Essay About Confidence Interval And Correct Answerthe Null Hypothesis

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Exam #3 Review1.(5 point(s))    A report on health care in the U.S. said that 28% of Americans have experienced times when they haven’t been able to afford health care. A news organization randomly sampled 801 black Americans, of whom 38% reported that there had been times in the last year when they had not been able to afford health care. Which hypotheses would be appropriate to test that this problem is more severe among black Americans?a)[pic 1] b)[pic 2] c)[pic 3] d)[pic 4] e) [pic 5][pic 7]a. (a) is the correct answer[pic 8]b. (b) is the correct answer[pic 9]c. (c) is the correct answer[pic 10]d. (d) is the correct answer[pic 11]e. (e) is the correct answerThe null hypothesis must have an = sign, that is why (a) and (c) are incorrect[pic 6]The null hypothesis is about the population (not sample) proportion, that is why (d) and (e) are incorrect[pic 12]2.(5 point(s))    A report on health care in the U.S. said that 28% of Americans have experienced times when they haven’t been able to afford health care. A news organization randomly sampled 801 black Americans, of whom 31% reported that there had been times in the last year when they had not been able to afford health care. When a test of significance was conducted the p-value was 0.031. What conclusion can be drawn from this data?[pic 13]a. There are 3.1% more Black Americans who have experience times when they cannot afford health care[pic 14]b. There is no evidence from this data that indicates that a higher proportion of Black Americans cannot afford health care (p < .031)[pic 15]c. A higher proportion of Black Americans have experienced times when they cannot afford health care (p < .031)[pic 16]d. There is a 3.1% probability that there is a higher percentage of Black Americans who have experienced times when they cannot afford health care[pic 17]e. The proportion of Black Americans who have experienced times when they could not afford health care is the same as the general populationWhen the p-value is less than .05 we have uncovered a significant difference and must mention the p-valueSince .03 < .05 this is significant so (b) and (e) are incorrect(a) and (e) incorrectly interpret the meaning of the p-value[pic 18]3.(5 point(s))    A coffee machine dispenses coffee into paper cups. You are supposed to get 10 ounces of coffee, but the amount varies slightly from cup to cup. A random sample of 20 cups produced a sample mean of 9.85 ounces with a standard deviation of .2 ounces. Which hypotheses could be used to see if this provides evidence that the machine is shortchanging customers?a) [pic 19]  b) [pic 20]  c) [pic 21]  d)  [pic 22]  e)  [pic 23]  [pic 24]a. (a) is the correct answer[pic 25]b. (b) is the correct answer[pic 26]c. (c) is the correct answer[pic 27]d. (d) is the correct answer[pic 28]e. (e) is the correct answerThe null hypothesis is about the population mean not the sample mean that is why (c) (d) and (e) are incorrect. The null hypothesis must include and = sign which is why (b) is incorrect [pic 29]4.(5 point(s))    A coffee machine dispenses coffee into paper cups. You are supposed to get 10 ounces of coffee, but the amount varies slightly from cup to cup. A random sample of 20 cups produced a sample mean of 9.85 ounces with a standard deviation of .2 ounces. When a test of significance was conducted the p-value was 0.818. What conclusion can be drawn from this data?[pic 30]a. There is no evidence that less than 10 ounces of coffee is dispensed from this machine[pic 31]b. 81.8% of the time the machine dispenses 10 ounces of coffee[pic 32]c. The average coffee dispensed from this machine is less than 10 ounces (p < .818)[pic 33]d. There is an 81,8% probability that this machine dispenses 10 ounces of coffee[pic 34]e. The average coffee dispensed from this machine is 10 ounces [pic 35]When the p-value exceeds .05 we fail to reject the null hypothesisThe null hypothesis is that the coffee machine does dispense 10  ounces of coffee on average in each cup.Since .818 > .05 we do not reject thisNo evidence the machine is under filling the cups. Be careful. The statement in (e) is too strong. We are not 100% sure there is  on average 10 ounces per cup, we just have nothing inthis data to contradict it. [pic 36]5.(5 point(s))    A coffee machine dispenses coffee into paper cups. You are supposed to get 10 ounces of coffee, but the amount varies slightly from cup to cup. A random sample of 20 cups produced a sample mean of 9.85 ounces with a standard deviation of 0.2 ounces. If a test of significance was conducted to see if the machine did on average dispense 10 ounces of coffee, what would be the value of the test statistic? [pic 37]a. -1.85[pic 38]b. -3.35[pic 39]c. -4.02[pic 40]d. -2.33[pic 41]e. -0.74[pic 42] [pic 43]6.(5 point(s))    The saturated fat content in grams of several pizzas sold by two national chains is given in the table. We want to know if the two chains have significantly different mean saturate fat content. Which hypotheses could be used to test that there is no difference in the mean saturate fat content?Valentinos17121088101051616Sebastians671194479113a)   [pic 44]  b)  [pic 45]   c) [pic 46]  d) [pic 47] e) [pic 48][pic 49]a. (a) is the correct answer[pic 50]b. (b) is the correct answer[pic 51]c. (c) is the correct answer[pic 52]d. (d) is the correct answer[pic 53]e. (e) is the correct answerThis is quantitative data, hence means not proportions – (a) (b) wrongNull hypotheses are about population parameters, not sample statistics –(b) and (d) wrongThis is two independent samples not paired data  – (e) wrongOften students think this is paired data. To be paired data there would have to be a good explanation for why the 17 and 6, for example, “go together” and there is not.Be careful not to confuse independent data with paired data. If it is paired the reason for the paired will be explained in the problem. [pic 54]7.(5 point(s))    Ten men volunteered to test an exercise devise advertised on television by using it 3 times a week for 20 minutes. Their resting pulse rates (beats per minute) were measured before the test began, and then 6 weeks later. Results are in the table. Which hypotheses could be used to test that there is no difference in the mean pulse rates of men before and after they used the exercise device as described?SubjectBeforeAfterAllen7373Brandon8379Carlos8581David8786Edwin9187Franco9991George8784Henry8583Ivan8384Dan7976 a)   [pic 55]  b)  [pic 56]   c) [pic 57]  d) [pic 58] e) [pic 59][pic 60]a. (a) is the correct answer[pic 61]b. (b) is the correct answer[pic 62]c. (c) is the correct answer[pic 63]d. (d) is the correct answer[pic 64]e. (e) is the correct answerThis is paired, quantitative data. Here the reason for pairing is that a “pair” consists of a before and after measurement on the same person.  [pic 65]8.(5 point(s))    A survey of 430 randomly chosen adults found that 21% of the 222 men and 18% of the 208 women had purchased books online. . Does the data indicate there is a difference in the proportion of men and women who purchase books online? Which hypotheses could be used to test that there is no difference in the percent of men and women who purchase books online?a)   [pic 66]  b)  [pic 67]   c) [pic 68]  d) [pic 69] e) [pic 70][pic 72]a. (a) is the correct answer[pic 73]b. (b) is the correct answer[pic 74]c. (c) is the correct answer[pic 75]d. (d) is the correct answer[pic 76]e. (e) is the correct answer[pic 71] Now we have categorical data, hence proportions not means(b) is wrong since null hypothesizes are about population parameters, not sample statistics [pic 77]9.(5 point(s))    The saturated fat content in grams of several pizzas sold by two national chains is given in the table. We want to know if the two chains have significantly different mean saturate fat content. The following StatCrunch output was obtained: ValentinosSebastians176127101189841041075916111613[pic 79][pic 80]a. the mean fat content of Sebastins pizza is higher than Valentinos [pic 81]b. the mean fat content of Valentinos pizza is higher than Sebastians [pic 82]c. the mean fat content of Sebastins pizza is higher than Valentinos (p < .06)[pic 83]d. There is no significant difference in the mean fat content Valentinos and Sebastians pizzas[pic 84]e. the mean fat content of Valentinos pizza is higher than Sebastians (p < .06)[pic 85]f. The mean fat content of Valentinos pizza is the ame as Sebastians[pic 86][pic 78][pic 87] [pic 88]10.(5 point(s))    The saturated fat content in grams of several pizzas sold by two national chains is given in the table. We want to know if the two chains have significantly different mean saturate fat content. ValentinosSebastians176127101189841041075916111613The following 95% confidence interval was obtained. Which of the following best interprets this interval?[pic 89][pic 90]a. With 95% confidence the mean fat content of Valentinos pizza is between .23 grams and 6.43 grams[pic 91]b. With 95% confidence the mean fat content of Sebastians pizza is between .23 grams and 6.43 grams[pic 92]c. With 95% confidence we cannot establish a difference in the mean fat content of Valentinos and Sebastians pizzas[pic 93]d. With 95% confidence we can say that the mean fat content of Sebastians pizza is higher than Valentinos[pic 94]e. With 95% confidence we can say that mean fat content of Valentinos pizza is higher than Sebastians Always check to see if the interval contains zero. If it does then the difference in means could be zero. Hence not able to establish a difference with 95% confidence.[pic 95]11(5 point(s))    Ten men volunteered to test an exercise devise advertised on television by using it 3 times a week for 20 minutes. Their resting pulse rates (beats per minute) were measured before the test began, and then 6 weeks later. Results are in the table. We want to know if the resting pulse rates have changed. The following StatCrunch output was obtained. Which statement best interprets this output? SubjectBeforeAfterAllen7373Brandon8379Carlos8581David8786Edwin9187Franco9991George8784Henry8583Ivan8384Dan7976[pic 97][pic 98][pic 99][pic 100][pic 96][pic 101]a. On average the resting pulse rates decreased after the exercise program (p < .0067)[pic 102]b. There is no evidence from this data of a change in the average pulse rates[pic 103]c. On average the resting pulse rates did no change as a result of the exercise program (p < .0067)[pic 104]d. There is no evidence from this data of a change in the average pulse rates (p < .0067)[pic 105]e. On average the resting pulse rate increased after the exercise program (p < .0067)When rejecting the null hypothesis must mention the p-value [pic 106]12.(5 point(s))    Ten men volunteered to test an exercise devise advertised on television by using it 3 times a week for 20 minutes. Their resting pulse rates (beats per minute) were measured before the test began, and then 6 weeks later. Results are in the table.  SubjectBeforeAfterAllen7373Brandon8379Carlos8581David8786Edwin9187Franco9991George8784Henry8583Ivan8384Dan7976The following 95% confidence interval was obtained. Which of the following best interprets this interval?[pic 107]Before – After positive → Before if higher → pulse rates decreasedUnits are beats per minute not percent[pic 108]a. with 95% confidence we can say that on average the pulse rate will increase by between .99 and 4.61 beats per minute following the exercise program[pic 109]b. with 95% confidence we can say that on average pulse rate will increase between 9.9% and 46.1% following the exercise program[pic 110]c. with 95% confidence we can say that on average will decrease between 9.9% and 46.1% following the exercise program[pic 111]d. with 95% confidence we can say that on average pulse rates will decrease between .99 and 4.61 beats per minute following the exercise program[pic 112]e. with 95% confidence we cannot detect a significant difference in pulse rates following the exercise program [pic 113]13.(5 point(s))    A survey of 430 randomly chosen adults found that 21% of the 222 men and 18% of the 208 women had purchased books online. . Does the data indicate there is a difference in the proportion of men and women who purchase books online? The following StatCrunch output was obtained. Which statement best interprets these results?[pic 115][pic 116]a. A significantly higher proportion of women buy books online than men (p < .37) [pic 117]b. There is no evidence from this data of a significant difference in the mean number of men and women who buy books online[pic 118]c. The mean number of men who purchase books online is significantly higher than the mean number of women who buy books online (p < .37)[pic 119]d. A significantly higher proportion of men buy books online than women (p < .37)[pic 120]e. There is no evidence from this data of a significant difference in the proportion of men and women who buy books onlineWhen you fail to reject the null hypothesis you do not mention the p-value[pic 121][pic 122][pic 114] [pic 123]14.(5 point(s))    A survey of 430 randomly chosen adults found that 21% of the 222 men and 18% of the 208 women had purchased books online. . Does the data indicate there is a difference in the proportion of men and women who purchase books online? The following 95% confidence interval was obtained. Which statement best interprets this interval? [pic 125]We notice that zero is included in the interval which means the difference in proportions could be zero. Hence no evidence of a significant difference here.[pic 124]General principle when making comparisons. If zero included in the interval no evidence of a significant difference. If zero not included then there is a significant difference and the boundary values tell us how much of a difference[pic 126]a. with 95% confidence we cannot establish a difference in the percent of men and women who buy books online[pic 127]b. with 95% confidence we can say that between 4% and 11% more women than men buy books on line[pic 128]c. with 95% confidence we can say that a higher mean number of men buy books online than women[pic 129]d. with 95% confidence we can say that between 4% and 11% more men than women buy books online[pic 130]e. with 95% confidence we cannot establish a difference in the mean number of men and women who buy books online [pic 131]15.(5 point(s))    A survey of 430 randomly chosen adults found that 47 of the 222 men and 37 of the 208 women had purchased books online. . Does the data indicate there is a difference in the proportion of men and women who purchase books online? Based on the data they construct a 95% confidence interval. Could this interval be used to test the null hypothesis? [pic 133][pic 132]Ho: p(men) – p(women) = 0 and zero is in the 95% interval so we should not reject Ho because it could be trueGeneral principle when making comparisons. If zero is included in the confidence interval then do not reject H0. If zero is not included then do reject H0.[pic 134]a. Yes, since zero falls within the confidence interval we decide not reject the null hypothesis meaning we do not have enough evidence to claim there is a difference in the percent of men and women who buy books online.[pic 135]b. No, because more men were sampled than women[pic 136]c. Yes, since zero falls within the confidence interval we can decide to reject the null hypothesis and decide a higher percent of women buy books online (p < .05)[pic 137]d. No, because sample proportions are not always normally distributed[pic 138]e. Yes, since zero falls within the confidence interval we can decide to reject the null hypothesis and decide a higher percent of men buy books online (p < .05) [pic 139]16.(5 point(s))    The saturated fat content in grams of several pizzas sold by two national chains is given in the table. We want to know if the two chains have significantly different mean saturate fat content. Based on the data a 95% confidence interval was constructed. Could this interval be used to test the null hypothesis.ValentinosSebastians176127101189841041075916111613[pic 140]The confidence interval includes zero → there may be do difference in the mean fat content → do not reject Ho[pic 142]a. Yes, since zero included in the interval we concluded there is not enough evidence to conclude there is a difference in the mean saturated fat content of Valentinos and Sebastians pizza[pic 143]b. Yes, since zero is included in the confidence interval we can conclude that the mean saturated fat content of Sebastians is higher than Valentinos (p < .05)[pic 144]c. No, since n < 30 the sample means are not normally distributed[pic 145]d. No, since we need to calculate the sample mean and standard deviation to test the hypothesis[pic 146]e. Yes, since zero is included in the confidence interval we can conclude that the mean saturated fat content of Valentinos pizza is higher than Sebastians (p < .05)[pic 141] [pic 147]17.(5 point(s))    Ten men volunteered to test an exercise devise advertised on television by using it 3 times a week for 20 minutes. Their resting pulse rates (beats per minute) were measured before the test began, and then 6 weeks later. Results are in the tableBased on the data a 95% confidence interval was constructed. Could this interval be used to test the null hypothesis?SubjectBeforeAfterAllen7373Brandon8379Carlos8581David8786Edwin9187Franco9991George8784Henry8583Ivan8384Dan7976[pic 149]Since zero is not included in the confidence interval we decide to reject Ho since it is unlikely the difference is zero. Since (before – after) is positive we conclude that the pulse rates decreased[pic 150]a. No, since n < 30 sample means are not normally distributed[pic 151]b. Yes, since zero is not included in the confidence interval we accept the null hypothesis (p < .05)[pic 152]c. Yes, since zero is not in the confidence interval we fail to reject the null hypothesis (p < .05)[pic 153]d. No, since zero is not included in the confidence interval the results are inconclusive[pic 154]e. Yes, since zero is not included in the confidence interval we reject the null hypothesis (p < .05)[pic 148][pic 155]18.(5 point(s))    A researcher investigating whether joggers are less likely to get colds than people who do not jog found a p-value of .03. This means that A p-value measures the probability that the null hypothesis was rejected when in fact it was true. A p-value of .03 would cause us to reject the null hypothesis of no difference and conclude that joggers have fewer colds. [pic 156]a. 3% of joggers get colds[pic 157]b. Joggers get fewer colds than non-joggers and there is only a 3% chance this is incorrect[pic 158]c. There is a 3% chance that joggers do not get fewer colds[pic 159]d. Joggers get 3% fewer colds than non-joggers[pic 160]e. There is a 3% chance that joggers get fewer colds [pic 161]19.(5 point(s))    We are about to test an hypothesis using data from a well designed study. Which is true? I: A large p-value will provide strong evidence against the null hypothesisII: A small p-value will provide strong evidence against the null hypothesisIII: The smaller the p-value the less likely the null hypothesis is trueIV: The larger the p-value the more likely it is we will accept the null hypothesisSmall p-values are evidence against the null hypothesis. The smaller the p-value the less likely it is that the null hypothesis is correct.!V is wrong since we never accept the null hypothesis …. We simply fail to reject it.[pic 162]a. III only[pic 163]b. II only[pic 164]c. I and IV only[pic 165]d. I only[pic 166]e. II and III only [pic 167]20.(5 point(s))    The p-value in a test of significance is [pic 168]a. The probability of incorrectly rejecting the null hypothesis[pic 169]b. The probability of incorrectly accepting the null hypothesis[pic 170]c. The probability the null hypothesis is correct[pic 171]d. The probability of incorrectly rejecting the alternative hypothesis[pic 172]e. The probability of accepting a false null hypothesis