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Assume that a simple random sample has been selected from anormally distributed population and test the given claim. Identifythe null and alternative hypotheses, te...

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Assume that a simple random sample has been selected from anormally distributed population and test the given claim. Identifythe null and alternative hypotheses, test statistic,P-value, and state the final conclusion that addresses the originalclaim.A safety administration conducted crash tests of child boosterseats for cars. Listed below are results from those tests,with the measurements given in hic (standard head injurycondition units). The safety requirement is that the hicmeasurement should

Assume that a simple random sample has been selected from a normally distributed population and test the given claim. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. A safety administration conducted crash tests of child booster seats for cars. Listed below are results from those tests, with the measurements given in hic (standard head injury condition units). The safety requirement is that the hic measurement should be less than 1000 hic. Use a 0.05 significance level to test the claim that the sample is from a population with a mean less than 1000 hic. Do the results suggest that all of the child booster seats meet the specified requirement? 693 715 1041 579 543 557 Identify the Test statistic. t= ____ (round to three decimal places as needed) Identify the P Value P = ______ ( Round to four decimal places as needed)



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Assume that a simple random sample has been selected and test the given claim. Unless specified by your instructor, use either the P-value method or the critical value method for testing hypotheses. 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. The National Highway Traffic Safety Administration conducted crash tests of child booster seats for cars. Listed below are results from those tests, with the measurements given in hic (standard head injury condition units). The safety requirement is that the hic measurement should be less than 1000 hic. Use a 0.01 significance level to test the claim that the sample is from a population with a mean less than 1000 hic. Do the results suggest that all of the child booster seats meet the specified requirement? $$774 \quad 649 \quad 1210 \quad 546 \quad 431 \quad 612$$

So we're looking at people wearing seatbelts and not wearing seatbelts Children and finding out how many days they stay in the I. C. U. And so we're going to assume that those students with a seat belt have equal stay in the hospital as those who do not wear a seatbelt. And alternately that the seat belted Children have a mean stay less than those who were not seat belted. And so we're assuming that the difference between these two seat belt minus not seatbelt is zero and we're actually getting something that's negative and this will be our p. Value. So let's look at the data and let's get our test statistic and we have sample sizes to sample sizes. One sample size was 1 23 and the other was to 90. So that was the for the seat belt and this is for the non seatbelt. And so we're going to use degrees of freedom of 122 to be conservative. And they're relatively large sample sizes anyway. And so let's find our test statistic and we have our 0.83 minus are 1.39 And then we're going to divide that by the standard deviation of the seat belted group squared, divided by the sample size, and then the standard deviation of the non seat belted Children divided by the sample size of it. And when we do that calculation, we get the test statistic comes out to be negative 2.330 And so now we want to find, and that's so that's what this value is. We want to find the likelihood, if the difference is actually zero or the means are equal, how likely is this number or more extreme to come up? And so I'm going to use my uh my T c D E f to find this mighty CDF. And I'm going to use negative uh minority has negative one times 10 to the 99th hour in it. And then our upper value is going to be that negative 2.330 My degrees of freedom is again the 1 22. And I will paste and let that do the calculation. And I have a value of a p value of about 1%. And that is less than my 5% significance level. So I definitely have sufficient evidence to reject the null meaning that we believe that the Children that had seat belts on have a smaller uh stay in the uh in ICU than uh those that are non seatbelt. And again you write my sons for that. So now we want to find the confidence interval and the appropriate confidence intervals. Since this was a one tailed test and we had 5% in this tale, the confidence interval has 5% down here, 5% up there. We would need to look at a 90% confidence interval. And so my table does not have a T. V. Star value for 122 degrees of freedom. So I'm going to use my inverse T. Value my inverse teeth. And on the inverse T. I want the area below it to the point oh five. So I'm going to put in the 0.5. It will be finding basically this low limit which will be symmetrical. What the upper one and then our degrees of freedom are that 1 22? And we get that value to be Yeah. Uh And this was 1 22 comes out to be negative 1.657 And so now let's get that confidence center, let's look at the difference first. So we need this difference the 0.83 minus 1.39 And that difference comes out to be negative 0.56 plus or minus. And we have that T star value 1.657 because we have one down here and went up here and then we have this same uh same standard deviation we had here. So let's see if I get 1.77 squared and that is a 3.6 squared mm And our sample sizes are 1 23 and 2 90. Okay, And let's get that margin of air, Yeah. Mhm And 1.657 times square root of 1.77 squared divided by 1 23 plus 3.6 square divided by 2 90. And that margin of air comes out to be 0.398 and two. And it keeps going on but I'm just going to store that is X. And so let's get those two values. We have the negative 20.56 minus the X. Value. And that comes out to be negative 0.958 And then we can change that into an addition sign. And we find out that that is still negative and notice that that does not include zero. So we would again from this interval see that the difference does not appear to be zero as he appears to be negative. So it does appear as though um that the seat belts make a difference seatbelts beautiful. They're good that they appear to have a lesser chance of a lesser stay in the hospital in the ICU for Children.

So we believe that the mean number of ounces is 12 ounces and alternately, it doesn't say anything about trying to find if they're under filling or over filling. So we're going to use the not equal to and do a two tailed test. Now they found that the mean was 12.19 ounce. And we had a standard deviation of 0.11 ounces. And we have a sample size of 36. And so again, we're going to assume that 12 is the center and we're getting 12.19 And since it's a two tailed tests are also going to be finding both of these together will be our P value. So, well now I'm going to do this finding the P value and not doing it with the finding the critical T value. So what's the likelihood of sampling If the mean is actually 12 and getting 12.19 for a mean? Now we want both ends of the tail, so we're going to double that. So we need to convert that to a T. Value and we would have 35 degrees of freedom and we'll take our 12.19 minus 12 and then divided by the standard deviation or the square root of an and when we do that calculation that numerator becomes 0.19 divided by and then 0.11 and that looks like that's divided by six. And that comes out to a test statistic holy moly that test statistic comes out to be uh make sure I didn't type anything in in incorrectly. We get a test statistic of 10.36 and if we find that p value that's going to be approximately zero. So we actually have strong evidence what evidence against the null since this is way less than 5% which is our significance level. And so we would say that the mean the main seems to be different is different. Mhm. From 12 ounces now and again our p value is very very close to zero. It's a little bit bigger than zero. Now, it asked, does this does this end up show evidence in the way they were to the question that does it say that the people are being cheated and no they're not being cheated. Not cheated. Uh It looks like they're actually getting more than 12 ounces. So we didn't do that significance test for more than but it doesn't appear as though they're getting cheated. What?

Properly. 23. Good. Today it's note new. It's smaller battery quickly. 45 and alternative than you. It's bigger than do you mind that having t is equal to X bar minus you note over s over square root off at which is equal to 48 minus 45/5 480.4 over a square root off 25 which is equal to 2.778 for could should be the The value is a number off or interval in the column of title table five containing the T value is in the row off the F equal toe number off size when this one is 20 five minus one, which is 24 so that we value his between oh point over five and a four point or one. So if they devalue smaller than significantly developing, the non hypothesis is rejected and they be value here is smaller than a 0.1. So we reject then on high processes. So we can say that there is sufficient evidence to support Beckley

In this problem, we're going to be testing the effectiveness off seat bells we have to simple random samples off to groups off people. The first group is off. People not wearing seatbelts on the second group is for people who are wearing seatbelts, and we have proportions off people who, um, were killed during a car crash. So P one represents the portion off people who were killed on not wearing seatbelts, and that is that you won out off 2823 and P two. Heart represents the proportion off people who were killed in a car crash. Yet they were wearing seatbelts, and that is 16 out of 7000 765. So in the in this problem, you're going to be testing the the clean that seat belts are effective introducing fatalities, and the first step would be testing the claim using hypothesis test. So, for the hypothesis, the null hypothesis behalf p one equals P two, and for the alternative hypotheses we have P one is greater than P two. This implies that not wearing, uh, does the proportion of people who are killed when they are not wearing seatbelts is much higher than the proportion of people who were killed when they are wearing signals. That means that, uh, the seat belts are effective introducing fatalities. So for the test statistic, we need to substitute the values that you just obtained here into the formula. And when we do that, the value off the calculated value of that is 6.49 Okay. And since this is a one tailed test, the critical value is going to be 1.645 at 0.1 level 0.5 level of significance. Now we can compare these two values off that for the critical value, we can shade from 1.645 on the right and we can see that 6.45 is greater than 6.49 is greater than 1.6 now 45 which means that our calculated value is with being the critical region. And for that reason we conclude that we need to reject the nal hypotheses. And by rejecting the null hypothesis, we conclude that there is sufficient evidence to support the claim that the fatality rate is higher for those not wearing seatbelts. Next, we're going to test the claim by constructing on ah confidence interval and in this case we're going to construct a 90% confidence interval and fast. We need to get the margin of error e by using the formula given. And once we do that, the value off E is 0.33 We then need to substitute it to subtract it to the difference, be one heart and be too hot, and then added the difference to create the range. So in this case, the off the fast part will be p one minutes p two hut minus e and that gives you 0.57 now should be less than P one minus p two. And that is less than than when the some off the difference be one hut on and be may not speak to heart plus e which is 0.123 So we notice that the confidence interval limits do not include zero. And that means that the two fatality rates are not equal. Uh huh. And since the confidence interval limits on Lee include positive values, we can conclude that the fatality rate is higher for those not wearing seat bells. So once again, the claim has Bean supported using that the confidence interval method. Next, we're going to give an explanation to what the results suggest about the effectiveness off seat bells. So as we have seen, the proportions are different, and even when we check the proportions will notice that he won are divided by 202,828 gives us 1.1% and never on the P P. Two hot gives us 0.2%. And as you can see, the fast proportion is much greater. It's significantly larger, 1.1% is significantly larger than 0.2%. So the results suggest that the use of seat belt is associate it with fatalities, um, fatality rates that are lower than those as she waited with no, not using seat belts. So if you're using seatbelt, you're much more secure than when you're not


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