Answer all numeric questions to 3 decimal places unless the number has fewer dig
ID: 3316960 • Letter: A
Question
Answer all numeric questions to 3 decimal places unless the number has fewer digits or you are otherwise instructed.
I just need answer. I do not need R code. I want to make sure if I have correct answers. Answer all. Thank you
5. + Question Details My Notes Ask Your Teacher Body temperatures of humans (in degrees Fahrenheit) have a known standard deviation of -0.75 degree. A random sample of 23 people yielded a mean of x-98.6 degrees with a sample standard deviation of s= 0.70 degrees. It is known that the human body temperatures has a normal distribution, we want to estimate the true average human body temperature, (in degrees Fahrenheit). a)What is the critical value for a 97.5% confidence interval for ? b) Create a 97.5% confidence interval for c) How many observations would we need to guarantee that the 97.5% confidence interval has has a length of 0.2 or less? d) Create a 97.5% prediction interval for the body temperature of a single human e) Assuming is not known, create a g7.5% confidence interval for using this data h) Copy your R script for the above into the text box hereExplanation / Answer
PART A.
level of significance, = 0.025
from standard normal table, two tailed z /2 =2.241
PART B.
TRADITIONAL METHOD
given that,
standard deviation, =0.75
sample mean, x =98.6
population size (n)=23
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 0.75/ sqrt ( 23) )
= 0.16
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.025
from standard normal table, two tailed z /2 =2.241
since our test is two-tailed
value of z table is 2.241
margin of error = 2.241 * 0.16
= 0.35
III.
CI = x ± margin of error
confidence interval = [ 98.6 ± 0.35 ]
= [ 98.25,98.95 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =0.75
sample mean, x =98.6
population size (n)=23
level of significance, = 0.025
from standard normal table, two tailed z /2 =2.241
since our test is two-tailed
value of z table is 2.241
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 98.6 ± Z a/2 ( 0.75/ Sqrt ( 23) ) ]
= [ 98.6 - 2.241 * (0.16) , 98.6 + 2.241 * (0.16) ]
= [ 98.25,98.95 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 97.5% sure that the interval [98.25 , 98.95 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 97.5% of these intervals will contains the true population mean
PART C.
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.025% LOS is = 2.241 ( From Standard Normal Table )
Standard Deviation ( S.D) = 0.75
ME =0.2
n = ( 2.241*0.75/0.2) ^2
= (1.68/0.2 ) ^2
= 70.62 ~ 71
PART D.
TRADITIONAL METHOD
given that,
sample mean, x =98.6
standard deviation, s =0.7
sample size, n =23
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 0.7/ sqrt ( 23) )
= 0.146
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.025
from standard normal table, two tailed value of |t /2| with n-1 = 22 d.f is 2.405
margin of error = 2.405 * 0.146
= 0.351
III.
CI = x ± margin of error
confidence interval = [ 98.6 ± 0.351 ]
= [ 98.249 , 98.951 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =98.6
standard deviation, s =0.7
sample size, n =23
level of significance, = 0.025
from standard normal table, two tailed value of |t /2| with n-1 = 22 d.f is 2.405
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 = [ 98.6 ± t a/2 ( 0.7/ Sqrt ( 23) ]
= [ 98.6-(2.405 * 0.146) , 98.6+(2.405 * 0.146) ]
= [ 98.249 , 98.249]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 97.5% sure that the interval [ 98.249 , 98.951 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 97.5% of these intervals will contains the true population mean
PART E.
PART B.
> a <- 98.6
> s <- 0.75
> n <- 23
> error <- qt(0.975,df=n-1)*s/sqrt(n)
> left <- a-error
> right <- a+error
> left
[1] 98.25
> right
[1] 98.95
PART D.
> a <- 98.6
> s <- 0.75
> n <- 23
> error <- qnorm(0.975)*s/sqrt(n)
> left <- a-error
> right <- a+error
> left
[1] 98.249
> right
[1] 98.249
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