Twelve different video games showing substance use were observed and the duratio

Question

Twelve different video games showing substance use were observed and the duration times of game play (in seconds) are listed below. The design of the study justifies the assumption that the sample can be treated as a simple random sample. Use the data to construct a 99% confidence interval estimate of , the mean duration of game play. 5001 4053 here to view a t distribution table 3882 3852 4036 4316 4822 4652 4027 4828 4340 4322 standard n standard normal distribution table distribution tab lick lick ew What is the confidence interval estimate of the population mean ? (Round to one decimal place as needed.)

Explanation / Answer

PART A.

TRADITIONAL METHOD

given that,

sample mean, x =4344.25

standard deviation, s =396.1265

sample size, n =12

I.

stanadard error = sd/ sqrt(n)

where,

sd = standard deviation

n = sample size

standard error = ( 396.1265/ sqrt ( 12) )

= 114.352

II.

margin of error = t /2 * (stanadard error)

where,

ta/2 = t-table value

level of significance, = 0.01

from standard normal table, two tailed value of |t /2| with n-1 = 11 d.f is 3.106

margin of error = 3.106 * 114.352

= 355.177

III.

CI = x ± margin of error

confidence interval = [ 4344.25 ± 355.177 ]

= [ 3989.073 , 4699.427 ]

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DIRECT METHOD

given that,

sample mean, x =4344.25

standard deviation, s =396.1265

sample size, n =12

level of significance, = 0.01

from standard normal table, two tailed value of |t /2| with n-1 = 11 d.f is 3.106

we use CI = x ± t a/2 * (sd/ Sqrt(n))

where,

x = mean

sd = standard deviation

a = 1 - (confidence level/100)

ta/2 = t-table value

CI = confidence interval

confidence interval = [ 4344.25 ± t a/2 ( 396.1265/ Sqrt ( 12) ]

= [ 4344.25-(3.106 * 114.352) , 4344.25+(3.106 * 114.352) ]

= [ 3989.073 , 4699.427 ]

[ 3989.1 , 4699.4 ]

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interpretations:

1) we are 99% sure that the interval [ 3989.073 , 4699.427 ] contains the true population mean

2) If a large number of samples are collected, and a confidence interval is created

for each sample, 99% of these intervals will contains the true population mean

PART B.

Given that,

possible chances (x)=94

sample size(n)=470

success rate ( p )= x/n = 0.2

success probability,( po )=0.25

failure probability,( qo) = 0.75

null, Ho:p=0.25  

alternate, H1: p!=0.25

level of significance, = 0.05

from standard normal table, two tailed z /2 =1.96

since our test is two-tailed

reject Ho, if zo < -1.96 OR if zo > 1.96

we use test statistic z proportion = p-po/sqrt(poqo/n)

zo=0.2-0.25/(sqrt(0.1875)/470)

zo =-2.5033 ~ -2.5

PART C.

Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)

Z a/2 at 0.1 is = 1.645

Sample Proportion = 0.1

ME = 0.06

n = ( 1.645 / 0.06 )^2 * 0.1*0.9

= 67.6506 ~ 68

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