1 sally wants to open a candy store and is trying to estimate the amount that st

Question

1 sally wants to open a candy store and is trying to estimate the amount that startup costs wi looks at a random sample of 9 comparable stores and finds (in thousands of dollars) a mean o standard deviation of 29.4. She has also discovered that the population is not normally What is the 95% confidence interval for the population mean estimating the startup costs . ll be. She distributed. cant do t 2. The Department of Agriculture wants to estimate the average price that farmers get for their watermelon crops. The estimate must be within S0.30 of the mean price per 100 pounds of watermelon. Assume the population standard deviation is $1.92 per 100 pounds. What is the minimum sample size necessary to construct a 90% confidence interval for the population mean? Over the past several months, an adult patient has been treated for severe muscle spasms and his doctor wants to determine if he still has a calcium deficiency. This condition is determined by tests measuring total calcium levels in milligrams per deciliters. A random sample of 10 tests has been taken from a patient that has a sample mean of 9.95 mg/dl and a sample standard deviation of 1.02 mg/dl. The population is normally distributed what is the 99% confidence interval for the population mean of total calcium in this patient's blood? 3.

Explanation / Answer

2.
Compute Sample Size  
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.1% LOS is = 1.64 ( From Standard Normal Table )
Standard Deviation ( S.D) = 1.92
ME =0.3
n = ( 1.64*1.92/0.3) ^2
= (3.15/0.3 ) ^2
= 110.17 ~ 111

3.
TRADITIONAL METHOD
given that,
sample mean, x =9.95
standard deviation, s =1.02
sample size, n =10
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 1.02/ sqrt ( 10) )
= 0.323
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.01
from standard normal table, two tailed value of |t /2| with n-1 = 9 d.f is 3.25
margin of error = 3.25 * 0.323
= 1.048
III.
CI = x ± margin of error
confidence interval = [ 9.95 ± 1.048 ]
= [ 8.902 , 10.998 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =9.95
standard deviation, s =1.02
sample size, n =10
level of significance, = 0.01
from standard normal table, two tailed value of |t /2| with n-1 = 9 d.f is 3.25
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 = [ 9.95 ± t a/2 ( 1.02/ Sqrt ( 10) ]
= [ 9.95-(3.25 * 0.323) , 9.95+(3.25 * 0.323) ]
= [ 8.902 , 10.998 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 99% sure that the interval [ 8.902 , 10.998 ] 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

4.
TRADITIONAL METHOD
given that,
standard deviation, =42.6
sample mean, x =138
population size (n)=30
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 42.6/ sqrt ( 30) )
= 7.78
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.1
from standard normal table, two tailed z /2 =1.64
since our test is two-tailed
value of z table is 1.64
margin of error = 1.64 * 7.78
= 12.76
III.
CI = x ± margin of error
confidence interval = [ 138 ± 12.76 ]
= [ 125.24,150.76 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =42.6
sample mean, x =138
population size (n)=30
level of significance, = 0.1
from standard normal table, two tailed z /2 =1.64
since our test is two-tailed
value of z table is 1.64
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 = [ 138 ± Z a/2 ( 42.6/ Sqrt ( 30) ) ]
= [ 138 - 1.64 * (7.78) , 138 + 1.64 * (7.78) ]
= [ 125.24,150.76 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 90% sure that the interval [125.24 , 150.76 ] 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
[ANSWERS]
best point of estimate = mean = 138
standard error =7.78
z table value = 1.64
margin of error = 12.76
confidence interval = [ 125.24 , 150.76 ]

5.
TRADITIONAL METHOD
given that,
sample mean, x =91
standard deviation, s =30.7
sample size, n =6
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 30.7/ sqrt ( 6) )
= 12.533
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.2
from standard normal table, two tailed value of |t /2| with n-1 = 5 d.f is 1.476
margin of error = 1.476 * 12.533
= 18.499
III.
CI = x ± margin of error
confidence interval = [ 91 ± 18.499 ]
= [ 72.501 , 109.499 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =91
standard deviation, s =30.7
sample size, n =6
level of significance, = 0.2
from standard normal table, two tailed value of |t /2| with n-1 = 5 d.f is 1.476
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 = [ 91 ± t a/2 ( 30.7/ Sqrt ( 6) ]
= [ 91-(1.476 * 12.533) , 91+(1.476 * 12.533) ]
= [ 72.501 , 109.499 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 80% sure that the interval [ 72.501 , 109.499 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 80% of these intervals will contains the true population mean

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