A of 100 12oz ketchup bottles was taken. 1. If mu=12.02 and sigma=0.01ZS

iohanetc

iohanetc

Answered question

2021-10-11

A of 100 12oz ketchup bottles was taken.
1. If μ=12.02andσ=0.01. With what confidence can we claim that 12.02 oz is within 0.001 of the true mu amount of ketchup in a bottle?
2. If bottles yielded σ=0.1oz., what is the 99% confidence interval for the standard deviation of the amount of ketchup in a bottle?

Answer & Explanation

Jeffrey Jordon

Jeffrey Jordon

Expert2021-10-26Added 2605 answers

Here we have given that,

sample size (n)=100

mean (X)=12.02

standard deviation (sd)=0.01

Margin of error (E)=0.001

And we have to find confidence level (C).

C we can find by using formula,

E=Zcsdn

where Zc is the critical value for normal distribution.

Putting all the values in this formula,

0.001=Zc0.01100

0.001=Zc0.0110

0.01=Zc0.01

Zc=0.010.01=1

Zc=1

C we can find by using EXCEL.

syntax :

=NORMSDIST(z)

where Z is 1.

C=0.84

COnfidence level in % =C100=0.84100=84

With 84% confidence can we assert that 12.02 oz. is within 0.001 of the true mean amount of ketchup in a 12 oz. bottle.

b. If bottles yielded a standard deviation of 0.1 oz ., what is a 99% confidence interval for the standard deviation of the amount of ketchup in a 12 oz bottle

Here we have to find 99% confidence interval for standard deviation.

sd(s)=0.1

99% confidence interval for sigma is,

(n1)s2X2u<σ<(n1)s2X2l

where X2u and X2l are critical value for CHi square distribution.

These values we can find by using EXCEL.

syntax :

For X2u :

=CHIINV(probability, deg_freedom)

where probability =1C2

C is confidence level = 99% = 0.99

deg_freedom = n-1 = 100-1 = 99

For X2l :

=CHIINV(probability, deg_freedom)

where probability =1+C2

X2u=138.99

X2l=66.51

lower limit =(n1)s2X2u=(1001)0.12138.99=0.007123=0.084

upper limit =(n1)s2X2l=(1001)0.1266.51=0.01488=0.122

99% confidence interval for for the standard deviation of the amount of ketchup in a 12 oz bottle is (0.084, 0.122).

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