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Conduchypothesis test and provide tne test siabistc and Ihe cribcal value_ ard state the concusion Ananon dniico and filad it wth a Icad weght, then proceeded to ro...

Question

Conduchypothesis test and provide tne test siabistc and Ihe cribcal value_ ard state the concusion Ananon dniico and filad it wth a Icad weght, then proceeded to roll ( 200 tines Here are the observed froquoncics for tho culcomas ol 2, 3, and 6, raspacrivaly: 29 28. 49.39, 28,27 . Use 0.10 =gnficance tost tn0 caim that 0e outcoros are not oqujty Ilkely: Does appaar that tna logded dia banutus direrenby lair diatClkkbere Loltach-squnnieuatttonunt 0na le st btah sbc IATing Ihre A necimalcacutnaad

Conduc hypothesis test and provide tne test siabistc and Ihe cribcal value_ ard state the concusion Ananon dniico and filad it wth a Icad weght, then proceeded to roll ( 200 tines Here are the observed froquoncics for tho culcomas ol 2, 3, and 6, raspacrivaly: 29 28. 49.39, 28,27 . Use 0.10 =gnficance tost tn0 caim that 0e outcoros are not oqujty Ilkely: Does appaar that tna logded dia banutus direrenby lair diat Clkkbere Loltach-squnnieuatttonunt 0 na le st btah sbc IATing Ihre A necimalcacut naaded ) Tha cbchl valur IRoundI tlyga dacImal placus neudud Binla Ua concunkn Tholo cufllceun uvksulcu Luodtt Chirt Vui ouicontutautun nolennunil Mu4 Iina ouicomnanh bu uqunlly Akaly; a loudnd die buhavu dllerunb Iat di.



Answers

Conduct the hypothesis test and provide the test statistic and the P-value and/or critical value, and state the conclusion. The author drilled a hole in a dic and filled it with a lead weight, then proceeded to roll it 200 times. Here are the observed frequencies for the outcomes of $1,2,3,4,5,$ and 6 respectively: $27,31,42,40,28,$ and $32 .$ Use a 0.05 significance level to test the claim that the outcomes are not equally likely. Does it appear that the loaded die behaves differently than a fair die?

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

This question. Dayton through pin support at sailing so to you one plus you want equals So loss to so zero minus 300 stand sign. 30 equals over tools. Three over 32.2 three over two, minus 300. I understand. So we, in this case equal 17.9 44 feet per second. X direction, which is equal 0.4 equals dash minus Seattle A what? Seattle minus ***. 17.944 So dash equals 7.4 78 feet per second. Also t two Los Rito is equality three Close recently. So we have half times 300 over 32.2. Supply 7.178 Squared minus 300 times 10 equals zero minus 300 times. Then sign Sita. So Sita is equal 66.9 degrees.

For this problem. We're told that the proportions football injuries that occurred during practice is 57.6%. So we can work our claim as he equals 0.576 because that is the cleaned proportion. Well, we think that this is too high. So are alternative. Hypothesis is that P is actually less than 0.576 We're testing this out of significant level 0.5 And since he is less than this is a left tailed test. So our critical value this negative 1.6 for the next one you to find her test values. So we the sample in studies that sample we took a sample with the size of 36 and we got a result that 17 out of those 36 footballing periods occurred during practice. So our sample proportion p hat and that's equal to X over end. So 17/36. It gives us a proportion off 0.4 seven. So now we need to find ever see value and C is equal to p hat minus claimed pee all over the standard deviation just p. Times one minus p over end. If we plug in the values that we have played for seven minus point 576 of a square root 0.576 times one minus 10.576 over 36. And that gives us a Z value of negative one point 26 So to make her decision, we're going to draw a normal curve with a mean of zero and a standard deviation of one. I mean, zero 12 Great standard deviations. One negative too. Three. And I'm gonna draw in our critical value negative 1.6 floors about here and shade in their critical region that I'm gonna shade in our noncritical region. Yeah, that I'm going to write in her test value and greens. That's negative. 1.26 lobe. You fell here. We shaved this way so we can see her test value falls within are non critical reach in meaning for So we failed to reject the claims, even know So to summarize, this means that we do not have enough evidence to conclude that the U troop person of football injuries that occurred during practice. I'm just gonna rate contacts year because that's where you would include that is less fan so we can conclude that he true proportion off football injuries that occur during practices less Stan 57.6% and that's our answer.

On the long hypothesis for this portion will be that the proportion of degree recipients is the same for a sample and that for the report, the alternative hypothesis would be that they're not the same. Even if a single proportion is violated, The little hypothesis won't be, too. We had a sample size of 800 degree recipients off which 155 ever seen associated Greece. 450 harassing bachelor's degrees. 20 of them have received first professional degrees, 60 of them are masters and 15 of them are doctorates. Now the proportions given by the report zero point do. 33015110.30 point 206 and 0.2 for the respective categories. Now the expected values can be found out by multiplying the total sample size by their proportions. This is what I have done away. Everybody called him the expected calling and the values that I have found our 1 81 was anyone. 86.4 408.8 24 won, 64.8 and 16. The guy squares statistic over here can be calculated by this formula that is, observed Verrilli when it's expected Family who squared, divided by the expected value. You do this for all the categories and some them all up, and the value that we get for our tested distinct is 10.311 The degree of freedom for our question will be four, since we have five categories on the formalized and minus one satisfied by this one with a secret for the Alfa, for our question is your 10.5 and hence the critical value from the Chi Square table is 9.4 big now, as I calculated, that statistic is 10.31 which is greater than 9.48 We will reject the null hypothesis, which means that the proportion of degree who lives in our sample the first from the proportions given by their board this is a ransom


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