Question

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

The average daily volume of a computer stock in 2011 was μ=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 30 trading days in​ 2014, he finds the sample mean to be 28.8 million​ shares, with a standard deviation of s=11.1 million shares. Test the hypotheses by constructing a 95​%

confidence interval. Complete parts​ (a) through​ (c) below.

Construct a

9595​%

confidence interval about the sample mean of stocks traded in 2014.

The lower bound is

nothing

million shares.

The upper bound is

nothing

million shares.

​(Round to three decimal places as​ needed.)

​(c) Will the researcher reject the null​ hypothesis?

A.

Do not rejectDo not reject

the null hypothesis because

muμequals=35.135.1

million shares

does not falldoes not fall

in the confidence interval.

B.

Do not rejectDo not reject

the null hypothesis because

muμequals=35.135.1

million shares

fallsfalls

in the confidence interval.

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

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

Homework Answers

Answer #1

Solution :

Given that,

Point estimate = sample mean = = 28.8

sample standard deviation = s = 11.1

sample size = n = 30

Degrees of freedom = df = n - 1 = 30-1 = 29

At 95% confidence level

= 1-0.95% =1-0.95 =0.05

/2 =0.05/ 2= 0.025

t/2,df = t0.025,29 = 2.05

t /2,df = 2.05

Margin of error = E = t/2,df * (s /n)

= 2.05 * (11.1/ 30)

Margin of error = E = 4.145

The 95% confidence interval estimate of the population mean is,

- E < <  + E

28.8 - 4.145 < < 28.8+ 4.145

24.655 < < 32.945

(24.655,32.945)

Lower bound = 24.655

Upper bound = 32.945

= 35.1

This is the two tailed test .

The null and alternative hypothesis is

H0 :   = 35.1

Ha :    35.1

Test statistic = t

= ( - ) / s / n

= (28.8-35.1) / 11.1 / 30

= -3.109

P (Z < -3.109 ) = 0.0042

P-value = 0.0042

= 0.05  

p=0.0042<0.05

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

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