1) A class survey in a large class for first-year college students asked, About
ID: 3151372 • Letter: 1
Question
1) A class survey in a large class for first-year college students asked, About how many hours do you study during a typical week? The mean response of the 463 students was x = 15.3 hours.
Suppose that we know that the study time follows a Normal distribution with standard deviation ? = 8.5 hours in the population of all first-year students at this university.
Step 1: Use the survey result to give a 99% confidence interval for the mean study time of all first-year students.
Step 2:
What condition not yet mentioned is needed for your confidence interval to be valid
2) Sulfur compounds cause off-odors in wine, so winemakers want to know the odor threshold, the lowest concentration of a compound that the human nose can detect. The odor threshold for dimethyl sulfide (DMS) in trained wine tasters is about 25 micrograms per liter of wine (?g/l). The untrained noses of consumers may be less sensitive, however. Here are the DMS odor thresholds for 10 untrained students:
Step 1:
Do the three simple conditions hold in this case? Use a stemplot to check the shape of the distribution.
a) Yes, since this is an SRS, a stemplot shows that the distribution is roughly symmetric with no outliers, and ? is known.
b) No, we can't be sure. Although the stemplot looks roughly Normal and ? is known, these 10 students are not an SRS of the population in general, and without any more information on how they were chosen, they may not even be an SRS of the student population.
c) No. Although this is a SRS and ? is known, a stemplot shows that the distribution is not symmetric and therefore cannot be Normal.
d) No. Although this is a SRS and ? is known, a stemplot shows that the distribution has many outliers and therefore cannot be Normal.
Step 2:
Following the four-step process, give a 95% confidence interval for the mean DMS odor threshold among all students.
a) 14.98 to 21.02 b) 14.28 to 16.32 c) 24.08 to 28.92 d) 17.64 to 18.36Explanation / Answer
1.
Note that
Margin of Error E = z(alpha/2) * s / sqrt(n)
Lower Bound = X - z(alpha/2) * s / sqrt(n)
Upper Bound = X + z(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.005
X = sample mean = 15.3
z(alpha/2) = critical z for the confidence interval = 2.575829304
s = sample standard deviation = 8.5
n = sample size = 463
Thus,
Margin of Error E = 1.017525987
Lower bound = 14.28247401
Upper bound = 16.31752599
Thus, the confidence interval is
( 14.28247401 , 16.31752599 ) [ANSWER, OPTION B]
************************
STEP 2:
What condition not yet mentioned is needed for your confidence interval to be valid?
SRS. [ANSWER, C]
**************************
*******************************************
Hi! Please submit the next part as a separate question. That way we can continue helping you! Please indicate which parts are not yet solved when you submit. Thanks!
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.