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stepan [7]
3 years ago
8

In a completely randomized experimental design involving five treatments, 13 observations were recorded for each of the five tre

atments (a total of 65 observations). The following information is provided.
SSTR = 200 (Sum Square Between Treatments)
SST = 800 (Total Sum Square)


Refer to Exhibit 13-4. The number of degrees of freedom corresponding to within treatments is:

a.60

b.59

c.5

d.4
Mathematics
1 answer:
WINSTONCH [101]3 years ago
3 0

Answer:

The number of degrees of freedom corresponding to within treatments is (a) 60.

Step-by-step explanation:

The number of treatments involved in the experiment is, <em>k</em> = 5.

Each treatment has 13 observations recorded.

The total number of observations is, 5 × 13 =65

The degrees of freedom for between treatment and within treatment are:

df_{between}=k-1\\df_{within} =N-k\\

Compute the degrees of freedom for within treatment as follows:

df_{within} =N-k\\=65-5\\=60

Thus, the degrees of freedom for within treatment are 60.

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The Nelson Company makes the machines that automatically dispense soft drinks into cups. Many national fast food chains such as
Vika [28.1K]

Answer:

a) There is a 2.28% probability that a new cup will overflow when filled by the automatic dispenser.

b) The mean amount dispensed by the machine should be set at 16.14 ounces to satisfy this wish.

Step-by-step explanation:

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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.

In this problem, we have that:

Normal model with mean 16 ounces and standard deviation 0.31 ounces. This means that \mu = 16, \sigma = 0.31.

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a. What is the probability that a new cup will overflow when filled by the automatic dispenser?

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

Z = \frac{16.62 - 16}{0.31}

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Z = 2 has a pvalue of 0.9772. This means that there is a 1-0.9772 = 0.0228 = 2.28% probability that a new cup will overflow when filled by the automatic dispenser.

b. The company wishes to adjust the dispenser so that the probability that a new cup will overflow is .006. At what value should the mean amount dispensed by the machine be set to satisfy this wish?

This is the value of \mu, with X = 16.62 when Z has a pvalue of 0.94. It is between Z = 1.55 and Z = 1.56, so we use Z = 1.555.

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

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\mu = 16.62 - 0.31*1.555

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