Explain what would happen to the length of the interval if the confidence level were decreased to 95%.
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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: 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…A study was conducted to investigate the effect of a new drug on blood pressure. A sample of 25 patients was randomly selected and their blood pressure was measured both before and after taking the drug. The sample mean blood pressure before taking the drug was 120 mmHg and the sample mean blood pressure after taking the drug was 118 mmHg. The sample standard deviation for the blood pressure before taking the drug was 4 mmHg and the sample standard deviation for the blood pressure after taking the drug was 5 mmHg. Test the hypothesis that the average blood pressure after taking the drug is lower than the average blood pressure before taking the drug using a 0.01 level of significance.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?
- A tea shop has relocated to a new area and wants to make sure that all tea cups are consistent in tea whitener quantity to maintain the quality. Tea shop owner believes that each tea cup has an average of 4 oz. of whitener. If this is not the case, they must increase the amount to make a quality ‘milk leaf tea’ type of tea. For this purpose, he took a random sample of 25 tea cups, which showed a mean of 4.6 oz. of tea whitener and a standard deviation of 0.8 oz. Can the tea shop owner have 90 % confidence that the mean value of the tea whitener is not above 4 oz? Assume that distribution of alternate hypothesis is centralized at mean equal to 4.6 oz. Now using same standard error of mean as in part (a), graph the null and alternate hypotheses distributions by shading Type 1 and Type 2 error regions on them, taking =0.01. NOTE: DO NOT SOLVE ON EXCEL CLASSIFY EACH STEPFran 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.A sample of 150 matched pairs of banks was formed in a study comparing banks in Turkey and Germany. Each pair contained one bank from Turkey and one from Germany. The pairings were made in such a way that the two members were as similar as possible in regard to such factors as size and age. The ratio of total loans outstanding to total assets was calculated for each of the banks. For this ratio, the sample mean difference (Turkey - Germany) was 0.0412, and the sample standard deviation of the differences was 0.3039. Test, against a two-sided alternative, the null hypothesis that the two population means are equal. (Assume a = 0.01.) Click the icon to view the critical values of the Student's t distribution. Which of the following cor greater than tn-1, a/2 O B. Ho: Hx - Hy = 0 0< 1 - *rt :H O A. Ho: Hx -Hy =0 greater than - tn - 1, a 0# 서-저 : H O D. Ho: Hx-Hy = 0 OC. Ho: Hx-Hy #0 H: Hx-Hy =0 greater than t,-1. g 05 1 - Xt :'H less than -tn-1, a O F. Ho: Hx - Hy = 0 O E. Ho: Hx -Hy =0…
- 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)…
