one is greater than you to warm you. A. is greater than YouTube pulled is usually zero or no unless they tell you otherwise. So we can go ahead and calculate and that's gonna give us everything we need. So the T. Value if you want you can put it down there. But really it's just this P. Value that I want to see. So 00.27 is the p value. So P value equals 0.27 And then we compare that with the alpha value and it's less than the alpha value. So any time it's less than the alpha value. That means we reject the no hypothesis. So we're rejecting the statement that these means are the same. And we are I guess you could say accepting the fact that these two means are in fact the water content and field A. Is greater than the water content and field be on average. So we can type this out as there is enough evidence to suggest that field A. Has on average a higher soil content soil. Sorry, water content. Banfield beat. Okay so this field A. Has on average a higher soil water content on field because since we are rejecting that null hypothesis and accepting the alternative hypothesis.

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.


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