Example 11.3 sample mean was found to be Rs 900 and the standard deviation Rs 200. Construct an interval estimate of the nonulation mean with the confidence level of 95.44 per cent. A random sample of 400 firms was taken to find out the average sale per customer. The
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- The mean score of a class of 30 students in their math test is 6.267. The standard deviation is 1.9 and at 95% confidence, the class interval was found to be CI = (5.587, 6.947). If the passing score is 7.5, did the class passed the test?1. For a group of 10 students subjected to a stress situation the mean number of heartbeats/minutes was 126 and the standard deviation was 4. Find the 95% confidence interval for the true mean.Find the Margin of Error, if we are asked for 90% confidence interval where the mean is 52.3, the standard deviation is 2.15 and the sample size is 136.
- If the sample mean is 37.7 and population standard deviation is = 9.2 and the sample size is 100 then find the 90% confidence interval.Suppose that a recent article stated the mean time spent in jail by a first-time convicted burglar is 2.7 years. A study was then done to see if the mean time has increased in the new century. A random sample of 27 first-time convicted burglars in a recent year was picked. The mean length of time in jail from the survey was 3.01 with a standard deviation of 1.78 years. Assume that the distribution of jail times is approximately normal. Test the claim at the level of significance 0.1. Select one: a) t_stat = 0.9 and t_cr = 1.31 : The mean time increased significantly in the new century b) t_stat = 0.9 and t_cr = 1.31 : The mean time didn’t increase in the new century c) t_stat = 0.99 and t_cr = 1.71 : The mean time didn’t increase in the new century d) t_stat = 0.99 and t_cr = 1.71 : The mean time increased significantly in the new centurySuppose a researcher wants to test if the average diastolic blood pressure in Irish-Americans is different than the average diastolic blood pressure in the general population. Note that the average diastolic blood pressure in the general population is 70 mmHG. The researcher recruits 30 Irish-Americans into this particular study and then measures each of their diastolic blood pressures (mmHG). The population of this sample has a standard deviation equal to 3.0. What is the two-sided alternative hypothesis? (in words) Group of answer choices The mean is different than 70 The mean is different than 73 The mean is greater than 70 The mean is greater than 73
- You would like to estimate the variance of sick days at a Fortune 500 company. You randomly select 15 employees and record how many sick days they have used in the past year and you get a sample standard deviation of 1.2. Provide a 95% confidence interval estimate of the population variance.On a measure of running speed on a 100 meter dash, students received timed scores ranging from 12 seconds to 24 seconds. The distribution of running times is normative with a mean of 16 seconds. Transferring this ratio data to standard scores for comparison with the results of classes from other departments, the mean score is assigned a standard score of 100 with a standard deviation of 10. From this example, a standard score of 99 would mean the actual value in seconds for an individual would be about a. 16 b. 12 c. 19 d. 24Cotinine levels were measured in a group of non-smokers exposed to tobacco smoke (n =40), the mean of the sample was 60.58 ng/Ml and the standard deviation was 138.08 ng/Ml. Also, cotinine levels were measured in a group of non-smokers not exposed to tobacco smoke(n =40), the mean of the sample was 16.35 ng/mL and the standard deviation was 62.53 ng/mL. Cotinine is a metabolite of nicotine, meaning that when nicotine is absorbed by the body, cotinine is produced. Use a 0.05 significance level to test the claim that non-smokers exposed to tobacco smoke have a higher mean nicotine level than nonsmokers not exposed to tobacco smoke. a)Define the parameter and state null and alternative hypothesis b)Test Statistics c) Critical Value Approach d) Conclusion
- 4. The cost of a daily newspaper varies from city to city. However, the variation among prices remains steady with a standard deviation of 18¢. A study was done to test the claim that the mean cost of a daily newspaper is $1.00. Twelve costs yield a mean cost of 93¢ with a standard deviation of 17¢. Do the data support the claim at the 1% level?The ages of a group of 138 randomly selected adult females have a standard deviation of 17.5 years. Assume that the ages of female statistics students have less variation than ages of females in the general population, so leto= 17.5 years for the sample size calculation. How many female statistics student ages must be obtained in order to estimate the mean age of all female statistics students? Assume that we want 95% confidence that the sample mean is within one-half year of the population mean. Does it seem reasonable to assume that the ages of female statistics students have less variation than ages of females in the general population? The required sample size is (Round up to the nearest whole number as needed.)Suppose that you are designing an instrument panel for a large industrial machine. The machine requires the person using it to reach 2 feet from a particular position. The reach from this position for adult women is known to have a mean of 2.7 feet with a standard deviation of 0.5. The reach for adult men is known to have a mean of 3.1 feet with a standard deviation of 0.7. Both women's and men's reach from this position is normally distributed. If this design is implemented, (a) what percentage of women will not be able to work on this instrument panel? (b) What percentage of men will not be able to work on this instrument panel? (c) Explain your answers to a person who has never had a course in statistics. Click here to view page 1 of the table. Click here to view page 2 of the table. Click here to view page 3 of the table. Click here to view page 4 of the table.
