Mathematics, 24.12.2019 02:31 2021arabellacorsino

Abottled water distributor wants to estimate the amount of water contained in 1 - gallon bottles purchased from a nationally known water bottling company. the water bottling company's specifications state that the standard deviation of the amount of water is equal to 0.02 gallon. a random sample of 50 bottles is selected, and the sample mean of the water per 1 - gallon bottle is 0.995 gallon.

a. construct a 99% confidence interval estimate for the population mean of water included in a 1 - gallon bottle.

b. on the basis of these results, do you think that the distributor has a right to complain to the water bottling company? why?

c. must you assume that the population amount of water per bottle is normally distributed here? explain.

d. construct a 95% confidence interval estimate. how does this change your answer to (b)?

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Answer from: Lingz

a) $0.995 - 2.58 \frac{0.02}{\sqrt{50}}=0.988$

$0.995 + 2.58 \frac{0.02}{\sqrt{50}}=1.00$

The 99% confidence interval is given by (0.988;1.00)

b) NO, because a 1-gallon of bottle contain exactly 1-gallon of water and this value is inside of the 99% confidence interval.

c) Yes we are assuming this (normality) in order to construct the confidence interval.

d) $0.995 - 1.96 \frac{0.02}{\sqrt{50}}=0.989$

$0.995 + 1.96 \frac{0.02}{\sqrt{50}}=1.00$

The 95% confidence interval is given by (0.989;1.00)

The confidence interval not changes too much, we got very similar answers.

Step-by-step explanation:

1) Notation and definitions

n=50 represent the sample size

$\bar X=0.995$ represent the sample mean

s=0.02 represent the sample standard deviation

m represent the margin of error

Confidence =99% or 0.99

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Part a

In order to find the critical value is important to mention that we know about the population standard deviation, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ .

We can find the critical values in excel using the following formulas:

"=NORM.INV(0.005,0,1)" for $z_{\alpha/2}=-2.58$

"=NORM.INV(1-0.005,0,1)" for $z_{1-\alpha/2}=2.58$

The critical value $zc=\pm 2.58$

The interval for the mean is given by this formula:

$\bar X \pm z_{c} \frac{\sigma}{\sqrt{n}}$

And calculating the limits we got:

$0.995 - 2.58 \frac{0.02}{\sqrt{50}}=0.988$

$0.995 + 2.58 \frac{0.02}{\sqrt{50}}=1.00$

The 99% confidence interval is given by (0.988;1.00)

Part b

NO, because a 1-gallon of bottle contain exactly 1-gallon of water and this value is inside of the 99% confidence interval.

Part c

Yes we are assuming this (normality) in order to construct the confidence interval.

Part d

In order to find the critical value is important to mention that we know about the population standard deviation, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ .

We can find the critical values in excel using the following formulas:

"=NORM.INV(0.025,0,1)" for $z_{\alpha/2}=-1.96$

"=NORM.INV(1-0.025,0,1)" for $z_{1-\alpha/2}=1.96$

$0.995 - 1.96 \frac{0.02}{\sqrt{50}}=0.989$

$0.995 + 1.96 \frac{0.02}{\sqrt{50}}=1.00$

The 95% confidence interval is given by (0.989;1.00)

The confidence interval not changes too much, we got very similar answers.

Answer from: Quest

That’s what i got when i looked up what the solutions were

Answer from: Quest

1/800 is the right answer

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