Question

In: Statistics and Probability

1. The average daily volume of a computer stock in 2011 was μ= 35.1million​ shares, according...

1. The average daily volume of a computer stock in 2011 was μ= 35.1million​ 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 30 trading days in​ 2014, he finds the sample mean to be 28.5 million​ shares, with a standard deviation of s=12million shares. Test the hypotheses by constructing a 95​% confidence interval. Complete parts​ (a) through​ (c) below.

a. State the hypotheses for the test.

H0: (p,o,μ) (<,>,=, ≠) 35.1 million shares

H1: (p,o,μ) (<,>,=, ≠) 35.1 million shares

b. Construct a 95​% confidence interval about the sample mean of stocks traded in 2014.

The lower bound is BLANK million shares.

The upper bound is BLANK million shares.

c. Will the researcher reject the null​ hypothesis?

A. Reject the null hypothesis because μ=35.1million shares falls in the confidence interval.

B.Do not reject the null hypothesis because μ=35.1million sharesdoes not fall in the confidence interval.

C.Do not reject the null hypothesis because μ=35.1million shares falls in the confidence interval.

D. Reject the null hypothesis because μ=35.1million shares does not fall in the confidence interval.

Solutions

Expert Solution

Given that the average daily volume of a computer stock in 2011 was μ= 35.1million​ shares, according to a reliable source and the stock analyst believes that the stock volume in 2014 is different from the 2011 level.

So, the hypotheses are:

a)

Ho: μ = 35.1 million shares

H1: μ ≠ 35.1 million shares

Now the given details are:

a random sample of n = 30 trading days in​ 2014, the sample mean is M =28.5 million​ shares, with a standard deviation of s=12million shares.

b) The confidence interval is calculated as:

μ = M ± t(sM)

where:

M = sample mean

n = Sample size

df = n-1
t = t statistic determined by the confidence level and the degree of freedom
sM = standard error = √(s2/n)

M = 28.5

N = 30

DF = 30-1= 29
t = 2.05 calculated using the excel formula for t-distribution which is =T.INV.2T(0.05, 29)
sM = √(122/30) = 2.19

μ = M ± t(sM)
μ = 28.5 ± 2.05*2.19
μ = 28.5 ± 4.481

Thus based on the calculation the Lower bound is: 24.019, and the Upper bound is: 32.981.

c) From the calculation and the above-stated confidence interval we can see that 35.1 does not lie in the confidence interval hence we can reject the null hypothesis as:

D. Reject the null hypothesis because μ=35.1million shares does not fall in the confidence interval.


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