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Agata [3.3K]
2 years ago
7

Defatted blackcurrant seeds (DBS) could be used as a dietary supplement in gluten-free bread but their use may affect texture. A

2011 study measured bread volume (in cm) using independent random samples of bread loaves baked with flours containing different DBS levels. The data showed no skew or outliers. Here are the summary statistics: Flour type z Mean StDev No DBS 00 552 18 5% DBS 00 525 23 15% DBS 00 485 24 We want to know if bread volume is significantly affected by the type of flour used.
Which inference procedure should be used?
1. chi-square for two-way tables
2. one sample or matched-pairs t procedure for a mean
3. z procedure for a proportion
4. two sample t procedure for two means
5. ANOVA for several means
Mathematics
1 answer:
dalvyx [7]2 years ago
5 0

Note: This is the correct table format.

Flour type                   z                      Mean                            StDev

No DBS                       00                      552                                 18

5% DBS                       00                      525                                 23

15% DBS                     00                      485                                  24

Answer:

5. ANOVA for several means

Step-by-step explanation:

In this question, three different means of the data are compared. ANOVA is used for comparing between two or more means to test for differences. It extends the z and t test that are only used for comparing two means.

Since three means are compared, the inference procedure to be used is ANOVA.

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stiv31 [10]
A. 70°

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Where 180 - 110 = 70

110+ 70 = 180
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2 years ago
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A practice law exam has 100 questions, each with 5 possible choices. A student took the exam and received 13 out of 100.If the s
Cloud [144]

Answer:

z=\frac{13-20}{4}=-1.75

Assuming:

H0: \mu \geq 20

H1: \mu

p_v = P(Z

Step-by-step explanation:

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest (number of correct answers in the test), on this case we now that:

X \sim Binom(n=100, p=0.2)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

We need to check the conditions in order to use the normal approximation.

np=100*0.2=20 \geq 10

n(1-p)=20*(1-0.2)=16 \geq 10

So we see that we satisfy the conditions and then we can apply the approximation.

If we appply the approximation the new mean and standard deviation are:

E(X)=np=100*0.2=20

\sigma=\sqrt{np(1-p)}=\sqrt{100*0.2(1-0.2)}=4

So we can approximate the random variable X like this:

X\sim N(\mu =20, \sigma=4)

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  The letter \phi(b) is used to denote the cumulative area for a b quantile on the normal standard distribution, or in other words: \phi(b)=P(z

The z score is given by this formula:

z=\frac{x-\mu}{\sigma}

If we replace we got:

z=\frac{13-20}{4}=-1.75

Let's assume that we conduct the following test:

H0: \mu \geq 20

H1: \mu

We want to check is the score for the student is significantly less than the expected value using random guessing.

So on this case since we have the statistic we can calculate the p value on this way:

p_v = P(Z

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Step-by-step explanation:

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