How many different simple random samples of size 5 can be obtained from a population whose size is 51? The number of simple random samples which can be obtained is (Type a whole number.)
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- One sample has n=7 and SS=35 and a second sample has n =17 and SS =45. What is the pooled variance for these two samples?an automobile engineer claims that 1in 10 automobile accidents are due to driver fatigue . if a random sample of 5 accidents observed , then the probabilty that 3 accidentsare due to drive fatigue isThe FAA is interested in knowing if there is a difference in the average number of on-time arrivals for four of the major airlines. The FAA believes that the number of on-time arrivals varies by airport, therefore, a randomized block design is used for the study. To control for this variation, they randomly select 100100 flights for each of the major airlines at each of four randomly selected airports and record the number of on-time flights. Can the FAA conclude that there is a significant difference among the average number of on-time arrivals for the four major airlines? The results of the study are as follows. Average Number of On-Time Arrivals Airport Airline A Airline B Airline C Airline D Airport A 9090 8282 8080 8383 Airport B 8282 7070 7474 7979 Airport C 8787 8484 7676 8888 Airport D 9090 7979 7979 9090 Copy DataANOVA Source of Variation SSSS dfdf MSMS Rows 181.6875181.6875 33 60.562560.5625 Columns 278.6875278.6875 33 92.895892.8958 Error…
- Generate 10000 instances of a poisson (10) random variable and then compute the sample mean, sample variance, and sample standard deviation.2. In a population it is estimated that 20% have a desired trait of interest for the researcher. The researcher wants to know how many people on average he has to draw from a population to get 2 people with the trait. Use rows 20-24 of the Random Number Table to carry out the simulation. Explain clearly how you set up the problem and report your findings. Answer:If samples of a fixed size N are drawn randomly and with replacement from the same population of scores, should we expect the sample means to differ across samples? why or why not?
- A random selection of volunteers at a research institute has been exposed to a weak flu virus. After the volunteers began to have flu symptoms, 20 of them were given multivitamin tablets daily that contained 1 gram of vitamin C and 3 grams of various other vitamins and minerals. The remaining 20 volunteers were given tablets containing 4 grams of vitamin C only. For each individual, the length of time taken to recover from the flu was recorded. At the end of the experiment the following data were obtained. Treated with multivitamin Treated with vitamin C Days to recover from flu 3.2, 7.3, 8.4, 6.1, 2.5, 6.7, 6.3, 3.7, 7.0, 7.7, 9.1, 9.9, 9.0, 2.6, 2.0, 5.1, 5.7, 2.5, 2.9, 4.5 2.7, 5.3, 3.7, 5.5, 4.5, 3.8, 7.6, 5.3, 3.6, 5.2, 2.3, 4.7, 3.6, 5.2, 2.3, 7.1, 3.3, 6.1, 7.9, 4.2 Send data to calculator. Send data to Excel Suppose that it is known that the population standard deviation of recovery time from the flu is 1.8 days when treated with multivitamins and that the population standard…A lawn mower company will produce 1,500 lawn mowers by 2020. In an effort to determine how much maintenance-free consumers will buy when they buy lawn mowers, the company decided to conduct a study over the past 1 year of these mowers. First, a random sample was taken by contacting 200 grass mowers. The company owner provides an 800 number and is asked to contact the company when the first repair is needed for the lawn mower. Owners who no longer use lawn mowers to cut their grass are either disqualified or not sampled. After 1 year, it turned out that 187 owners had reported for the first repair. However, the other 74 disqualified themselves. The average number per year until the first improvement was 5.3 times for samples deemed feasible. It is believed that the standard deviation is 1.28 per year. If the company wants to advertise the lawnmower and with a confidence level of 95%, what is the estimated annual mean value of repair-free lawn mower for this lawn mower? What is the…One cable company claims that it has excellent customer service. In fact, the company advertises that a technician will arrive within 35 minutes after a service call is placed. One frustrated customer believes this is not accurate, claiming that it takes over 35 minutes for the cable technician to arrive. The customer asks a simple random sample of 4 other cable customers how long it has taken for the cable technician to arrive when they have called for one. The sample mean for this group is 39.l minutes with a standard deviation of 2.6 minutes. Assume that the population distribution is approximately normal. Test the customer's claim at the 0.10 level of significance. Step 3 of 3: Draw a conclusion and interpret the decision.
- A random selection of volunteers at a research institute has been exposed to a weak flu virus. After the volunteers began to have flu symptoms, 20 of them were given multivitamin tablets daily that contained 1 gram of vitamin C and 3 grams of various other vitamins and minerals. The remaining 20 volunteers were given tablets containing 4 grams of vitamin C only. For each individual, the length of time taken to recover from the flu was recorded. At the end of the experiment the following data were obtained. Days to recover from flu Treated with 8.5, 5.6, 8.8. 6.1. 6.1. 3.3. 8.9, 6.0, 3.3, 9.7, 6.0, 9.3, 2.3, 4.6. 5.5.6.7. 6.6. 4.9. 10.0.6.6 multivitamin 6.5, 5.0, 3.2, 4.7, 3.5, 2.7, 3.8, 4.4, 4.6, 3.6, 4.0, 2.7, 3.2, 4.8, 6.3, 7.5, 6.9, 4.7, 4.8, 1.3 Treated with vitamin C Send data to calculator Send data to Excel Suppose that it is known that the population standard deviation of recovery time from the flu is 1.8 days when treated with multivitamins and that the population standard…Assume a population with 5 elements: X1-0, X2-1, X3-2. X4-3,X5-4. Calculate mean and variance of X for random samples of size 3 taken with replacement Provide a numerical answer only.A biologist argues that a certain species of fish found in Lake Tanzania in Africa should be classified in a different genus than the currently accepted genus for the species. A simple random sample of standard lengths of fully grown fish from the species in Lake Tanzania is given below: {116.2, 100.3, 118, 152.8, 141.3, 132.8, 95.6, 111.5, 95.2, 95.7} The population of alls such lengths is assumed to be close to normally distributed. Calculate the following interval estimates using this sample (note that the only one of the methods below can be used for a single sample in practice). Round answers to one decimal place accuracy. A 99% confidence interval for the mean of all fully grown fish of this type: type your answer... A 99% prediction interval for the length of one fully grown fish of this type is: type your answer... type your answer... type your answer...