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Stetistica prodrameJerommeidedAccording ancle subn erced ar Kelding (SAW) process commonly used for jo ning thick plates anx plpes The heat affecred zone IHAZI obse...

Question

Stetistica prodrameJerommeidedAccording ancle subn erced ar Kelding (SAW) process commonly used for jo ning thick plates anx plpes The heat affecred zone IHAZI obsenyations cepth (mm) the HAZ botn when tne cumene setting was high and when i was loteDano createdKhin ThedasiGnnnoTialla intereSTnvesngatonsK7-hiqh1.93 2,49 2.711.39 2.09 2,52 2,61 2.81 1.53 2.01 2.05 2.34 2.56 2.92 2.93 2.96HighCalculate the test statistic and P-value: (Round your test statistic to two decimal places and your P-value

stetistica prodrame Jerommeided According ancle subn erced ar Kelding (SAW) process commonly used for jo ning thick plates anx plpes The heat affecred zone IHAZI obsenyations cepth (mm) the HAZ botn when tne cumene setting was high and when i was lote Dano createdKhin Thedasi Gnnno Tialla intereST nvesngatons K7-hiqh 1.93 2,49 2.71 1.39 2.09 2,52 2,61 2.81 1.53 2.01 2.05 2.34 2.56 2.92 2.93 2.96 High Calculate the test statistic and P-value: (Round your test statistic to two decimal places and your P-value to three decimal places ) 1.96 P-value 061



Answers

Either use technology to find the P-value or use Table A-3 to find a range of values for the $P$-value. technology. Use a $0.05$ significance level. Identify the null and alternative hypotheses, test statistic, P-value (or range of P-values), or critical value(s), and state the final conclusion that addresses the original claim.
Data Set 3 “Body Temperatures” in Appendix B includes 93 body temperatures measured at 12 am on day 1 of a study, and the accompanying XLSTAT display results from using those data to test the claim that the mean body temperature is equal to $98.6°F$ . Conduct the hypothesis test using these results.

Now, In this case, I have my P statistic as minus 2.503 and I am testing for the claim that new is less than 98.6. So this becomes my alternative hypothesis. So what will be my null hypothesis? Mine l hypothesis will be that new is greater than equal to 98.6 degrees Fahrenheit. And in this sample story in this case is given to me as four. So what is going to be my degrees of freedom? Degrees of freedom is given by the formula and minus one, and this becomes three. So my degrees of freedom is three and P is minus 2.3 Now, these are the value that I need to find a p value. We can use a table, but using the tea table will only get an approximate value. And if you want an exact value, were will be used a statistical tool like I'm doing over here. What is my peace school? It is minus 2.503 and my degrees of freedom is three significant level is 0.5 and I am checking form you less than some value. So this is going to be a left hand test left tail test. So this is of until test and I calculate the P value and the value that I get is 0.437 My p value that I get is 0.437 This is my answer. And looking at this p value, what we have done is we have used significant level of 0.0 fight. Okay, which means 95% confidence level and I can see that my P is less than 0.5 So I will reject minor hypothesis in this case. So I can say that Yes, I have enough statistical evidence to support my claim that new is less than ideal 0.6. This would be my answer and conclusion.

Okay, so here we are asked if there is a difference between this year's and last year's uh households. National Household Travel survey, if they travel more last year, before beer before that and were given that in the first year, it was 16.23 versus the second year, 17.69 with standard deviations of 4.6 and 4.42 respectively. Okay, so at first glance, it doesn't appear that they change significantly, but we will be evaluating this problem at the 5% significance level, which means that we are looking at a two tailed tests since we want to find out if they are equal or if they are not equal. So, I'm gonna do a two tailed test here. That means that we're looking at our normal distribution like this chop it off here, chop it off here. Yeah, I'd have to 5% here and the other half of the 5% there. If our statistic falls into one of these shaded regions, we will reject the null hypothesis. So what is our equation? It's going to be this blue equation right here, we'll go ahead and substitute those values of X one minus x to what we have is 16.23 minus 17.69 And the next thing we need to find out is what is S P. S. P. S are pooled. Uh Standard deviations, we can do this because our standard deviations are very close to each other and the sample sizes are large enough what our sample sizes. So I got to look up at the table. Okay, so we got 15 households and 14 households. All right, so first sample sizes 15 15 minus one is 14 and the standard deviation for that one is 4.6. So we square that and do the same thing with the other sample of 50. Okay. Of 14 minus ones. We have 13 times the standard deviation of 4.42 squared. And then we add those two. And what we get for RSP is 4.24 Great value. Okay, now we take one over each of the population or sample sizes, so that's gonna be 1/15 and won over 14. And then we and that gets us zero point negative 0.9 three. And then looking in the back of the book for our degrees of freedom, we have 27 degrees of freedom and a significance level is your point 0 to 5? Yeah, a critical value of 2.52 So obviously zero negative zero point and three is less than 2.52 And we're gonna take the absolute value since it's from both sides. Remember we're looking at that is a horrible normal graph, but are negative 0.93 lands right here and are shaded regions are right there and there. So what we do is we say that we fail to reject the null hypothesis.

So we're giving a problem that we have to find at the 5% significance level. The sample size is told to be of 10. So and is equal to 10. Which implies that are Significant value or critical value is at five and 10 and we're losing one till test. So we have our critical value of W. C. is equal to 44. Now when we take a look at this test that we're supposed to do it's one tailed. However which tail is it? We are asked to say that one is more effective than the other. So mm everything is mostly positive here and we find what are our values? Mhm. Use the following data set which gives the additional sleep in hours obtained by 10 patients who used love level high school. So I mean high bromide. So that's gonna be so the control group minus the other one or not. The control group. So what we'll just do here is calculate the both ends since I'm not sure if it's a To a left tail or right tail. So we'll go ahead and find out that right tail is at 44 and then the other tail is going to be 10 times 11 divided by two. So that's gonna be 55 minutes 44 Which is gonna be 11. Very handy. Okay. So if it's less than 11 won't reject and if it's greater than 44 will also reject. Sound good? So um we got 10 students at the positive values so 1.9 plus 0.8 plus 1.1 plus 0.1 was 4.4 plus 5.5 plus 1.6 plus 4.6 plus 3.4. And we get 23.4 for our w nut. Okay. And or critical value that just leads us straight in the middle. So at alpha equals .05 We failed to reject the null hypothesis and this was just a right tail test that are evaluation is insignificant. So now if we tried it again at the 1% significance level, mhm at point a one that changes our values to be mhm uh 50. So we got 50 right here and then we do 10 tons 11 divided by two. So 55 -50, which is going to be five. And once again 23.4 still lands us right in there. So we failed to reject the my hypothesis once more.


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