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olga2289 [7]
3 years ago
12

Granola Crunch cereal is packaged in 1 pound boxes. Susan Torres, a quality control analyst who works for the manufacturer of th

e cereal, wants to check if the packaging process is working improperly: boxes are being over-filled or under-filled. A sample of 40 cereal boxes of Granola Crunch yields a mean weight of 1.02 pounds of cereal per box. Assume that the weight is normally distributed with a population standard deviation of 0.04 pound. Conduct a hypothesis test at the 5% significance level.
Mathematics
1 answer:
sineoko [7]3 years ago
5 0

Answer:

z=\frac{1.02-1}{\frac{0.04}{\sqrt{40}}}=3.162    

p_v =2*P(z>3.162)=0.0016    

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to to reject the null hypothesis, so we can conclude that the true mean is different from 1 pound at 5% of signficance.

Step-by-step explanation:

Data given and notation    

\bar X=1.02 represent the sample mean

\sigma=0.04 represent the population standard deviation    

n=40 sample size    

\mu_o =1 represent the value that we want to test  

\alpha=0.05 represent the significance level for the hypothesis test.    

z would represent the statistic (variable of interest)    

p_v represent the p value for the test (variable of interest)    

State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to check if the true mean is equal to 1 pound or no :    

Null hypothesis:\mu =1    

Alternative hypothesis:\mu \neq 1    

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:    

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)    

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

Calculate the statistic    

We can replace in formula (1) the info given like this:    

z=\frac{1.02-1}{\frac{0.04}{\sqrt{40}}}=3.162    

P-value    

Since is a two-sided test the p value would be:    

p_v =2*P(z>3.162)=0.0016    

Conclusion    

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to to reject the null hypothesis, so we can conclude that the true mean is different from 1 pound at 5% of signficance.    

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zimovet [89]

Answer:

CLASS     FREQUENCIES     RELATIVE FREQUENCIES

A                        60                                 0.5

B                        12                                  0.1

C                        48                                 0.4

TOTAL              120                                  1

Step-by-step explanation:

Given that;

the frequencies of there alternatives are;

Frequency A = 60

Frequency B = 12

Frequency C = 48

Total = 60 + 12 + 48 = 120

Now to determine our relative frequency, we divide each frequency by the total sum of the given frequencies;

Relative Frequency A = Frequency A / total = 60 / 120 = 0.5

Relative Frequency B = Frequency B / total = 12 / 120 = 0.1

Relative Frequency C = Frequency C / total = 48 / 120 = 0.4

therefore;

CLASS     FREQUENCIES     RELATIVE FREQUENCIES

A                        60                                 0.5

B                        12                                  0.1

C                        48                                 0.4

TOTAL              120                                  1

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Step-by-step explanation:

The domain of a log is the baseline is greater than 0.

Set x + 3 greater than zero and solve

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Answer:

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Step-by-step explanation:

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y = - \frac{1}{3} x + 4 ← is i slope- intercept form

with slope m = - \frac{1}{3}

Parallel lines have equal slopes, thus

y = - \frac{1}{3} x + c ← is the partial equation

To find c substitute (6, 5) into the partial equation

5 = - 2 + c ⇒ c = 5 + 2 = 7

y = - \frac{1}{3} x + 7 ← equation of parallel line

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Murljashka [212]

I got the answer 18.79

I believe this is correct, Please tell me if im wrong. :)

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