Based oil interviews with 96 SARS patients, researchers found that the mean incu

Question

Based oil interviews with 96 SARS patients, researchers found that the mean incubation period was 5 6 days, with a standard deviation of 14.5 day. Based on this information, construct a 95% confidence interval for the mean incubation period of the SAKS virus Interpret the interval The lower bound is days. (Round to two decimal places as needed) The upper bound is days. (Round to two decimal places as needed) Interpret the interval Choose the correct answer below. There is a 95% probability that the mean incubation period lies between the lower and upper bounds of the interval There is 95% confidence that the mean incubation period is greater than the upper bound of the interval There is 95% confidence that the mean incubation period lies between the lower and upper bounds of the interval There is 95% confidence that the mean incubation period is less than the lower bound of the interval

Explanation / Answer

Note that              
Margin of Error E = t(alpha/2) * s / sqrt(n)              
Lower Bound = X - t(alpha/2) * s / sqrt(n)              
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    5.6          
t(alpha/2) = critical t for the confidence interval =    1.985251004          
s = sample standard deviation =    14.5          
n = sample size =    96          
df = n - 1 =    95          
Thus,              
Margin of Error E =    2.937973065          
Lower bound =    2.662026935          
Upper bound =    8.537973065          
              
Thus, the confidence interval is              
              
(   2.662026935   ,   8.537973065   ) [ANSWER]

****************************************

Hence,

OPTION C: There is 95% confidence that the mean incubation time lies between the lower and upper bounds of the interval. [ANSWER]

*************************************

Hi! If you use z distribution because n = 96 is large enough, this is the alternative solution:

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 =    0.025          
X = sample mean =    5.6          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    14.5          
n = sample size =    96          
              
Thus,              
Margin of Error E =    2.900550804          
Lower bound =    2.699449196          
Upper bound =    8.500550804          
              
Thus, the confidence interval is              
              
(   2.699449196   ,   8.500550804   ) [ALTERNATIVE ANSWER]

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.