Question

One random sample of the hours of one type of user watch TV per day was recorded: 2.9, 3.5, 3, 3.2, 3.3, 3.5, 3.3, 3, 3.5.

1. Apply the backward empirical rule to check normality of the data, and conclude if any evidence of non-normality.

2. Assume that the underlying population is normal. Find a two-sided 99% confidence interval for the the true population average hours.

3. Assume that population variance ?2=0.09σ2=0.09 and the bound on the error of estimation of 99% confident interval to be 0.2. Find the minimum sample size ?n.

Answer 1

1.
given data is
the Empirical Rule states that 99.7% of data observed following a normal distribution lies within 3 standard deviations of the mean. Under this rule,
68% of the data falls within one standard deviation, 95% percent within two standard deviations, and 99.7% within three standard deviations from the mean.
so that here it is 3 standard deviations
2.
TRADITIONAL METHOD
given that,
sample mean, x =3.244
standard deviation, s =0.221
sample size, n =9
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 0.221/ sqrt ( 9) )
= 0.074
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.01
from standard normal table, two tailed value of |t α/2| with n-1 = 8 d.f is 3.355
margin of error = 3.355 * 0.074
= 0.247
III.
CI = x ± margin of error
confidence interval = [ 3.244 ± 0.247 ]
= [ 2.997 , 3.491 ]
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DIRECT METHOD
given that,
sample mean, x =3.244
standard deviation, s =0.221
sample size, n =9
level of significance, α = 0.01
from standard normal table, two tailed value of |t α/2| with n-1 = 8 d.f is 3.355
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 = [ 3.244 ± t a/2 ( 0.221/ Sqrt ( 9) ]
= [ 3.244-(3.355 * 0.074) , 3.244+(3.355 * 0.074) ]
= [ 2.997 , 3.491 ]
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interpretations:
1) we are 99% sure that the interval [ 2.997 , 3.491 ] 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
3.
Given data,
population variance = 0.09
standard deviation = sqrt(0.09)=0.3
confidence level is 99%
margin of error =0.2
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Zα/2 at 0.01% LOS is = 2.576 ( From Standard Normal Table )
Standard Deviation ( S.D) = 0.3
ME =0.2
n = ( 2.576*0.3/0.2) ^2
= (0.773/0.2 ) ^2
= 14.93 ~ 15