Use either the critical-value approach or the P-value approach to perform the required hypothesis test. For several years, evidence had been mounting that folic acid reduces major birth defects. A. Czeizel and I. Dudas of the National Institute of Hygiene in Budapest directed a study that provided the strongest evidence to date. Their results were published in the paper “Prevention of the First Occurrence of Neural-Tube Defects by Periconceptional Vitamin Supplementation” (New England Journal of Medicine, Vol. 327(26), p. 1832). For the study, the doctors enrolled women prior to conception and divided them randomly into two groups. One group, consisting of 2701 women, took daily multivitamins containing 0.8 mg of folic acid, the other group, consisting of 2052 women, received only trace el

Use either the critical-value approach or the P-value approach to perform the required hypothesis test. For several years, evidence had been mounting that folic acid reduces major birth defects. A. Czeizel and I. Dudas of the National Institute of Hygiene in Budapest directed a study that provided the strongest evidence to date. Their results were published in the paper “Prevention of the First Occurrence of Neural-Tube Defects by Periconceptional Vitamin Supplementation” (New England Journal of Medicine, Vol. 327(26), p. 1832). For the study, the doctors enrolled women prior to conception and divided them randomly into two groups. One group, consisting of 2701 women, took daily multivitamins containing 0.8 mg of folic acid, the other group, consisting of 2052 women, received only trace el

Question
Study design
asked 2020-11-26
Use either the critical-value approach or the P-value approach to perform the required hypothesis test. For several years, evidence had been mounting that folic acid reduces major birth defects. A. Czeizel and I. Dudas of the National Institute of Hygiene in Budapest directed a study that provided the strongest evidence to date. Their results were published in the paper “Prevention of the First Occurrence of Neural-Tube Defects by Periconceptional Vitamin Supplementation” (New England Journal of Medicine, Vol. 327(26), p. 1832). For the study, the doctors enrolled women prior to conception and divided them randomly into two groups. One group, consisting of 2701 women, took daily multivitamins containing 0.8 mg of folic acid, the other group, consisting of 2052 women, received only trace elements. Major birth defects occurred in 35 cases when the women took folic acid and in 47 cases when the women did not. a. At the 1% significance level, do the data provide sufficient evidence to conclude that women who take folic acid are at lesser risk of having children with major birth defects? b. Is this study a designed experiment or an observational study? Explain your answer. c. In view of your answers to parts (a) and (b), could you reasonably conclude that taking folic acid causes a reduction in major birth defects? Explain your answer.

Answers (1)

2020-11-27

Step 1 a) \(\displaystyle{H}_{{{0}}}:{p}_{{{1}}}={p}_{{{2}}}\)
\(\displaystyle{h}_{{{a}}}:{p}_{{{1}}}{<}{p}_{{{2}}}\) The sample proportion is the number of successes divided by the sample size: \(\displaystyle{w}{i}{d}{e}\hat{{{p}}}_{{{1}}}={\frac{{{x}_{{{1}}}}}{{{n}_{{{1}}}}}}={\frac{{{35}}}{{{2701}}}}\approx{0.013}\)
\(\displaystyle{w}{i}{d}{e}\hat{{{p}}}_{{{2}}}={\frac{{{x}_{{{2}}}}}{{{n}_{{{2}}}}}}={\frac{{{47}}}{{{2052}}}}\approx{0.023}\)
\(\displaystyle{w}{i}{d}{e}\hat{{{p}}}_{{{p}}}={\frac{{{x}_{{{1}}}+{x}_{{{2}}}}}{{{n}_{{{1}}}+{n}_{{{2}}}}}}={\frac{{{35}+{47}}}{{{2701}+{2052}}}}={\frac{{{82}}}{{{4753}}}}\approx{0.017}\) Determine the value of the test statistic: \(z=\frac{\widehat{p}_{1}-\widehat{p}_{2}}{\sqrt{\widehat{p}_{p}(1-\widehat{p}_{p})}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}}=\frac{0.013-0.023}{\sqrt{0.017(1-0.017)\sqrt{\frac{1}{2701}+\frac{1}{2052}}}}\approx -2.64\) Determine the P-value using table 2: \(\displaystyle{P}={0.0041}\) If the P-value is smaller than the significance level, reject the null hypothesis: \(\displaystyle{P}{<}{0.01}={1}\%\Rightarrow\) Reject \(\displaystyle{H}_{{{0}}}\) Step 2 b) An experiment deliberately imposes some trearment on individuals in order to observe their responses. An observational study tries to gather information without disturbing the scene they are observing. Designed Experiment c) Yes, because we rejected the null hypothesis in (a). Answer: a. Yes b. Designed experiment c. Yes

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Relevant Questions

asked 2021-03-06
Use either the critical-value approach or the P-value approach to perform the required hypothesis test. Approximately 450,000 vasectomies are performed each year in the United States. In this surgical procedure for contraception, the tube carrying sperm from the testicles is cut and tied. Several studies have been conducted to analyze the relationship between vasectomies and prostate cancer. The results of one such study by E. Giovannucci et al. appeared in the paper “A Retrospective Cohort Study of Vasectomy and Prostate Cancer in U.S. Men” (Journal of the American Medical Association, Vol. 269(7), pp. 878-882). Of 21,300 men who had not had a vasectomy, 69 were found to have prostate cancer, of 22,000 men who had had a vasectomy, 113 were found to have prostate cancer. a. At the 1% significance level, do the data provide sufficient evidence to conclude that men who have had a vasectomy are at greater risk of having prostate cancer? b. Is this study a designed experiment or an observational study? Explain your answer. c. In view of your answers to parts (a) and (b), could you reasonably conclude that having a vasectomy causes an increased risk of prostate cancer? Explain your answer.
asked 2021-05-05

A random sample of \( n_1 = 14 \) winter days in Denver gave a sample mean pollution index \( x_1 = 43 \).
Previous studies show that \( \sigma_1 = 19 \).
For Englewood (a suburb of Denver), a random sample of \( n_2 = 12 \) winter days gave a sample mean pollution index of \( x_2 = 37 \).
Previous studies show that \( \sigma_2 = 13 \).
Assume the pollution index is normally distributed in both Englewood and Denver.
(a) State the null and alternate hypotheses.
\( H_0:\mu_1=\mu_2.\mu_1>\mu_2 \)
\( H_0:\mu_1<\mu_2.\mu_1=\mu_2 \)
\( H_0:\mu_1=\mu_2.\mu_1<\mu_2 \)
\( H_0:\mu_1=\mu_2.\mu_1\neq\mu_2 \)
(b) What sampling distribution will you use? What assumptions are you making? NKS The Student's t. We assume that both population distributions are approximately normal with known standard deviations.
The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations.
The standard normal. We assume that both population distributions are approximately normal with known standard deviations.
The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations.
(c) What is the value of the sample test statistic? Compute the corresponding z or t value as appropriate.
(Test the difference \( \mu_1 - \mu_2 \). Round your answer to two decimal places.) NKS (d) Find (or estimate) the P-value. (Round your answer to four decimal places.)
(e) Based on your answers in parts (i)−(iii), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level \alpha?
At the \( \alpha = 0.01 \) level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
At the \( \alpha = 0.01 \) level, we reject the null hypothesis and conclude the data are statistically significant.
At the \( \alpha = 0.01 \) level, we fail to reject the null hypothesis and conclude the data are statistically significant.
At the \( \alpha = 0.01 \) level, we reject the null hypothesis and conclude the data are not statistically significant.
(f) Interpret your conclusion in the context of the application.
Reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver. (g) Find a 99% confidence interval for
\( \mu_1 - \mu_2 \).
(Round your answers to two decimal places.)
lower limit
upper limit
(h) Explain the meaning of the confidence interval in the context of the problem.
Because the interval contains only positive numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, we can not say that the mean population pollution index for Englewood is different than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains only negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is less than that of Denver.
asked 2021-01-17
A new thermostat has been engineered for the frozen food cases in large supermarkets. Both the old and new thermostats hold temperatures at an average of \(25^{\circ}F\). However, it is hoped that the new thermostat might be more dependable in the sense that it will hold temperatures closer to \(25^{\circ}F\). One frozen food case was equipped with the new thermostat, and a random sample of 21 temperature readings gave a sample variance of 5.1. Another similar frozen food case was equipped with the old thermostat, and a random sample of 19 temperature readings gave a sample variance of 12.8. Test the claim that the population variance of the old thermostat temperature readings is larger than that for the new thermostat. Use a \(5\%\) level of significance. How could your test conclusion relate to the question regarding the dependability of the temperature readings? (Let population 1 refer to data from the old thermostat.)
(a) What is the level of significance?
State the null and alternate hypotheses.
\(H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}>?_{2}^{2}H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}\neq?_{2}^{2}H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}?_{2}^{2},H1:?_{1}^{2}=?_{2}^{2}\)
(b) Find the value of the sample F statistic. (Round your answer to two decimal places.)
What are the degrees of freedom?
\(df_{N} = ?\)
\(df_{D} = ?\)
What assumptions are you making about the original distribution?
The populations follow independent normal distributions. We have random samples from each population.The populations follow dependent normal distributions. We have random samples from each population.The populations follow independent normal distributions.The populations follow independent chi-square distributions. We have random samples from each population.
(c) Find or estimate the P-value of the sample test statistic. (Round your answer to four decimal places.)
(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?
At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant. At the ? = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.
(e) Interpret your conclusion in the context of the application.
Reject the null hypothesis, there is sufficient evidence that the population variance is larger in the old thermostat temperature readings.Fail to reject the null hypothesis, there is sufficient evidence that the population variance is larger in the old thermostat temperature readings. Fail to reject the null hypothesis, there is insufficient evidence that the population variance is larger in the old thermostat temperature readings.Reject the null hypothesis, there is insufficient evidence that the population variance is larger in the old thermostat temperature readings.
asked 2020-10-23
The table below shows the number of people for three different race groups who were shot by police that were either armed or unarmed. These values are very close to the exact numbers. They have been changed slightly for each student to get a unique problem.
Suspect was Armed:
Black - 543
White - 1176
Hispanic - 378
Total - 2097
Suspect was unarmed:
Black - 60
White - 67
Hispanic - 38
Total - 165
Total:
Black - 603
White - 1243
Hispanic - 416
Total - 2262
Give your answer as a decimal to at least three decimal places.
a) What percent are Black?
b) What percent are Unarmed?
c) In order for two variables to be Independent of each other, the P \((A and B) = P(A) \cdot P(B) P(A and B) = P(A) \cdot P(B).\)
This just means that the percentage of times that both things happen equals the individual percentages multiplied together (Only if they are Independent of each other).
Therefore, if a person's race is independent of whether they were killed being unarmed then the percentage of black people that are killed while being unarmed should equal the percentage of blacks times the percentage of Unarmed. Let's check this. Multiply your answer to part a (percentage of blacks) by your answer to part b (percentage of unarmed).
Remember, the previous answer is only correct if the variables are Independent.
d) Now let's get the real percent that are Black and Unarmed by using the table?
If answer c is "significantly different" than answer d, then that means that there could be a different percentage of unarmed people being shot based on race. We will check this out later in the course.
Let's compare the percentage of unarmed shot for each race.
e) What percent are White and Unarmed?
f) What percent are Hispanic and Unarmed?
If you compare answers d, e and f it shows the highest percentage of unarmed people being shot is most likely white.
Why is that?
This is because there are more white people in the United States than any other race and therefore there are likely to be more white people in the table. Since there are more white people in the table, there most likely would be more white and unarmed people shot by police than any other race. This pulls the percentage of white and unarmed up. In addition, there most likely would be more white and armed shot by police. All the percentages for white people would be higher, because there are more white people. For example, the table contains very few Hispanic people, and the percentage of people in the table that were Hispanic and unarmed is the lowest percentage.
Think of it this way. If you went to a college that was 90% female and 10% male, then females would most likely have the highest percentage of A grades. They would also most likely have the highest percentage of B, C, D and F grades
The correct way to compare is "conditional probability". Conditional probability is getting the probability of something happening, given we are dealing with just the people in a particular group.
g) What percent of blacks shot and killed by police were unarmed?
h) What percent of whites shot and killed by police were unarmed?
i) What percent of Hispanics shot and killed by police were unarmed?
You can see by the answers to part g and h, that the percentage of blacks that were unarmed and killed by police is approximately twice that of whites that were unarmed and killed by police.
j) Why do you believe this is happening?
Do a search on the internet for reasons why blacks are more likely to be killed by police. Read a few articles on the topic. Write your response using the articles as references. Give the websites used in your response. Your answer should be several sentences long with at least one website listed. This part of this problem will be graded after the due date.
asked 2021-03-05
A new vaccine was tested to see if it could prevent the ear infections that many infants suffer from. Babies about a year old were randomly divided into two groups. One group received vaccinations, and the other did not. The following year, only 328 of 2460 vaccinated children had ear infections, compared to 508 of 2453 unvaccinated children. Complete parts a) through c) below. a) Are the conditions for inference satisfied? A. Yes. The data were generated by a randomized experiment, less than 10% of the population was sampled, the groups were independent, and there were more than 10 successes and failures in each group. B. No. It was not a random sample. C. No. The groups were not independent. D. No. More than 10% of the population was sampled.
asked 2021-02-02
Potential buyers for a new car were randomly divided into two groups. One group was shown the "A" version of an ad for the car, while the other group was shown the "B" version of the ad. All were then tested on their recall of key points made in the ad. The researcher should run a hypothesis test based upon a comparison of means for ?
In another study, a healthcare insurance company took measures of subscribers’ cardiac (heart) health. The people were then provided an app for their phones which provided "nudges" and reminders about heart-healthy behaviors, such as eating more vegetables and less fried or fatty food, taking walks and breaks from sitting too long, and getting enough sleep. After 4 months of having the app, the cardiac health measures were taken again, with the objective of seeing if nudges from the app would result in decreased cardiac risk. The researcher should run a hypothesis test based on a comparison of means for?
asked 2021-03-02
In the 1970s a study was conducted in Philadelphia in which 500 cases were randomly assigned to treatments for the common cold: 250 subjects received the medication and 250 received a placebo. A total of 383 patients improved within 24 hours. Of those who received the medication 241 improved within 24 hours and of those who received the placebo 142 improved within 24 hours. A test of significance was conducted on the following hypotheses.
\(\displaystyle{H}_{{o}}\): The rates for the two treatments are equal.
\(\displaystyle{H}_{{a}}\): The treatment of medication has a higher improvement rate.
This test resulted in a p-value of 0.0761.
a.) Interpret what this p-value measures in the context of this study.
b.) Based on this p-value and study design, what conclusion should be drawn in the context of this study? Use a significance level of \(\displaystyle\alpha={0.05}\).
c.) Based on your conclusion in part (b), which type of error, Type I or Type II, could have been made? What is one potential consequence of this error?
asked 2021-03-07
The American Journal of Political Science (Apr. 2014) published a study on a woman's impact in mixed-gender deliberating groups. The researchers randomly assigned subjects to one of several 5-member decision-making groups. The groups' gender composition varied as follows: 0 females, 1 female, 2 females, 3 females, 4 females, or 5 females. Each group was the n randomly assigned to utilize one of two types of decision rules: unanimous or majority rule. Ten groups were created for each of the \(\displaystyle{6}\ \times\ {2}={12}\) combinations of gender composition and decision rule. One variable of interest, measured for each group, was the number of words spoken by women on a certain topic per 1,000 total words spoken during the deliberations. a) Why is this experiment considered a designed study? b) Identify the experimental unit and dependent variable in this study. c) Identify the factors and treatments for this study.
asked 2020-11-08
Testing for a Linear Correlation. In Exercises 13–28, construct a scatterplot, and find the value of the linear correlation coefficient r. Also find the P-value or the critical values of r from Table A-6. Use a significance level of \(\alpha = 0.05\). Determine whether there is sufficient evidence to support a claim of a linear correlation between the two variables. (Save your work because the same data sets will be used in Section 10-2 exercises.) Lemons and Car Crashes Listed below are annual data for various years. The data are weights (metric tons) of lemons imported from Mexico and U.S. car crash fatality rates per 100,000 population [based on data from “The Trouble with QSAR (or How I Learned to Stop Worrying and Embrace Fallacy),” by Stephen Johnson, Journal of Chemical Information and Modeling, Vol. 48, No. 1]. Is there sufficient evidence to conclude that there is a linear correlation between weights of lemon imports from Mexico and U.S. car fatality rates? Do the results suggest that imported lemons cause car fatalities? \(\begin{matrix} \text{Lemon Imports} & 230 & 265 & 358 & 480 & 530\\ \text{Crashe Fatality Rate} & 15.9 & 15.7 & 15.4 & 15.3 & 14.9\\ \end{matrix}\)
asked 2021-04-25
The unstable nucleus uranium-236 can be regarded as auniformly charged sphere of charge Q=+92e and radius \(\displaystyle{R}={7.4}\times{10}^{{-{15}}}\) m. In nuclear fission, this can divide into twosmaller nuclei, each of 1/2 the charge and 1/2 the voume of theoriginal uranium-236 nucleus. This is one of the reactionsthat occurred n the nuclear weapon that exploded over Hiroshima, Japan in August 1945.
A. Find the radii of the two "daughter" nuclei of charge+46e.
B. In a simple model for the fission process, immediatelyafter the uranium-236 nucleus has undergone fission the "daughter"nuclei are at rest and just touching. Calculate the kineticenergy that each of the "daughter" nuclei will have when they arevery far apart.
C. In this model the sum of the kinetic energies of the two"daughter" nuclei is the energy released by the fission of oneuranium-236 nucleus. Calculate the energy released by thefission of 10.0 kg of uranium-236. The atomic mass ofuranium-236 is 236 u, where 1 u = 1 atomic mass unit \(\displaystyle={1.66}\times{10}^{{-{27}}}\) kg. Express your answer both in joules and in kilotonsof TNT (1 kiloton of TNT releases 4.18 x 10^12 J when itexplodes).
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