Listening to music has long been thought to enhance intelligence, especially dur

Question

Listening to music has long been thought to enhance intelligence, especially during infancy and childhood. To test whether this is true, a researcher records the number of hours that eight high-performing students listened to music per day for 1 week. The data are listed in the table. Music Listening Per Day (in hours) 4.1 4.7 4.9 3.7 4.3 5.4 4.1 4.4 (a) Find the confidence limits at a 95% CI for this one-independent sample. (Round your answers to two decimal places.) to hours per day (b) Suppose the null hypothesis states that students listen to 3.5 hours of music per day. What would the decision be for a two-tailed hypothesis test at a 0.05 level of significance? Retain the null hypothesis because the value stated in the null hypothesis is within the limits for the 95% CI. Reject the null hypothesis because the value stated in the null hypothesis is outside the limits for the 95% CI. Reject the null hypothesis because the value stated in the null hypothesis is within the limits for the 95% CI. Retain the null hypothesis because the value stated in the null hypothesis is outside the limits for the 95% CI.

Explanation / Answer

a.
TRADITIONAL METHOD
given that,
sample mean, x =4.45
standard deviation, s =0.5345
sample size, n =8
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 0.5345/ sqrt ( 8) )
= 0.19
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 7 d.f is 2.365
margin of error = 2.365 * 0.19
= 0.45
III.
CI = x ± margin of error
confidence interval = [ 4.45 ± 0.45 ]
= [ 4 , 4.9 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =4.45
standard deviation, s =0.5345
sample size, n =8
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 7 d.f is 2.365
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 = [ 4.45 ± Z a/2 ( 0.5345/ Sqrt ( 8) ]
= [ 4.45-(2.365 * 0.19) , 4.45+(2.365 * 0.19) ]
= [ 4 , 4.9 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 4 , 4.9 ] 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

b.

Given that,
population mean(u)=3.5
sample mean, x =4.45
standard deviation, s =0.5345
number (n)=8
null, Ho: =3.5
alternate, H1: !=3.5
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.365
since our test is two-tailed
reject Ho, if to < -2.365 OR if to > 2.365
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =4.45-3.5/(0.5345/sqrt(8))
to =5.027
| to | =5.027
critical value
the value of |t | with n-1 = 7 d.f is 2.365
we got |to| =5.027 & | t | =2.365
make decision
hence value of | to | > | t | and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 5.0271 ) = 0.0015
hence value of p0.05 > 0.0015,here we reject Ho
ANSWERS
---------------
null, Ho: =3.5
alternate, H1: !=3.5
test statistic: 5.027
critical value: -2.365 , 2.365
decision: reject Ho
p-value: 0.0015

we have enough evidence to support the claim

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