Explain what would happen to the length of the interval if the confidence level were decreased to 95%.
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A: The provided information are:
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A: The hypotheses can be constructed as: H0: µ = 6.1 H1: µ ≠ 6.1 Sample size (n) = 40 Assume level of…
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A: From the given informationThe population mean is, µµ=36 minutesThe sample size is, n=28The sample…
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A:
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A: Given that, He knows that three years ago, the mean number of hours per week students spent of…
Q: It currently takes users a mean of 44 minutes to insta made to the program, the company executives…
A: Sample size=51 Sample mean=47.8 Population mean=44 Standard deviation =10.2 Level of…
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A: Sample of size(n)=21Sample mean()=96.3Sample standard deviation(s)=15.2Population mean()=98.3and…
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A: The objective of this question is to perform a hypothesis test to determine if the mean installation…
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A: The hypothesized average time (in minutes) to install the program, μ0=37.The sampled number of…
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A: Given μ=12.3, n=21 , x̄=12.9, s=1.3 We want to test the hypothesis.
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A: n1=n2=28 X1-bar=15.2 X2-bar=12.3 S1=2.5 S2=1.9
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In a recent year, a sample of the South American Redeye Piranha was collected by fishermen on a particular river. Wildlife biologists regard this sample as a random sample of all Redeye Piranha in that river. The
Explain what would happen to the length of the interval if the confidence level were decreased to 95%.
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- Without external cues such as the sun, people attempting to walk in a straight line tend to walk in circles. It has been suggested that this tendency is due to internal asymmetries between individuals, or that individuals' legs differ in length or strength. Souman et al. (2009) tested for differences in individuals' tendencies to change direction by blindfolding 15 participants and asking them to walk in a straight line in an empty field. The numbers in this dataset represent median change in direction (or turning angle) of each of the 15 participants, measured in degrees per second. Negative angles = left turns; positive angles = right turns.Begin by producing a frequency distribution graph of the dataset for yourself. Do the data appear to be normally distributed? Answer "yes" or "no" Which direction are people more likely to turn in based on your graph? Answer "right" or "left" Is this a 1- or a 2- tailed t-test? (enter either "1" or "2" for your answer) Conduct a…A large number of bags of cement have been manufactured. You sample 10 of the bags and find that they have a mean weight of 91.5 pounds with a standard deviation of 8.8 pounds. Find the 90 % confidence limit for the mean of the weight of all of the cement bags.An important measure in the study of contagious infectious diseases is the number of cases directly generated by one previous case. Jessica is an epidemiologist studying the spread of an infectious disease in her country. She claimed that the mean number of cases directly generated by one previous case is now greater than 1.2. A study of 12 randomly selected cases of the disease is conducted and finds the sample mean number of cases directly generated by one previous case to be 1.5 with a sample standard deviation of 0.7. Assume that the population of the number of cases directly generated by one previous case is approximately normally distributed. Complete the parts below to perform a hypothesis test to see if there is enough evidence, at the 0.10 level of significance, to support the claim that u, the mean number of cases directly generated by one previous case, is greater than 1.2. (a) State the null hypothesis H, and the alternative hypothesis H, that you would use for the test.…
- A certain type of battery supercharger was found to take a mean of 15.4 minutes to charge a standard electric vehicle (EV) battery. A firmware update for this type of supercharger was recently released, but the update has received negative reviews from users. A researcher suspects the mean time the updated superchargers take to charge a standard EV battery is greater than 15.4 minutes. To test his claim, he chooses 15 of the updated superchargers at random and measures the time it takes each one to charge a standard EV battery. He finds that the superchargers take a sample mean of 16.1 minutes to charge a standard EV battery, with a sample standard deviation of 1.8 minutes. Assume that the population of amounts of time to charge a standard EV battery using the updated superchargers is approximately normally distributed. Complete the parts below to perform a hypothesis test to see if there is enough evidence, at the 0.05 level of significance, to support the claim that u, the mean time…IQ tests are approximately normally distributed with a mean of 100 for adults. A tech company CEO wants to conduct a hypothesis test to see if the average IQ of employees at high-tech firms in California is higher than average. What is Type I error in the context of this problem?Fran is training for her first marathon, and she wants to know if there is a significant difference between the mean number of miles run each week by group runners and individual runners who are training for marathons. She interviews 42 randomly selected people who train in groups and finds that they run a mean of 47.1 miles per week. Assume that the population standard deviation for group runners is known to be 4.4 miles per week. She also interviews a random sample of 47 people who train on their own and finds that they run a mean of 48.5 miles per week. Assume that the population standard deviation for people who run by themselves is 1.8 miles per week. Test the claim at the 0.01 level of significance. Let group runners training for marathons be Population 1 and let individual runners training for marathons be Population 2. Step 2 of 3 : Compute the value of the test statistic. Round your answer to two decimal places.
- The University of Montana admission standards require students to have an ACT score of at least 22.We know that Montana ACT scores are normally distributed with a mean of 20.1 and standarddeviation of 4. If we ask a random sample of 100 students who took the ACT, how many would be expected toqualify for admission to UM?According to the U.S. Department of Transportation’s Air Travel Consumer Report, the nation’s 12 largest airlines recorded an on-time arrival percentage of 77.4% in 2001. Of interest is to estimate the mean delay time for the 22.6% of all flights that did not arrive on time during 2013. A simple random sample of 28 late arriving flights was selected, and the mean delay time of this sample of 28 flights was 14.2 minutes, with a sample standard deviation of s= 6.4 minutes. Use this information to calculate and interpret a 98% confidence interval for the mean delay time for all flights that did not arrive on time during 2013.A cell phone manufacturer tests the battery lifetimes of its cell phones by recording the time it takes for the battery charges to run out while testers are streaming videos on the phones continuously. The manufacturer claims that the population mean of the battery lifetimes of all phones of their latest model is 5.26 hours. As a researcher for a consumer information service, you want to test that claim. To do so, you select a random sample of 45 cell phones of the Español manufacturer's latest model and record their battery lifetimes. Assume it is known that the population standard deviation of the battery lifetimes for that cell phone model is 2.42 hours. Based on your sample, follow the steps below to construct a 99% confidence interval for the population mean of the battery lifetimes for all phones of the manufacturer's latest model. Then state whether the confidence interval you construct contradicts the manufacturer's claim. (If necessary, consult a list of formulas.) 00 (a)…
- If one sample with a mean of M = 10 is combined with a second sample with a mean of M = 20, then the mean for the combined sample will be between 10 and 20. True FalseA team of researchers found that age distributions in a population of 100,000 people was not Normal. However, after they took a large number of simple random samples, each with (N=100), they observed that the sampling distribution of mean tended toward Normality. Really? Why is it possible?According to the National Health and Nutrition Examination Survey (NHANES) sponsored by the U.S. government, a random sample of 712 males between 20 and 29 years of age and a random sample of 1,001 males over the age of 75 were chosen and the weight of each of the males were recorded (in kg). Do the data provide evidence that the younger male population weighs more (on average) than the older male population? Use “Y” for ages 20-29 and “S” for ages 75+. It was found that x̅Y=83.4, sY=18.7, x̅S=78.5, and sS=19.0. a)Suppose the test statistic is t = 2.398. What is the associated p-value? Group of answer choices 0.001 < p-value < 0.002 0.005 < p-value < 0.01 0.01 < p-value < 0.02 0.0005 < p-value < 0.001 b) Suppose the p-value is 0.02 < p-value < 0.04. At α = 0.10 what is the appropriate conclusion to make? Group of answer choices Fail to reject H0 and conclude that the mean weight of all males ages 20-29 is greater than the mean weight of all…
