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tensa zangetsu [6.8K]
2 years ago
6

The soup produced by a company has a salt level that is normally distributed with a mean of 5.4 grams and a standard deviation o

f 0.3 grams. The company takes readings of every 10th bar off the production line. The reading points are 5.8, 5.9, 4.9, 5.2, 5.0, 4.9, 6.2, 5.1, 5.7, 6.1. Is the process in control or out of control and why?
Mathematics
1 answer:
grigory [225]2 years ago
4 0

Answer:

Step-by-step explanation:

The mean of the reading points is

Mean = (5.8 + 5.9 + 4.9 + 5.2 + 5.0 + 4.9 + 6.2 + 5.1 + 5.7 + 6.1)/10 = 5.48

The process is out of control if the mean salt level of the readings is greater than 5.4

For the null hypothesis,

µ = 5.4

For the alternative hypothesis,

µ > 5.4

This is a right tailed test.

Since the population standard deviation is given, z score would be determined from the normal distribution table. The formula is

z = (x - µ)/(σ/√n)

Where

x = sample mean

µ = population mean

σ = population standard deviation

n = number of samples

From the information given,

µ = 5.4

x = 5.48

σ = 0.3

n = 10

z = (5.48 - 5.4)/(0.3/√10) = 0.84

Looking at the normal distribution table, the probability corresponding to the z score is 0.7996

The probability value to the right of the z score is 1 - 0.7996 = 0.2

Assuming a significance level of 0.05

Since alpha, 0.05 < than the p value, 0.2, then we would fail to reject the null hypothesis. Therefore, At a 5% level of significance, we can conclude that the process is not out of control. If we had rejected the null hypothesis, then our conclusion would be that the process is out of control.

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A doctor at a local hospital is interested in estimating the birth weight of infants. How large a sample must she select if she
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Answer:

The minimum sample size needed is n = (\frac{1.96\sqrt{\sigma}}{4})^2. If n is a decimal number, it is rounded up to the next integer. \sigma is the standard deviation of the population.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.9}{2} = 0.05

Now, we have to find z in the Z-table as such z has a p-value of 1 - \alpha.

That is z with a pvalue of 1 - 0.05 = 0.95, so Z = 1.645.

Now, find the margin of error M as such

M = z\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

How large a sample must she select if she desires to be 90% confident that her estimate is within 4 ounces of the true mean?

A sample of n is needed, and n is found when M = 4. So

M = z\frac{\sigma}{\sqrt{n}}

4 = 1.96\frac{\sigma}{\sqrt{n}}

4\sqrt{n} = 1.96\sqrt{\sigma}

\sqrt{n} = \frac{1.96\sqrt{\sigma}}{4}

(\sqrt{n})^2 = (\frac{1.96\sqrt{\sigma}}{4})^2

n = (\frac{1.96\sqrt{\sigma}}{4})^2

The minimum sample size needed is n = (\frac{1.96\sqrt{\sigma}}{4})^2. If n is a decimal number, it is rounded up to the next integer. \sigma is the standard deviation of the population.

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3 years ago
A popular soft drink is sold in 2-liter (2000 milliliter) bottles. Because of variation in the filling process, bottles have a m
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Answer:

a) 0.27% probability that the mean is less than 1995 milliliters.

b) 2002.3 milliliters or more will occur only 10% of the time for the sample of 100 bottles.

Step-by-step explanation:

To solve this problem, it is important to understand the normal probability distribution and the central limit theorem

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 2000, \sigma = 18

A) If the manufacturer samples 100 bottles, what is the probability that the mean is less than 1995 milliliters?

So n = 100, s = \frac{18}{\sqrt{100}} = 1.8

This probability is the pvalue of Z when X = 1995. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{1995 - 2000}{1.8}

Z = -2.78

Z = -2.78 has a pvalue of 0.0027

So 0.27% probability that the mean is less than 1995 milliliters.

B) What mean overfill or more will occur only 10% of the time for the sample of 100 bottles?

This is the value of X when Z has a pvalue of 1-0.1 = 0.9.

So it is X when Z = 1.28

Z = \frac{X - \mu}{s}

1.28 = \frac{X - 2000}{1.8}

X - 2000 = 1.8*1.28

X = 2002.3

2002.3 milliliters or more will occur only 10% of the time for the sample of 100 bottles.

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