All answers must include the rationale of using a test, the hypotheses, the crit
ID: 3315638 • Letter: A
Question
All answers must include
the rationale of using a test,
the hypotheses,
the critical-value,
the test-statistics,
the p-value,
the test result,
the distribution chart, and
the conclusion.
Calcium is essential to tree growth because it promotes the formation of wood and maintains cell walls. In 1990, the concentration of calcium in precipitation in a certain area was 0.12 (mg/L). A random sample of 10 precipitation dates in 2007 results in the data table. Construct a 90% confidence interval about the mean age and test the hypothesis at the level. Complete your answer precisely following the step by step processes.
0.077
0.076
0.088
0.251
0.118
0.178
0.125
0.224
0.328
0.114
Calcium is essential to tree growth because it promotes the formation of wood and maintains cell walls. In 1990, the concentration of calcium in precipitation in a certain area was 0.12 (mg/L). A random sample of 10 precipitation dates in 2007 results in the data table. Construct a 90% confidence interval about the mean age and test the hypothesis at the level. Complete your answer precisely following the step by step processes.
0.077
0.076
0.088
0.251
0.118
0.178
0.125
0.224
0.328
0.114
Explanation / Answer
a.
t test for single mean because they given data is sample data
TRADITIONAL METHOD
given that,
sample mean, x =0.1579
standard deviation, s =0.0851
sample size, n =10
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 0.0851/ sqrt ( 10) )
= 0.027
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.1
from standard normal table, two tailed value of |t /2| with n-1 = 9 d.f is 1.833
margin of error = 1.833 * 0.027
= 0.049
III.
CI = x ± margin of error
confidence interval = [ 0.1579 ± 0.049 ]
= [ 0.109 , 0.207 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =0.1579
standard deviation, s =0.0851
sample size, n =10
level of significance, = 0.1
from standard normal table, two tailed value of |t /2| with n-1 = 9 d.f is 1.833
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 = [ 0.1579 ± t a/2 ( 0.0851/ Sqrt ( 10) ]
= [ 0.1579-(1.833 * 0.027) , 0.1579+(1.833 * 0.027) ]
= [ 0.109 , 0.207 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 90% sure that the interval [ 0.109 , 0.207 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population mean
b.
Given that,
population mean(u)=0.12
sample mean, x =0.1579
standard deviation, s =0.0851
number (n)=10
null, Ho: =0.12
alternate, H1: !=0.12
level of significance, = 0.1
from standard normal table, two tailed t /2 =1.833
since our test is two-tailed
reject Ho, if to < -1.833 OR if to > 1.833
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =0.1579-0.12/(0.0851/sqrt(10))
to =1.408
| to | =1.408
critical value
the value of |t | with n-1 = 9 d.f is 1.833
we got |to| =1.408 & | t | =1.833
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 1.4083 ) = 0.1926
hence value of p0.1 < 0.1926,here we do not reject Ho
ANSWERS
---------------
null, Ho: =0.12
alternate, H1: !=0.12
test statistic: 1.408
critical value: -1.833 , 1.833
decision: do not reject Ho
p-value: 0.1926
we do not have enough evidence to support the claim that Calcium is essential to tree growth because it promotes the formation of wood and maintains cell walls
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