The number of grams of carbohydrates contained in 1-ounce servings of randomly selected chocolate and nonchocolate candy is listed here. Is there sufficient evidence to conclude that the difference in the means is significant? Use a = 0.01. Chocolate: 29 25 17 36 41 25 32 29 38 34 24 27 29 Nonchocolate: 41 41 37 29 30 38 39 10 29 55 29
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- For a sample of 12 offenders convicted of weapons violations, the length of prison sentence in months was recorded as: 6 6 2 12 36 48 60 24 24 20 18 15 What percentage of the prison sentence would be expected to count less than 24 months?A sample of 21 cities were selected from around the continental United States. The prices of whole milk (in units of dollars per gallon) in these cities are seen below. 2.52 2.92 4.64 3.39 3.47 2.96 3.52 4.16 2.70 3.59 4.84 3.55 4.12 4.25 3.99 3.02 2.86 3.84 3.42 3.55 4.02 Calculate the mean milk cost in these cities. Round to the nearest cent. What is the median milk cost in these cities? Determine the first quartile Q1 for this set of data. Determine the third quartile Q3 for this set of data. Calculate the interquartile range (IQR) for these milk prices. Express your answer to the nearest cent. Do not include a dollar sign in your answer. Calculate the standard deviation for these milk prices.A researcher collected how many grams of fat people on a specific diet consumed at breakfast and at lunch. The data for a sample of 8 people is shown in the table below. xx 11.3 8.2 7.2 3.6 2.3 2.5 2.7 0.4 yy 13.7 10.7 10.2 7.2 6.3 6.5 6.1 4.1 xx = grams of fat in breakfastyy = grams of fat in lunchThis data can be modeled by the equation ˆy=0.84x+4.07.y^=0.84x+4.07.Given this, how many grams of fat in lunch would be expected for a person who consumed 1.2 grams of fat at breakfast? Round to 1 decimal place.Answer =
- Summary: This study aimed to determine whether exposure to Artificial Light At Night (ALAN) while sleeping is associated with the prevalence and risk of obesity, as previous research has shown a link between short sleep and obesity. The study involved analyzing data from 43,722 women aged 35 to 74 years in the US who had no history of cancer or cardiovascular disease and were not shift workers, daytime sleepers, or pregnant. The women reported their ALAN exposure at enrollment, and the data were analyzed from 2017 to 2018. ALAN exposure was categorized as 'No ALAN Exposure', and 'Any ALAN Exposure' (small nightlight, light outside the room, and light or television in the room). The presence of obesity was determined based on the measurement of general obesity, defined as having a body mass index (BMI) (calculated as weight in kilograms divided by height in meters squared) of 30.0 or higher. Research Question: Is artificial light at night while sleeping associated with obesity? Results:…In a study conducted in Italy, 10 patients with hypertriglyceridemia were placed on a low-fat, high-carbohydrate diet. Before the start of the diet, cholesterol and triglyceride measurements were recorded for each subject. Patient Cholesterol Level (mmol/l) Triglyceride level (mmol/l) 1 5.12 2.30 2 6.18 2.54 3 6.77 2.95 4 6.65 3.77 5 6.36 4.18 6 5.90 5.31 7 5.48 5.53 8 6.02 8.83 9 10.34 9.48 10 8.51 14.20 a. Construct a two-way scatter plot for these data. b. Use STATA to calculate Pearson’s Correlation Coefficient for these data. c. Test, α = 0.05, whether or not the population correlation, ρ, equals 0.A certain virus affects 0.7% of the population. A test used to detect the virus in a person is positive 87% of the time if the person has the virus (true positive) and 14% of the time if the person does not have the virus (false positive). Fill out the remainder of the following table and use it to answer the two questions below based on a total sample of 100,000 people. Virus No Virus TotalPositive Test Negative Test Total 100,000a) Find the probability that a person has the virus given that they have tested positive. Round your answer to the nearest hundredth of a percent and do not include a percent sign. % b) Find the probability that a person does not have the virus given that they test negative. Round your answer to the nearest hundredth of a percent and do not include a percent sign. %
- 2. The following data was collected from 1 bag of Hershey Kisses®. Each Kiss® was weighed in grams with the wrapper and recorded in the table below. Hershey claims that there are 368 grams of chocolate in one bag. Hershey Kiss Weights in Grams 4.76 4.72 4.74 4.55 4.91 4.74 4.78 4.71 4.80 4.78 4.78 4.75 4.79 4.82 4.91 4.83 4.68 4.74 4.70 4.80 4.70 4.76 4.70 4.83 4.93 4.74 4.84 4.82 4.76 4.77 4.72 4.78 4.83 4.75 4.74 4.68 4.84 4.71 4.71 4.76 4.66 4.78 4.73 4.74 4.92 4.77 4.80 4.79 4.86 4.64 4.78 4.70 4.75 4.78 4.76 4.83 4.66 4.77 4.83 4.78 4.69 4.81 4.68 4.78 4.88 4.72 4.85 4.85 4.81 4.74 4.80 4.82 4.84 4.70 4.85 4.70 4.81 4.72 4.79 4.73 4.61 What could be some reasons for variation in the weights of the Kisses®? Note: Take time answering this one. There are lots of things to consider here and I'll be looking for a well thought out answer with several given reasons contributing to the variation. Of course, the wrappers and tags could vary but what about…Several studies showed that after eating a low-fat cereal for two meals a day, subjects had lost some weight.A cereal companyfinanced this research. Identify what is wrong. Choose the correct answer below. A. Since the research is composed of voluntary response samples, there may be key data points missing. B. The data used in the studies is not reliable because it was not measured by the administrator. C. It is questionable that the sponsor is a cereal company because this sponsor can be greatly affected by the conclusion. D. It is not possible to take accurate measurements.Several methods of estimating the number of seeds in soil samples have been developed by ecologists. An article gave the accompanying data on the number of seeds detected by the direct method and by the stratified method for 27 soil specimens. Specimen 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Direct 21 33 0 60 20 61 40 8 13 93 1 68 77 21 Stratified 8 37 8 57 53 61 28 8 8 100 0 57 68 53 Specimen 15 Find the test statistic. (Round your answer to two decimal places.) t = 16 17 18 19 20 21 22 23 24 25 26 27 Direct Find the df. (Round your answer down to the nearest whole number.) df = 33 0 37 17 93 1 40 21 0 8 13 Use technology to find the P-value. (Round your answer to four decimal places.) P-value = 17 40 Stratified 28 0 37 13 93 13 48 21 0 13 40 Do the data provide sufficient evidence to conclude that the mean number of seeds detected differs for the two methods? Test the relevant hypotheses using a = 0.05. (Use direct stratified.) 13 77
- The Department of the Census provides the following data table. single married widowed divorced total Male .129 .298 .013 .040 .48 Female .305 .057 .054 .52 Total .233 .603 .070 .094 1.0 a) Fill in the missing word for each of the totals. (What is being totaled in each column/row?) b) Find P(Widowed and Male). = .013 c) Find P(Widowed)=.070 d) Find P(Married or Female). Write out the formula you use so I know that you know what you are doing. e) Find P(Widowed and Divorced). Explain your answer. f) Are Divorced and Male Mutually Exclusive events. Why or why not?18. The governor of New York has banned selling or purchasing "large" soft drinks, where a "large" is defined as being more than 16 oz. Sonic, famous for its Route 44 drinks is interested in estimating the proportion of New Yorkers that favor this ban. From a random survey of 800 New Yorkers, 375 answered no to the question, “Do you favor the governor's ban on 'large' soft drinks?" a. Find a 97.5% confidence interval for the proportion of New Yorkers who favor the ban on "large" soft drinks. b. Interpret the confidence interval. c. Can you conclude with 97.5% confidence that a majority of New Yorkers favor the ban on "large" soft drinks? Why or why not. d. Without doing any other calculations, can you be 99% confident that a majority of New Yorkers favor the ban on “large" soft drinks?The average spent nationally on Christmas gifts in 2020 was reported to be $650 per person. Below is a random sample of gift givers from Iowa and what they spent in 2020. Is there sufficient evidence to conclude that Iowans spend less than the national average? Estimate the average amount spent by Iowans on Christmas gifts in 2020 with a 90%-CI. 700 450 400 525 550 550 600 575 650 625 500