Question

*1. The national norm for third graders on a standardized test of reading achievement is a mean score of 27 σ 4 . Rachel determines the mean score on this test for a random sample of third graders from her school district. (a) Phrase a question about her population mean that could be answered by testing a hypothesis. (b) Phrase a question for which an estimation approach would be appropriate.

Needing problem 2 answered, but question 1 goes along with the problem.

*2. The results for Rachel’s sample in Problem 1 is X=33.10 (n=36). (a) Calculate σX . (b) Construct the 95% confidence interval for her population mean score. (c) Construct the 99% confidence interval for her population mean score. (d) What generalization is illustrated by a comparison of your answers to Problems 2b and 2c? Explain in precise terms the meaning of the interval you calculated in Problem 2b.

Answer 1

1.
Given that,
population mean(u)=27
standard deviation, sigma =4
sample mean, x =33.1
number (n)=36
null, Ho: μ=27
alternate, H1: μ!=27
level of significance, alpha = 0.05
from standard normal table, two tailed z alpha/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) = x-u/(s.d/sqrt(n))
zo = 33.1-27/(4/sqrt(36)
zo = 9.15
| zo | = 9.15
critical value
the value of |z alpha| at los 5% is 1.96
we got |zo| =9.15 & | z alpha | = 1.96
make decision
hence value of | zo | > | z alpha| and here we reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != 9.15 ) = 0
hence value of p0.05 > 0, here we reject Ho
ANSWERS
---------------
null, Ho: μ=27
alternate, H1: μ!=27
test statistic: 9.15
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that population mean

2.
a.
sample mean = 33.10
population standard deviation = 4

b.
TRADITIONAL METHOD
given that,
standard deviation, σ =4
sample mean, x =33.1
population size (n)=36
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 4/ sqrt ( 36) )
= 0.667
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 0.667
= 1.307
III.
CI = x ± margin of error
confidence interval = [ 33.1 ± 1.307 ]
= [ 31.793,34.407 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, σ =4
sample mean, x =33.1
population size (n)=36
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
value of z table is 1.96
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 33.1 ± Z a/2 ( 4/ Sqrt ( 36) ) ]
= [ 33.1 - 1.96 * (0.667) , 33.1 + 1.96 * (0.667) ]
= [ 31.793,34.407 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [31.793 , 34.407 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean

c.
TRADITIONAL METHOD
given that,
standard deviation, σ =4
sample mean, x =33.1
population size (n)=36
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 4/ sqrt ( 36) )
= 0.667
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
since our test is two-tailed
value of z table is 2.576
margin of error = 2.576 * 0.667
= 1.717
III.
CI = x ± margin of error
confidence interval = [ 33.1 ± 1.717 ]
= [ 31.383,34.817 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, σ =4
sample mean, x =33.1
population size (n)=36
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
since our test is two-tailed
value of z table is 2.576
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 33.1 ± Z a/2 ( 4/ Sqrt ( 36) ) ]
= [ 33.1 - 2.576 * (0.667) , 33.1 + 2.576 * (0.667) ]
= [ 31.383,34.817 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 99% sure that the interval [31.383 , 34.817 ] 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

d.
95% sure that the interval [31.793 , 34.407 ]
99% sure that the interval [31.383 , 34.817 ]
slight difference both the intervals