(20pts) 2. Phonics is an instructional method in which children are taught to co
ID: 3066843 • Letter: #
Question
(20pts) 2. Phonics is an instructional method in which children are taught to connect sounds with letters or groups of letters. A sample of 134 first-graders who were learning English were asked to kdenlis as mnany trter soumds as posibie in poriod fone minute. The average numbor f erter Construct a 98% confidence interval for the mean number of letter sounds identified in one minute. a) b) If a 95% confidence interval were constructed with these data, would it be wider or narrower than the interval constructed in part a? Explain If a sample of 150 students had been studied, would you expect the confidence interval to be wider or narrower than the interval constructed in part a? c)Explanation / Answer
PART A.
TRADITIONAL METHOD
given that,
sample mean, x =34.06
standard deviation, s =23.83
sample size, n =134
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 23.83/ sqrt ( 134) )
= 2.059
II.
margin of error = t ?/2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, ? = 0.02
from standard normal table, two tailed value of |t ?/2| with n-1 = 133 d.f is 2.355
margin of error = 2.355 * 2.059
= 4.848
III.
CI = x ± margin of error
confidence interval = [ 34.06 ± 4.848 ]
= [ 29.212 , 38.908 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =34.06
standard deviation, s =23.83
sample size, n =134
level of significance, ? = 0.02
from standard normal table, two tailed value of |t ?/2| with n-1 = 133 d.f is 2.355
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 = [ 34.06 ± t a/2 ( 23.83/ Sqrt ( 134) ]
= [ 34.06-(2.355 * 2.059) , 34.06+(2.355 * 2.059) ]
= [ 29.212 , 38.908 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 98% sure that the interval [ 29.212 , 38.908 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 98% of these intervals will contains the true population mean
PART B.
Using a 95% CI
confidence interval = [ 34.06 ± t a/2 ( 23.83/ Sqrt ( 134) ]
= [ 34.06-(1.978 * 2.059) , 34.06+(1.978 * 2.059) ]
= [ 29.988 , 38.132 ]
the interval is narrower than the interval constructed in part a
PART C.
sample size, n =150
confidence interval = [ 34.06 ± t a/2 ( 23.83/ Sqrt ( 150) ]
= [ 34.06-(1.976 * 1.946) , 34.06+(1.976 * 1.946) ]
= [ 30.215 , 37.905 ]
using the larger sample size the interval is wider than usual
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.