Question

In testing a proposed method of raising cognitive performance, assume results are collected as IQ test...

In testing a proposed method of raising cognitive performance, assume results are collected as IQ test scores. When deciding to accept or reject the null hypothesis, one should use:

Group of answer choices

A one-sided (left) hypothesis

A two sided or two tailed test

A no-sided (null) hypothesis

A one-sided (right) hypothesis

Math   Statistics-And-Probability   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

In the above problem focus on the term rising cognitive performance, which implies population parameter is greater than certain value.

Therefore, for the above problem we can use, one sided test. As population parameter is increasing. We go for one sided right hypothesis test

Last option is correct.

One sided right hypothesis

Dear student,
I am waiting for your feedback. I have given my 100% to solve your queries. If you satisfied with my answer then please please like this.
Thank You

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
Identifying individuals with a high risk of Alzheimer’s disease usually involves a long series of cognitive...
Identifying individuals with a high risk of Alzheimer’s disease usually involves a long series of cognitive tests. However, researchers have developed a 7-Minute Screen, which is a quick and easy way to accomplish the same goal. The question is whether the 7-Minute Screen is as effective as the complete series of tests. To address this question, Ijuin et al. (2008) administered both tests to a group of patients and compared the results. The following data represent results similar to those...
To determine whether training in a series of workshops on creative thinking increases IQ scores, a...
To determine whether training in a series of workshops on creative thinking increases IQ scores, a total of 70 students are randomly divided into treatment and control groups of 35 each. After two months of training, the sample mean IQ (–X1) for the treatment group equals 110, and the sample mean IQ (–X2) for the control group equals 108. The estimated standard error equals 1.80. Question: Steps: Calculations or Logic: Answer: Using t , test the null hypothesis at the...
Identifying individuals with a high risk of Alzheimer’s disease usually involves a long series of cognitive...
Identifying individuals with a high risk of Alzheimer’s disease usually involves a long series of cognitive tests. However, researchers have developed a 7-Minute Screen, which is a quick and easy way to accomplish the same goal. The question is whether the 7-Minute Screen is as effective as the complete series of tests. To address this question, Ijuin et al. (2008) administered both tests to a group of patients and compared the results. The following data represent results similar to those...
Assume we are conducting a hypothesis test at the 5% significance level. Find the p-value for...
Assume we are conducting a hypothesis test at the 5% significance level. Find the p-value for a left-tailed test if z0= -1.92. Is there enough evidence to reject the null hypothesis? Find the p-value for a right-tailed test if z0= 2.53. Is there enough evidence to reject the null hypothesis? Find the p-value for a two-tailed test if z0= -2.91. Is there enough evidence to reject the null hypothesis? Find the p-value for a two-tailed test if z0= 2.88. Is...
A cognitive researcher wonders whether massed studying (studying all at once) or spaced studying leads to...
A cognitive researcher wonders whether massed studying (studying all at once) or spaced studying leads to better test performance. He randomly selects 40 participants and, believing that IQ might be a confounding variable, matches participants on IQ scores and then divides the matched pairs into two groups. The first group studies the test material for 5 hours in one night, while the other group studies for 1 hour a night for 5 nights. All 40 people are then given on...
The test statistic of z= "1.43" is obtained when testing claims that p>0.4 using a significance...
The test statistic of z= "1.43" is obtained when testing claims that p>0.4 using a significance level of a=0.01, should we reject H0 or should we fail to reject H0. identify the hypothesis test as being two-tailed, left-tail, or right-tailed. find the p-value
The test statistic of z=1.21 is obtained when testing the claim that p?0.653. a. Identify the...
The test statistic of z=1.21 is obtained when testing the claim that p?0.653. a. Identify the hypothesis test as being? two-tailed, left-tailed, or? right-tailed. b. Find the? P-value. c. Using a significance level of ?=0.01?, should we reject Upper H 0H0 or should we fail to reject Upper H 0?
The test statistic of zequals=1.32 is obtained when testing the claim that pnot equals≠0.492 a. Identify...
The test statistic of zequals=1.32 is obtained when testing the claim that pnot equals≠0.492 a. Identify the hypothesis test as being​ two-tailed, left-tailed, or​ right-tailed. b. Find the​ P-value. c. Using a significance level of alphaαequals=0.10,should we reject Upper H 0 or should we fail to reject Upper H0​?
Use the t-distribution and the sample results to complete the following test of the hypotheses. Use...
Use the t-distribution and the sample results to complete the following test of the hypotheses. Use a 5% significance level. Test Ho : μ = 500 vs Ha : μ ≠ 500 using the sample results x ¯ = 432, s = 118, with n = 75. a) Discuss whether it is OK to use the t-distribution to determine the results of the test. The underlying distribution is skewed to the right with several outliers. b) How many degrees of...
Use a 0.05 significance level to test the claim that IQ scores of people with low...
Use a 0.05 significance level to test the claim that IQ scores of people with low lead levels vary more than IQ scores of people with high lead levels. Use the following statistics: Low Lead Level: n=78, x=92.88462, s=15.34451 High Lead Level: n=21, x=86,90476, s=8.988352 a)Write down the claim b) Write down the null hypothesis and the alternative hypothesis. Indicate which one is the claim. c) Draw the probability distribution curve. Write down the testing statistic and the P-Value on...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT