Choose the correct answer: We would like to test the hypothesis that p= 0.5 versus the alternative that p < 0.5 using data from a random sample. We calculate the standard score to be -1.8. The P value would then be A. 0.6915 B. 0.0359 C. 0.9641 D. 0.5000

You are conducting a study to see if the proportion of voters who prefer Candidate A is significantly more than 0.46. Your sample data produce the test statistic z=1.883. Find the p-value accurate to 4 decimal places.

In an experiment, two different species of flowers were crossbred. The resulting 217 flowers from this crossbreeding experiment were categorized by the color of the flower and the stigma. The table provides the study results. 1: Magenta flower with Green stigma 2: Magenta flower with Red stigma 3: Red flower 4: Red flower Flower Type with Green stigma with Red stigma Observed # Flowers 115 49 32 21 According to genetic theory, the four resulting flower types should follow the ratio of 9:3:3:1, that is, should follow the following population proportions of 0.5625, 0.1875, 0.1875, and 0.0625. The biologist is interested in assessing whether the distribution for flower type for the experiment matches these genetic theory rates. The hypotheses to be tested are Họ: P1 = 0.5625, p2 = 0.1875, P3 = 0.1875, p4 = 0.0625, where p; represents the proportion of flowers in the population in group i HA: at least one p; differs from the stated proportions

A random sample of 10,000 bottles of cola was taken to see whether the mean weight was 16 fluid ounces, as marked on the container. The null hypothesis is that the population mean is 16 ounces. Use the read-out below to test the hypothesis that the colas do not have a population mean of 16 fluid ounces. You do not have to do any calculations; just interpret the given data clearly and thoroughly in the context of the problem. | One-Sample T: ounces Test of u = 16 vs = 16 Variable Mean StDev SE Mean 95 CI T P ounces 10000 15.9988 0.0985 0.0010 (15.9969, 16.0007) -1.21 0.225

Q3.3 Normal, but why? A meal package for use at the space station is to be prepared by combining three separate components: protein, fruit/vegetable, and grains. Each of the components is to be randomly selected. Summary values for the caloric content for each of the components are given in the table. Component Protein (P) Fruit/vegetable (F) Grain (G) Mean (calories) 120 55 84 Stdv (calories) 6.4 Save Answer 1.0 3.6 Assume that the distributions of caloric content for each of the components are approximately normally distributed. Since the meal package consists of protein, fruit/vegetable, and grains, the total caloric content of a meal package is the random variable T, where T = P + F + G. Briefly explain why it is appropriate to say that T is approximately normally distributed. Q3.4 Probability Less than 275 Calories On your own piece of paper find the approximate probability that a randomly selected meal package will have less than 269 calories? (Draw an appropriate diagram, find a…

A company that has a 10% market share (they sell their product to 10% of all potential buyers) launches a marketing campaign. At the end of the campaign period, the company conducts a survey in order to assess whether its market share has increased. The result of the survey issued to 500 customers was 12.5%.h0= p=0.10, h1= p>0.10test statistic= 1.86P value= 0.0314 a. State your conclusion using ΅ = 0.05.b. What would be the potential Type I and Type II errors in this problem? Explain these possible errors in context.

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Solve this problem to find the sample standard deviation

K A magazine provided results from a poll of 500 adults who were asked to identify their favorite pie. Among the 500 respondents, 14% chose chocolate pie, and the margin of error was given as ±4 percentage points. Describe what is meant by the statement that "the margin of error was given as ±4 percentage points." 身 Choose the correct answer below. OA. The statement indicates that the study is 100% -4% = 96% confident that the true population percentage of people that prefer chocolate pie is 14%. OB. The statement indicates that the true population percentage of people that prefer chocolate pie is in the interval 14% ±4%. OC. The statement indicates that the interval 14% ± 4% is likely to contain the true population percentage of people that prefer chocolate OD. The statement indicates that the study is only 4% confident that the true population percentage of people that prefer chocolate pie is exactly 14%.

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