The average daily volume of a computer stock in 2011 was u = 35.1 million shares, according to a reliable source. A stock analyst believes that the stock volume in 2014 is different from the 2011 level. Based on a random sample of 30 trading days in 2014, he finds the sample mean to be 31.2 million shares, with a standard deviation of s = 14.2 million shares. Test the hypotheses by constructing a 95% confidence interval. Complete parts (a) through (c) below. (a) State the hypotheses for the test. 35.1 million shares 35.1 million shares Ho: (b) Construct a 95% confidence interval about the sample mean of stocks traded in 2014. The lower bound is million shares. The upper bound is O million shares. (Round to three decimal places as needed.) (c) Will the researcher reject the null hypothesis? O A. Reject the null hypothesis because u = 35.1 million shares falls in the confidence interval. O B. Do not reject the null hypothesis because u = 35.1 million shares falls in the confidence interval. OC. Do not reject the null hypothesis because u = 35.1 million shares does not fall in the confidence interval. O D. Reject the null hypothesis because u = 35.1 million shares does not fall in the confidence interval.

The average daily volume of a computer stock in 2011 was u = 35.1 million shares, according to a reliable source. A stock analyst believes that the stock volume in 2014 is different from the 2011 level. Based on a random sample of 30 trading days in 2014, he finds the sample mean to be 31.2 million shares, with a standard deviation of s = 14.2 million shares. Test the hypotheses by constructing a 95% confidence interval. Complete parts (a) through (c) below. (a) State the hypotheses for the test. 35.1 million shares 35.1 million shares Ho: (b) Construct a 95% confidence interval about the sample mean of stocks traded in 2014. The lower bound is million shares. The upper bound is O million shares. (Round to three decimal places as needed.) (c) Will the researcher reject the null hypothesis? O A. Reject the null hypothesis because u = 35.1 million shares falls in the confidence interval. O B. Do not reject the null hypothesis because u = 35.1 million shares falls in the confidence interval. OC. Do not reject the null hypothesis because u = 35.1 million shares does not fall in the confidence interval. O D. Reject the null hypothesis because u = 35.1 million shares does not fall in the confidence interval.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

Health insurers and the federal government are both putting pressure on hospitals to shorten the average length of stay (LOS) of their patients. The average LOS in Canada is 4.6 days. A random sample of 18 hospitals in one province had a mean LOS of 3.8 days and a standard deviation of 1.7 days. Assume that the sampled population is approximately normal. You are to construct a 95% confidence interval to see if the population mean of 4.6 days is true. Answer the followings: a. What is/are the critical value(s)? b. What is the margin of error? c. Identify the 95% confidence interval for the mean. d. Interpret the interval in terms of this application. Circle the letter of the correct answer below. A. The state can be 95% confident that its mean LOS is within this interval. B. The state can be 5% confident that its mean LOS is within this interval. C. The state can be 95% confident that its mean LOS is outside this interval. D. The state can be 95% confident that its mean LOS is 3.8 days.…
The number of needed repairs while under warranty for a Sure Cut lawnmower has a population distribution as shown in the probability histogram below. The population mean is also given. Population 0.6 - 0.5- 0.4- population mean: u=1.15 0.3- 0.2- Number of repairs Probability
If a sample of the lives of 18 car batteries has a standard deviation of 0.722 years, what is the LOWER bound for a 80% confidence interval for ? what is the UPPER bound for a 80% confidence interval for ?
A study by the International Coffee Organisation found that persons aged 30 – 59 drank an average of 0.57 cups of coffee per day. Assuming that the sample size is 200 with a sample standard deviation of 0.4 cups, construct a 95% confidence interval for the population mean cup of coffee. Interpret your answer.
Suppose that your sample size is 30 and that you wish to construct a 95% confidence interval. Your sample standard deviation is 40.9. Estimate your margin of error (MOE) using this information and the t-distribution table. Estimate your response to the first decimal point.
The average daily volume of a computer stock in 2011 was μ=35.1million shares, according to a reliable source. A stock analyst believes that the stock volume in 2018 is different from the 2011 level. Based on a random sample of 30 trading days in 2018, he finds the sample mean to be 29.1 million shares, with a standard deviation of s=13 million shares. Test the hypotheses by constructing a 95%confidence interval. Complete parts (a) through (c) below. ( B ) Construct a 95% confidence interval about the sample mean of stocks traded in 2018. With 95% confidence, the mean stock volume in 2018 is between
Eden obtains a mark of 67 on her final year exam. Assuming the reliability of the exam is 0.66 and the standard deviation is 10. What is the 95% confidence interval of her mark? What does the interval imply?
A random sample of high school seniors were asked whether they were applying for college. The resulting confidence interval for the proportion of students applying for college is (0.65,0.69). What is the margin of error?
A survey of 73 randomly selected homeowners finds that they spend a mean of $74 per month on home maintenance. Construct a 95% confidence interval for the mean amount of money spent per month on home maintenance by all homeowners. Assume that the population standard deviation is $16 per month. Round to the nearest cent.
Use the following information to answer the question. According to the website ww.costofwedding.com, the average cost of flowers for a wedding is $698. Recently, in a random sample of 40 weddings in the U.S. it was found that the average cost of the flowers was $734, with a standard deviation of $102. On the basis of this, a 95% confidence interval for the mean cost of flowers for a wedding is $701 to $767.Does the confidence interval provide evidence that the mean cost of flowers for a wedding has increased? a) Yes b) No
You measure 37 textbooks' weights, and find they have a mean weight of 69 ounces. Assume the population standard deviation is 12.9 ounces. Based on this, construct a 80% confidence interval for the true population mean textbook weight.Report your answers to three decimal places.
In a sample of 45 students, the mean scholarship amount received was $854, with a standard deviation of $17.25. Contruct a 90% confidence interval for the mean scholarship award for all students. Round your answer to the nearest cent.