A simple random sample of heights of 6400 Englishmen has a mean of 67.85 inches and SD 2.56 inches, while a simple random sample of heights of 1600 Australians has a mean of 68.55 inches and SD of 2.52 inches. Do the data indicate that Australians are, on the average, taller than Englishmen?
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A simple random sample of heights of 6400 Englishmen has a
67.85 inches and SD 2.56 inches, while a simple random sample of heights
of 1600 Australians has a mean of 68.55 inches and SD of 2.52 inches. Do
the data indicate that Australians are, on the average, taller than
Englishmen?
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- It currently takes users a mean of 2222 minutes to install the most popular computer program made by RodeTech, a software design company. After changes have been made to the program, the company executives want to know if the new mean is now different from 2222 minutes so that they can change their advertising accordingly. A simple random sample of 9191 new customers are asked to time how long it takes for them to install the software. The sample mean is 23.423.4 minutes with a standard deviation of 5.55.5 minutes. Perform a hypothesis test at the 0.050.05 level of significance to see if the mean installation time has changed. Step 1 of 3 : State the null and alternative hypotheses for the test. Fill in the blank below. H0Ha: μ=22: μ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯22H0: μ=22Ha: μ00000000_22 Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places. Step 3 of 3: Draw a conclusion and interpret the decision.2. The lengths (in feet) of a random sample of suspension bridges in the United States, Europe, and Asia are shown. At a 0.05, is there sufficient evidence to conclude that there is a difference in mean lengths? United States Europe Asia 4260 5238 6529 3500 4626 4543 2300 4347 3668 2000 3300 3379 1850 2874Consider the example from the previous problem, in part (b). The sizes (in grams) of Latavia's 14 oranges are 124, 126, 129, 130, 134, 136, 139, 142, 146, 149, 152, 154, 155, and 157 g. a. Find the sample mean of Latavia's data. b. Make a boxplot or a histogram of Latavia's data with a bin size of 10. c. If even a small sample is graphed and has no major skew or outliers, it is generally reasonable to assume that the population “could reasonably be normal," and therefore that the sampling distribution of the mean is approximately normal. This is the "normal population" condition. Does this sample appear to satisfy this condition? d. Latavia reads online that medium-sized navel oranges have a mean mass of 131 g with a standard deviation of 18 g. She is interested to test if the Fresh Foods oranges match those parameters. State a null and alternative hypothesis for this situation using symbols. Note that Latavia is making two assumptions in her test. e. Calculate a test statistic: what…
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- A boat capsized and sank in a lake. Based on an assumption of a mean weight of 133 lb, the boat was rated to carry 70 passengers (so the load limit was 9,310 lb). After the boat sank, the assumed mean weight for similar boats was changed from 133 ib to 170 1b. Complete parts a and b below. a. Assume that a similar boat is loaded with 70 passengers, and assume that the weights of people are normally distributed with a mean of 181.5 lb and a standard deviation of 35.7 Ib. Find the probability that the boat is overloaded because the 70 passengers have a mean weight greater than 133 lb. The probability is (Round to four decimal places as needed.) b. The boat was later rated to carry only 17 passengers, and the load limit was changed to 2,890 Ib Find the probability that the boat is overloaded because the mean welght of the passengers is greater than 170 (so that their total weight is greater than the maximum capacity of 2,890 Ib). The probability is (Round to four decimal places as…An adventure company runs two obstacle courses, Fundash and Coolsprint. The designer of the courses suspects that the mean completion time of Fundash is not equal to the mean completion time of Coolsprint. To test this, she selects 250 Fundash runners and 285 Coolsprint runners. (Consider these as random samples of the Fundash and Coolspring runners.) The 250 Fundash runners complete the course with a mean time of 76.7 minutes and a standard deviation of 5.7 minutes. The 285 Coolsprint runners complete the course with a mean time of 77.5 minutes and a standard deviation of 5.6 minutes. Assume that the population standard deviations of the completion times can be estimated to be the sample standard deviations, since the samples that are used to compute them are quite large. At the 0.05 level of significance, is there enough evidence to support the claim that the mean completion time, u, of Fundash is not equal to the mean completion time, µa, of Coolsprint? Perform a two-tailed test.…In a population of 150 pea plants, there are 89 tall-purple plants, 25 short-purple plants, 28 tall-white plants, and 8 short-white plants. In order to test if the two traits are experiencing independent assortment researchers would perform a chi squared test. What is your calculated Chi Squared statistic? What is the corresponding P value?
