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.46 Null hypotheses and research each of the toloevingd epearchhehypotheshc For same as those Exercise 7.45),state the null hypothesis and the In research hypothesi...

Question

.46 Null hypotheses and research each of the toloevingd epearchhehypotheshc For same as those Exercise 7.45),state the null hypothesis and the In research hypothesis, in both words and symbolic notation: a. A researcher is interested in studying the relation between the use of antibacterial products and the dryness of peoples skin. He thinks these produce might alter the moisture in skin differently from other products that are not antibacterial. b. A student wonders if grades in a class are in

.46 Null hypotheses and research each of the toloevingd epearchhehypotheshc For same as those Exercise 7.45),state the null hypothesis and the In research hypothesis, in both words and symbolic notation: a. A researcher is interested in studying the relation between the use of antibacterial products and the dryness of peoples skin. He thinks these produce might alter the moisture in skin differently from other products that are not antibacterial. b. A student wonders if grades in a class are in any way related to where student sits in the classroom: In particular; do students who sit in the front IOW get better grades, on average, than the general popula- tion of students? Cell phones are everywhere, and we are now avail- able by phone almost all of the time Does this translate into change in the closeness of our long distance relationships?



Answers

Some of the following statements refer to the null hypothesis, some to the alternate hypothesis.
State the null hypothesis, Ho, and the hypothesis. $H_{a},$ in terms of the appropriate parameter $(\mu \text { or } p)$
a. The mean number of years Americans work before retiring is $34 .$
b. At most 60$\%$ of Americans vote in presidential elections.
c. The mean starting salary for San Jose State University graduates is at least $\$ 100,000$ per year.
d. Twenty-nine percent of high school seniors get drunk each month.
e. Fewer than 5 $\%$ of adults ride the bus to work in Los Angeles.
f. The mean number of cars a person owns in her lifetime is not more than ten.
g. About half of Americans prefer to live away from cities, given the choice.
h. Europeans have a mean paid vacation each year of six weeks.
i. The chance of developing breast cancer is under 11$\%$ for women.
j. Private universities' mean tuition cost is more than $\$ 20,000$ per year.

In this problem, We're going to be looking at the study that was conducted to investigate the association between cell phone use and the hemispheric brain dominance. So we have two samples. The first sample has, uh, 216 subjects who preferred to use their left year on bond, and and yet 166 are right 100. So the first proportion off right 100 people is 166 out off the 216 people who preferred to use their left ear. The second sample was coming from 452 subjects would prefer to use their right here. So this is from the left here. And this is for those who prefer to use their right ear on. For those who prefer to use your there, Right here are 400 36 were right handed out off 452. So we're going to be conducting tests, uh, to test the claim that the writ off right handedness for those who prefer to use their left here, it's less than the rich off right handedness for those who prefer to use their right here. In other words, the fast proportion is much greater than the second proportion. Okay, so let's compare and fast to do a hypothesis test, we will need to right there. Now hypothesis on the alternative hypothesis. So they're not. Hypothesis is p one he calls p two on the second hypothesis, the alternative hypotheses p one iss less than p two. Which is to say that this fraction this proportion is much smaller. That when you're and you're right, when you prefer to use your left here, then fewer people will be, uh, right handed. No for the zed, uh, calculate the value of that. We need to go to substitute the values into the formula. And this gives us the value of that off negative. 7.933 Now we need to get a critical value for this being a one tailed test and the level of significance being 0.1 then the critical value offset would be equal to negative 2.33 Because this time we're saying that p one IHS less than p two. So therefore it's going to be negative and not positive. Now we can compare the two by during the critical region and as you can see this would be negative 2.33 and the the left Month will be the critical region. And with that, we would have to notice that the calculated Valley offset is within the critical region. And for that reason we reject the null hypothesis. Rejecting the null hypotheses means that there is sufficient evidence to support the claim that the rate off right handedness for those who profound to use their left here is less than the rate off right handedness for those who prefer to use your right ear for cell phones. So it's much easier for you to use your your right hand if you're right handed. In other words. So we tested clean using, uh ah confidence interval. And for that, we need to complete the value off the margin of error has shown in this formula, and when you substitute the values you obtained e us zero point 0698 Next you substitute the values into the confidence interval expression and you find the limit will be negative 0.26 six aunt, the other limit his negative 0.1 to 16. So notice that all these values are negative, meaning zero is not contained within the confidence limits. And that means that there is a significant difference between the two proportions. And because the interval consists of negative numbers only it appears that they claim is supported. Therefore, we can see that the difference between the populations does appear to have practical significance.

In this problem, we're going to be testing the effectiveness off. ECON ASIA In treating calls, we have two groups off subjects. The first subject was given a kidney Asia. The first group was given magnesia, and the second group was given a placebo. So we can say P one hunt represents the proportion off the people who developed the retrovirus infections after being given a condition, and that is 40 45. So that's a fraction off. Those who developed renew various infections after being given akin Asia. And for those who developed in various infections after being given the possible are 88 out off a total of 103 subjects. And to test the effectiveness off back in Asia for Coles, we're going to use two approaches. The fast approach is going to be a hypothesis. Test on the second one is going to be the confidence interval. So we're going to use the 0.5 significance level, and we're testing the clean that back in Asia has an effect on rhinovirus infections. We're not giving, uh, we're not saying that one has ah, one is more effective than they ever were. Just saying that there is no effect on grain of virus infections that makes these tests are two tails test on the critical value on that is plus or minus 1.96 So let's go ahead and get the test statistic, which is that obtained by substituting the values obtained into the formula. And when we do so, we get the calculated value of that zero point 573 and when we compare the calculated value of that and the critical value of that in this case we have it as 1.96 positive and negative 1.96 So the calculated value of that is zero 0.573 and it is not within the critical region and for that reason we fail to reject the narrow hypothesis. Failure to reject an al hypotheses means that there is not sufficient evidence to support the claim that back in Asia has a NIF effect. So we move on to the second test by constructing an appropriate confidence interval and to do so to get a 95% confidence interval. We need to use the formula given and fast work out the margin of error e, and when you substitute the values off into the formula, we get that. The margin of error e is 0.1143 and when we substitute this into their confidence interval, we get that. The intervals limits are negative. 0.7 93 less than P 1 may not be too, and 0.1493 So we noticed that the confidence interval limits do contain zero, so zero is included within the confidence interval limits and thes shows that there is not significance thing. There is not a significant difference between the two proportions because when zero is included, WAY can see that there is not a significant difference between the proportions. In other words, there is no sufficient evidence to support the claim that back in Asia treatments has an effect. And in the last part of the question see, we're answering the question. Does echinacea appear to have any effect on the infection rate and according to the results, we see that back in Asia does not appear to have a significant effect on the infection rate and because it does not appear to have an effect, a significant effect, it should not be recommended because it's a safe for those two proportions do not have any significant difference.

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

In this problem we're going to be looking at to smoking cessation programs. One of them is the sustained cab, and the other one is the standard camp in the sustained care. We had 198 smokers going through the program on out of those 198 smokers, 51 no longer smoke after six months and for this standard care program among 189 smokers, fatty one no longer smoking after six months. So we're going to use the 0.1 significance level to test the claim that the rate of success for smoking cessation is greater with the sustained care program are compared to them standard care program. So the first part of the question we're going to test the claim using a hypothesis test. After that, we're going to test it using the confidence interval. And then we're going to see whether there's a difference between the two programs, uh, in terms off the proportions. Okay, so let's begin and state the hypothesis on. In this case, the non hypothesis is P one equals P two, which implies that the two proportions have are equal and the alternative hypotheses is P one is greater than P two. So this means that according to the hypothesis that we have a higher proportion off people who no longer smokey the sustained care program compared to the standard care program. So this being, uh, one tales test, the critical value will be 2.33 So we need to substitute the values into the test statistic, and we have the following proportion. For those in the sustained care, we have 51 longest, smoking out off 188. And for the standard care, we have 30 the longest, walking out off the 199. And to get the calculated value of that, we need to substitute these values into the formula. And here p one hot in decimal form is 0.258 and we need to subtract p too hot, which is 0.151 Okay, then the difference. We subtract zero from the difference, which is assumed to be, uh, zero, according to the Nile hypothesis. So then, from there we get the square it off. P one p bar Cuba, developed by N one plus p. Baquba Weber. And to So PBA is given by the sum of all the, uh all the X X one and next to divided by anyone and into So what we need to have here is the sum of 51 and that he divided by the summer of 188 and 199. And when you walk that out together, P bob is 0.204 divided by N one, which is 188 on, we have to multiply the PBA, Cuba and Cuba is one minute 0.204 which is going to be 0.796 So the new Morita is repeated in the next fraction. So it's 0.204 time 0.796 divided by N to an end to is 199. And when we simplify the value off the test statistic that he's 2.6 five now, we can compare the calculated value of that and the critical value off that and you can see. But the critical value is 233 so we can share the right side and the calculated value of that is within the critical region, which is 2.65 And since that is the case, we have to make the conclusion to reject the null hypothesis. No, this means that there is sufficient evidence to support the clean, that the rate of success is for the smoking cessation is greater with the sustained care program compared to them the standard care program. Next, we're going to test the same claim by constructing, uh, confidence interval. And in this case, we're going to construct 98% confidence interval. And for that we need to get the margin of error. E is in the formula given, and when you walk that out you will get e equals 0.9 0.9 35 And when you substitute into this formula for P one heart minus B p one hut minus be too hot minus e, you will obtain the following confidence interval. So to be 0.1 35 it's less than P one main US P two, which is less than zero point 2005 now. In this case, we noticed that the confidence interval limits do not contain zero. And that means that there is significant, a significant difference between the two proportions, as we have had really seen in the test off hypothesis, meaning we could reject the null hypothesis off having the proportions being the same. So because the interval consists off positive numbers only, it appears that the success rate for the sustained care program is greater than the success rate for the standard care program. Lastly, in Passy, we're going to see whether the difference between the two programs have practical significance. So if you check the the percentages, for example, for P one it is 25.8%. And for P two hot, this is 15 0.1%. And based on this sample, the success rates off the programs our 5th 25 points, 8% and 15.1%. And that difference does appear to be substantial. There's a there's a difference. So the difference between the programs does appear to have practical significance, because it's about a difference off 10% between p of P one heart and P to heart


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