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UF STA 6166 - Fall 2008 – Exam 1

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STA 6166 – Fall 2008 – Exam 1 PRINT Name __________True or False- In conducting a hypothesis test regarding a mean, the larger the P-value, the stronger the evidence against the null hypothesis. False- We generally use histograms to display results corresponding to quantitative (as opposed to qualitative) variables. True- If breaking strengths of wood boards are normally distributed with a mean of 150,and standard deviation 20; then 95% of such boards have breaking strengths between 130 and 170. False - A sample that is conducted by taking a random number between 1 and 100, then taking that item and every 100th item after that on a listing of a population is a systematic sample. True- The (large-sample) sampling distribution for the sample mean is normally distributed only if the underlying population of measurements is normal. False- All else being equal, as we increase the sample size, we increase the width of a Confidence Interval for -----FalseProblems- Among male birds of a species, 20% have a particular gene. Among females of the species, 10% have the gene. The males comprise 40% of all the birds of the species (thus, females comprise 60%).o What is the probability a randomly selected bird of this species has the gene? P(G) = P(G&M) + P(G&F) = P(M)P(G|M) + P(F)P(G|F) = 0.40(0.20) + 0.60(0.10) = 0.08+0.06 = 0.14o What is the probability the bird is male, given it has the gene? P(M|G) = P(M&G) / P(G) =0.08/0.14 = 0.57- You are going to a foreign nation to conduct your research. On a weather website you see that the average high temperature during the period you will be there has been historically 20 degrees Celsius. What is the average in degrees Fahrenheit? Hint: F=32+(9/5)CE(F) = E(32+(9/5)C) = 32+(9/5)E(C) = 32+(9/5)(20) = 32+36 = 68- In a population of 100 watt light bulbs manufactured by a company, 80% (0.80 as a proportion) have lifetimes exceeding 800 hours. An inspector samples 10 bulbs at random. What is the probability that all 10 bulbs’ lifetimes exceed 800 hours? P(Y=10) = [10!/(0!10!)](.80)10(.20)0 = .810 = .1074- A method of spraying for rust mites has been conducted on a random sample of 100 10-acre plots in citrus groves in Florida. The sample mean and standard deviation for the yield of fruit is 750 and 200, respectively. Construct a 95% confidence interval for -, the population mean yield per 10-acre plot. 2.397501002001.96750:for %95100200750 CInsy- Among students taking a standardized exam, scores are normally distributed with a mean of 550 and standard deviation 100. What proportion of the students score above 700?0668.5.1100550700)700(  ZPYP- A researcher wishes to estimate the population mean amount of bacteria per plate of culture when a fixed amount of antibiotic is applied. Previous work has shown that the standard deviation of the amount of bacteria present is approximately 18 cm2. Use this to determine the number of cultures to be run to estimate the true mean within 2 cm2 with 95% confidence. (That is, the half-with of a 95% confidence interval will be 2).31217.3112)18(96.1222/Eszn- An airplane manufacturer will only purchase aluminum plates of a given size from a supplier if the supplier can demonstrate that the true average pressure needed to crimp the plates exceed 2000 psi. They set up a test by randomly selecting n=64 plates from the supplier and measuring the pressure needed to crimp the plates. Let - represent the true mean crimping strength. Give the null and alternative hypotheses (note that the airplane manufacturer is putting the burden on the supplier to demonstrate that they meet the specification).o Null Hypothesis H0: ≤ Alternative HA: > 2000o They observe sample mean=2050 and standard deviation=160. Compute the test statistic. 50.26416020002050obszo If they conduct this test at the -=0.05 significance level (Probability of Type I Error), they will conclude that the true mean meets the specification if the test statistic falls in what range of values ≥ 1.645o Compute the P-value for this test P(Z ≥ 2.50) = .0062o Has the supplier demonstrated at this level of significance that they meet the specification? Yes or


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UF STA 6166 - Fall 2008 – Exam 1

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