Question

1. "Adults Who Stutter: Responses to Cognitive Stress," published in the Journal of Speech and Hearing Research in 1994, reported that speech rate, in words per minute, for 9 stutterers under conditions of stress had mean 52.6 and standard deviation 3.2.

Use the fact that the t multiplier for 8 degrees of freedom at 99% confidence is 3.36 to find the margin of error for the 99% confidence interval. Report to one decimal place.

2. Refer to question 1.

If a lower level of confidence were desired, would the confidence interval become wider (select YES) or narrower (select NO)?

3. Refer to question 1.

If the standard deviation would have been larger than 3.2, would the confidence interval become wider (select YES) or narrower (select NO)?

4. Refer to question 1.

If the sample size would have been smaller than 9, would the confidence interval become wider (select YES) or narrower (select NO)?

5. Refer to question 1.

If 3.2 was the known population standard deviation instead of the sample standard deviation, would the confidence interval become wider (select YES) or narrower (select NO)?

Answer 1

1) Margin of error

    

   

2) The confidence interval would become narrower

If the confidence level, decreases, the margin of error also decreases. So that the confidence interval becomes narrower.

3) The confidence interval would become wider

If the standard deviation increases, the margin of error also increases. So that the confidence interval becomes wider.

4) The confidence interval would become wider

If the sample size decreases, the standard error increases. Which increases the margin of error . So that the confidence interval becomes wider.

5) The confidence interval would become narrower

If the population standard deviation is known, then we will use z-distribution instead of t-distribution. The z critical value will be less than the t critical value. So the margin of error will decrease and the interval will become narrower.