A simple random sample of 100 postal employees is used to test if the average timepostal employees have worked for the postal service has changed from the value of 7.5years recorded 20 years ago. The sample mean was xdeviation of s = 2 years. Assume the distribution of the time the employees have workedfor the postal service is approximately Normal. The hypotheses being tested are Ho: u=7.5 versus Ha: µ±7.5. A one-sample t test will be used. What is a 95% confidenceinterval for u, the population mean time the postal service employees have spent withthe postal service?A) 7+2B) 7+1.984C) 7+0.4D) 7±0.27 years with a standard||

It is given that Sample size n = 100 Sample mean M = 7 Sample SD s = 2 The critical value of t at df…

Answered: A simple random sample of 100 postal…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Hawaiians are known to have an active lifestyle, enjoying the fresh air and natural beauty the islands have to offer. The life expectancy of the US population across all 50 states is µ=78.62 years with a standard deviation σ=16.51 years. A randomly selected sample of 100 Hawaiians show an average life expectancy x̅=82.52 years. Is Hawaiians’ average life expectancy significantly different from 78.62, assuming α=0.01?
In a sample of 12 randomly selected high school seniors, the mean score on a standardized test was 1182 and the standard deviation was 161.6.Further research suggests that the population mean score on this test for high school seniors is 1012. Does the t-value for the original sample fall between −t0.95 and t0.95? Assume that the population of test scores for high school seniors is normally distributed. What is The t-value? And does or does not it not fall between −t0.95 and t0.95 because t0.95=? (Round to two decimal places as needed.)
Attenuation is the reduction of the amplitude of a signal, electric current, or other oscillation. With maximum attenuation, a scanner will not be able to read a bar code at a fixed distance. After substantial amount of data collection, a group of instrumentation engineers decided to model a variation of maximum attenuation of scanners as a normal distribution with a mean of 10.1 decibels and standard deviation of 2.7, same unit. What proportion of the products have maximum attenuation greater than 15.1 dB? 0.0718 none of these 0.0322 0.0214
what is the value of (M-m)?
An obstetrician read that a newborn baby loses on average 7 ounces in the first two days of his or her life. He feels that in the hospital where he works, the average weight loss of a newborn baby is less than 7 ounces. A random sample of 33 newborn babies has a mean weight loss of 6.2 ounces. The population standard deviation is 1.5 ounces. Is there enough evidence at =α0.01 to support his claim? Assume that the variable is normally distributed. Use the P -value method with tables. hello there are five parts to the question it asks to state the hypothessis and identify the claim it asks to compute the test value find the p value and choose the null hypothesis
The average zinc concentration recovered from a sample of measurements taken in 40 different locations in a river is found to be 2.5 grams per milliliter. How large a sample is required if we want to be 95% confident that our estimate of ? is off by less than 0.05. Assume that the population standard deviation is 0.25 gram per milliliter.
An obstetrician read that a newborn baby loses on average 7 ounces in the first two days of his or her life. He feels that in the hospital where he works, the average weight loss of a newborn baby is less than 7 ounces. A random sample of 30 newborn babies has a mean weight loss of 6.4 ounces. The population standard deviation is 1.6 ounces. Is there enough evidence at =α0.01 to support his claim? Assume that the variable is normally distributed. Use the critical value method with tables. hello the question askas to find the critical value compute the test value and select the hypothesis
The level of calcium in the blood of healthy young adults follows a normal distribution with mean u = 10 milligrams per deciliter and standard deviation s = 0.4. A clinic measures the blood calcium of 100 healthy pregnant young women at their first visit for prenatal care. The sample mean of these 100 measurements is X = 9.8. Is this evidence that the mean calcium level in the population from which these women come is less than 10? To answer this question, we perform the following hypothesis test: H0: u = 10, Ha: u < 10. What is the test statistic? (use two decimal places in your answer) What is the p-value equal to? At the 10% significance level, do you accept or reject the null hypothesis? Answer ACCEPT or REJECT t the 5% significance level, do you accept or reject the null hypothesis? Answer ACCEPT or REJECT
75 specimens of the new chip were tested and 67 specimens of the old chip were tested. The average speed, in MHz, for the new chips was 494.9 and the standard deviation was 19.3. The average speed of the old chips was 485.8 and the standard deviation was 16.7. Suppose we wish to test Ho : Hx - Hy = 0 versus H1 : Hx - Hy # 0. Calculate the z-score. (answers are considered correct if they're within 5% of the correct value)
Suppose the police have set up automatic speed surveillance on a street in an area. During one week they get 987 measurements. The average in the data set is 50,7 kh/h, with a standard deviation of 3,2 km/h. Calculate the z-value for an observation with the value 53,8 km/h.
An experiment is being conducted to see if one type of animal feed is better than another. The experiment used 120 animals. The weight of the animals was a mean of 70.05 kg with the new food vs 66.61 with the current food being used. The standard deviation of the samples was 11.16lbs. Write your hypotheses in statistics form. Attached is a table of T values. Determine if the new food is contributing to larger weight. Assume an alpha 10%. What would you conclude about the new feed if the alpha were 5%

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