The average daily volume of a computer stock in 2011 was mu = 35.1 million share

Question

The average daily volume of a computer stock in 2011 was mu = 35.1 million shares, according to a reliable source A stock analyst believes that the stock volume in 2014 is different from the 2011 level. Based on a random sample of 40 trading days in 2014, he finds the sample mean to be 31.7 million shares, with a standard deviation of s = 11 8 million shares Test the hypotheses by constructing a 95% confidence interval Complete parts 0 through (c) below State the hypotheses for the test. H_0: 35.1 million shares H_1: million shares (b) Construct a 95% confidence interval about the sample mean of stocks traded in 2014 (million shares, million shares) use ascending order Round to three decimal places, as needed) Will the researcher reject the null hypothesis? Reject the null hypothesis because mu = 35.1 million shares does not tall in the confidence interval Reject the hypothesis because mu = 35.1 million shares falls in the confidence interval Do not reject the null hypothesis because mu = 35 1 million shares does not fall in the confidence interval Do not reject the null hypothesis because mu = 35.1 million shares falls in the confidence interval.

Explanation / Answer

a)

"The analyst believes 2014 level is different from 2011 level."

Hence, by "different", it is two tailed, so

Ho: u = 35.1
Ha: u =/= 35.1 [ANSWER]

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

b)

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 =    31.7          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    11.8          
n = sample size =    40          
              
Thus,              
Margin of Error E =    3.656790691          
Lower bound =    28.04320931          
Upper bound =    35.35679069          
              
Thus, the confidence interval is              
              
(   28.04320931   ,   35.35679069   ) [ANSWER]

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

c)

As 35.1 is inside the interval, we do not reject Ho.

Hence,

OPTION D. [ANSWER]

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

Hi! If you use t distribution (even when n = 40 is large), this is the alternative for part b:

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 =    31.7          
t(alpha/2) = critical t for the confidence interval =    2.02269092          
s = sample standard deviation =    11.8          
n = sample size =    40          
df = n - 1 =    39          
Thus,              
Margin of Error E =    3.773823083          
Lower bound =    27.92617692          
Upper bound =    35.47382308          
              
Thus, the confidence interval is              
              
(   27.92617692   ,   35.47382308   ) [ALTERNATIVE ANSWER, B]

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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