Question

Time perception. Randomly selected college students participated in an experiment that tested their ability to determine...

Time perception. Randomly selected college students participated in an experiment that tested their ability to determine the course of 1 minute (or 60 seconds). Forty students produced a sample mean of 58.3 seconds. Assuming that the standard deviation is 9.5 seconds, use a significance level of 0.05 to test the assertion that the population mean is less than 60 seconds.

Homework Answers

Answer #1

solution

this is the left tailed test .  

The null and alternative hypothesis is ,

H0 :   = 60

Ha : <60

Test statistic = z

= ( - ) / / n

= (58.3-60) / 9.5 / 40

= -1.13

P(z < -1.13 ) = 0.1292

P-value = 0.1292

= 0.05  

P-value >

do not Reject the null hypothesis .

There is no sufficient evidence to suggest that   

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
Randomly selected statistics students participated in an experiment to test their ability to determine when 1...
Randomly selected statistics students participated in an experiment to test their ability to determine when 1 minute (or 60 seconds) has passed. Forty students yielded a sample mean of 58.3 sec with a standard deviation is 9.5 sec, construct a 95% confidence interval estimate of the population mean of all statistics students. A. 50.4 sec < mu  < 77.8 B. 54.5 sec < mu  < 63.2 C. 56.3 sec < mu  < 62.5 D. 55.4 sec < mu  < 61.2 Based on the Confidence...
Randomly selected students participated in an experiment to test their ability to determine when one minute​...
Randomly selected students participated in an experiment to test their ability to determine when one minute​ (or sixty​ seconds) has passed. Forty students yielded a sample mean of 59.7 seconds. Assuming that sigma =10.1 ​seconds, construct and interpret a 90​% confidence interval estimate of the population mean of all students. What is the 90​% confidence interval for the population mean μ​?
Randomly selected students participated in an experiment to test their ability to determine when one minute​...
Randomly selected students participated in an experiment to test their ability to determine when one minute​ (or sixty​ seconds) has passed. Forty students yielded a sample mean of 62.4 seconds. Assuming that σ=8.5 seconds, construct and interpret a 99​% confidence interval estimate of the population mean of all students. What is the 99​% confidence interval for the population mean μ​? ____<μ<_____
Randomly selected statistics students of the author participated in an experiment to test their ability to...
Randomly selected statistics students of the author participated in an experiment to test their ability to determine when 60 seconds has passed. Forty students yielded a sample mean of 57.3 sec. Assume that the population standard deviation is σ = 8.5 sec. a) Find the critical value zα /2 for a 82% confidence interval b) Construct a 82% confidence interval estimate of the population mean of all statistics students. Use the z-table method. c) Use TI84/83 calculator method. d) Write...
Actual times​ (in seconds) recorded when statistics students participated in an experiment to test their ability...
Actual times​ (in seconds) recorded when statistics students participated in an experiment to test their ability to determine when one minute​ (60 seconds) had passed are shown below. Find the​ mean, median, and mode of the listed numbers. 55  50  74  61  67  60  47  47
one sample showed that among 780 randomly selected students who completed four years of college 20%...
one sample showed that among 780 randomly selected students who completed four years of college 20% smoke. Use a=0.02 significance level to test the claim that the rate of smoking among those four years of college is less than 27% rate for the general population
15% of all college students volunteer their time. Is the percentage of college students who are...
15% of all college students volunteer their time. Is the percentage of college students who are volunteers larger for students receiving financial aid? Of the 332 randomly selected students who receive financial aid, 53 of them volunteered their time. What can be concluded at the α= 0.05 level of significance? The test statistic is ? The p-value = (Please show your answer to 4 decimal places.)
The scores of individual students on the American College Testing (ACT) program composite college entrance examination...
The scores of individual students on the American College Testing (ACT) program composite college entrance examination have a normal distribution with mean 18.6 and standard deviation 6.0. Forty-nine randomly selected seniors take the ACT test. What is the probability that their mean score is less than 18? Round your answer to 4 decimal places.
The length of time it takes college students to find a parking spot in the library...
The length of time it takes college students to find a parking spot in the library parking lot follows a normal distribution with a mean of 6.5 minutes and a standard deviation of 1 minute. a. Find the probability that it will take a randomly selected college student more than 4.0 minutes to find a parking spot in the library parking lot. b. Find the probability that a randomly selected college student will find a parking spot in the library...
Do college students enjoy playing sports less than watching sports? A researcher randomly selected ten college...
Do college students enjoy playing sports less than watching sports? A researcher randomly selected ten college students and asked them to rate playing sports and watching sports on a scale from 1 to 10 with 1 meaning they have no interest and 10 meaning they absolutely love it. The results of the study are shown below. Playing Vs. Watching Sports Play 6 2 4 7 5 10 5 1 7 9 Watch 5 3 6 6 6 10 7 2...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT