8. The principal at Riverside High School would Iie to estimate the mean length
ID: 3360885 • Letter: 8
Question
8. The principal at Riverside High School would Iie to estimate the mean length of time each day that it takesall the buses to arrive ad unload students. How large a sample needed fthe propal would like to assert with 90% confidence that te sample mean is off by, at most 7 minutes. Assume ·14 minutes. Find the critical t-value that corresponds to 90% confidence and n 15. 10. A random sample of 10 parking meters in a resort community showed the following incomes for a day US $3.60 $450 $2.80 $6 30 $2.60 $5.20 $6.75 $4.25 $8.00 $3.00 Assume the incomes are normally distributed. Find the 95% confidence interval fr te true mean. (Use t) st en nbExplanation / Answer
8.
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.1% LOS is = 1.645 ( From Standard Normal Table )
Standard Deviation ( S.D) = 14
ME =7
n = ( 1.645*14/7) ^2
= (23.03/7 ) ^2
= 10.824 ~ 11
9.
Given that,
population mean(u)=7
sample mean, x =7
standard deviation, s =14
number (n)=15
null, Ho: =7
alternate, H1: <7
level of significance, = 0.1
from standard normal table,left tailed t /2 =1.345
since our test is left-tailed
reject Ho, if to < -1.345
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =7-7/(14/sqrt(15))
to =0
| to | =0
critical value
the value of |t | with n-1 = 14 d.f is 1.345
we got |to| =0 & | t | =1.345
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :left tail - Ha : ( p < 0 ) = 0.5
hence value of p0.1 < 0.5,here we do not reject Ho
ANSWERS
---------------
null, Ho: =7
alternate, H1: <7
test statistic: 0
critical value: -1.345
decision: do not reject Ho
p-value: 0.5
10.
TRADITIONAL METHOD
given that,
sample mean, x =4.7
standard deviation, s =1.8338
sample size, n =10
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 1.8338/ sqrt ( 10) )
= 0.5799
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 = 9 d.f is 2.262
margin of error = 2.262 * 0.5799
= 1.3117
III.
CI = x ± margin of error
confidence interval = [ 4.7 ± 1.3117 ]
= [ 3.3883 , 6.0117 ]
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DIRECT METHOD
given that,
sample mean, x =4.7
standard deviation, s =1.8338
sample size, n =10
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 9 d.f is 2.262
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 = [ 4.7 ± t a/2 ( 1.8338/ Sqrt ( 10) ]
= [ 4.7-(2.262 * 0.5799) , 4.7+(2.262 * 0.5799) ]
= [ 3.3883 , 6.0117 ]
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interpretations:
1) we are 95% sure that the interval [ 3.3883 , 6.0117 ] 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
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