Question

III Question Help 4.80 5.19 5.70 The following data represent the pH of rain for a random sample of 12 rain dates. A normal probability plot suggests the data could come from a population that is normally distributed. A boxplot indicates 5.05 5.72 5.24 there are no outliers. Complete parts a) through d) below. 5.02 4,58 4.74 5.34 4.76 4.56 Click the icon to view the table of critical t-values. (a) Determine a point estimate for the population mean. A point estimate for the population mean is I (Round to two decimal places as needed.) (b) Construct and interpret a 95% confidence interval for the mean pH of rainwater. Select the correct choice below and fill in the answer boxes to complete your choice (Use ascending order. Round to two decimal places as needed) O A There is a 95% probability that the true mean pH of rain water is between and OB. There is 95% confidence that the population mean pH of rain water is between OC. I repeated samples we taken, 95% of them will have a sample pH of rain water between and Click to select your answer(s) and in course a summer i based on an informed Decisions Using ata, te

Answer 1

a.
point estimate for population mean is 5.0583 = 5.06
b.
TRADITIONAL METHOD
given that,
sample mean, x =5.0583
standard deviation, s =0.3948
sample size, n =12
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 0.3948/ sqrt ( 12) )
= 0.11
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 11 d.f is 2.201
margin of error = 2.201 * 0.11
= 0.25
III.
CI = x ± margin of error
confidence interval = [ 5.0583 ± 0.25 ]
= [ 4.81 , 5.31 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =5.0583
standard deviation, s =0.3948
sample size, n =12
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 11 d.f is 2.201
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 = [ 5.0583 ± t a/2 ( 0.3948/ Sqrt ( 12) ]
= [ 5.0583-(2.201 * 0.11) , 5.0583+(2.201 * 0.11) ]
= [ 4.81 , 5.31 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 4.81 , 5.31 ] 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,
sample mean, x =5.0583
standard deviation, s =0.3948
sample size, n =12
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 0.3948/ sqrt ( 12) )
= 0.11
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.01
from standard normal table, two tailed value of |t α/2| with n-1 = 11 d.f is 3.106
margin of error = 3.106 * 0.11
= 0.35
III.
CI = x ± margin of error
confidence interval = [ 5.0583 ± 0.35 ]
= [ 4.7 , 5.41 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =5.0583
standard deviation, s =0.3948
sample size, n =12
level of significance, α = 0.01
from standard normal table, two tailed value of |t α/2| with n-1 = 11 d.f is 3.106
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 = [ 5.0583 ± t a/2 ( 0.3948/ Sqrt ( 12) ]
= [ 5.0583-(3.106 * 0.11) , 5.0583+(3.106 * 0.11) ]
= [ 4.7 , 5.41 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 99% sure that the interval [ 4.7 , 5.41 ] 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