It is claimed that the average age of working students in a certainuniversity is 35 years. A researcher selected a random sample of25 working students. The computation of their ages resultedyears.to an average of 32 years with a standard deviation of 10Does this mean that the average age of the working studentsis different from 35 years? Use 0.05 level of significance andassume normality in the population.CSScanned with CamScanner

Name: It is claimed that the average age of working students in a certainun
Uploaded: 2024-03-20T07:07:12.000Z
Channel: Bartleby
Description: It is claimed that the average age of working students in a certainuniversity is 35 years. A researcher selected a random sample of25 working students. The computation of their ages resultedyears.to an average of 32 years with a standard deviation o

we have given that n=25, xbar=32 , mu=35 and sigma =10 , alpha =0.05

Math

Statistics

It is claimed that the average age of working students in a certain university is 35 years. A researcher selected a random sample of 25 working students. The computation of their ages resulted to an average of 32 years with a standard deviation of 10 Does this mean that the average age of the working students is different from 35 years? Use 0.05 level of significance and assume normality in the population. years.

It is claimed that the average age of working students in a certain university is 35 years. A researcher selected a random sample of 25 working students. The computation of their ages resulted to an average of 32 years with a standard deviation of 10 Does this mean that the average age of the working students is different from 35 years? Use 0.05 level of significance and assume normality in the population. years.

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

Suppose the scores of students on an exam are normally distributed with a mean of 253 and a standard deviation of 80. According to the empirical rule, what percentage of students scored between 13 and 493 on the exam?
Suppose that IQ scores have a bell-shaped distribution with a mean of 103 and a standard deviation of 14. Using the empirical rule, what percentage of IQ scores are at least 145?
A researcher wanted to know whether losing one night sleep can affect problem solving. A sample of n = 25 college students was given a problem-solving task at noon on one day and again at noon on the following day. The students were not permitted any sleep between the two tests. For each student, the difference between the first and the second score was recorded: MD = 5 and standard deviation of the difference score was s = 10. Do the data indicate a significant change in problem solving ability? Use a two-tailed test with α = .01. a. The alternative hypothesis in words is b. The null hypothesis in symbols is c. The critical t-values are d. The t-statistic is e. Your decision is
in a test of a lifelong brand of tires , a survey revealed that the mean life was 35,000 mikes and the standard deviation was 7083 mikes. if the shape of the population distribution is known to be normal, approximately what percentage of the tires lasts less than 20834 miles ?
A psychologist obtains a random sample of 30 mothers in the first trimester of their pregnancy. The mothers are asked to play Mozart in the house at least 45 minutes each day until they give birth. We know that IQs are normally distributed with a mean of 100. A few years after their births the children were given an IQ test and the IQs of the 30 children in the study resulted in a sample mean of 104.2. Assume the population standard deviation for IQ of children is 15. Is there evidence that the children who listen to Mozart for at least 45 minutes each day during pregnancy have higher IQs at the 0.05 level of significance? Step 1: Define the parameter & setup the testStep 2: State the Level of SignificanceStep 3: Find the value of the Test StatisticsStep 4: Find P-Value OR Find Critical ValueStep 5: State Conclusion and why
Derby Leicester is a city planner preparing for a meeting with the mayor. He would like to show that the population mean age of the houses on Lincoln Street is less than the population mean age of the houses on Maple Street so that more resources are allotted to repair Maple Street. Derby uses information from a previous study to assume that the population standard deviation for the ages of the houses on Lincoln Street is 7.72 years and 8.39 years for the houses on Maple Street. Due to limited time, Derby randomly selects the houses on Lincoln Street and the houses on Maple Street from the city’s property records and then records the age of each house in years. The results of the samples are shown in the table below. Assuming the conditions needed for the hypothesis test have been met, what is the z test statistic for this hypothesis test, rounded to two decimal places? Lincoln Street Maple Street x¯¯¯1=50.91 years x¯¯¯2=59.27 years n1=41 n2=37
A survey among freshmen at a certain university revealed that the number of hours spent studying the week before final exams was normally distributed with mean 25 and standard deviation 9. Between what two values are the middle 80% of the hours spent studying?
In Professor Krugman’s economics course, the correlation between the students’ total scores prior to the final examination and their final‑examination scores is ?=0.5.r=0.5. The pre‑exam totals for all students in the course have mean 280280 and standard deviation 40.40. The final‑exam scores have mean 7575 and standard deviation 8.8. Professor Krugman has lost Julie’s ﬁnal exam but knows that her total before the exam was 300.300. He decides to predict her ﬁnal‑exam score from her pre‑exam total. (a) Which of the choices is the least‑squares regression line of ﬁnal‑exam scores on pre‑exam total scores in this course?
Suppose that the duration of a particular type of criminal trial is known to be normally distributed with a mean of 24 days and a standard deviation of 6 days. 80% of all trials of this type are completed within how many days?