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amid [387]
2 years ago
13

A woman's weekly salary is increased from $350 to $380. The percent of increase is most nearly

Mathematics
1 answer:
Kay [80]2 years ago
3 0

Find the amount of the difference and then divide it by the original amount:


380 - 350 = 30


30 / 350 = 0.0857 = 8.6% increase.  Round the answer as needed.



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A liquid dietary product implies in its advertising that use of the product for one month results in an average weight loss of a
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Answer:

Following are the responses to the given question:

Step-by-step explanation:

Please find the table in the attached file.

mean and standard deviation difference: \bar{d}=\frac{\Sigma d}{n} =\frac{-4-6-.......-4-4}{8}=-4.125 \\\\S_d=\sqrt{\frac{\Sigma (d-\bar{d})^2 }{n-1}}=\sqrt{\frac{(-4 + 4.125)^2 +.......+(-4 +4.125)^2 }{8-1}}= 1.246

For point a:

hypotheses are:

H_0 : \mu_d \geq -3\\\\H_a : \mu_d < -3\\\\

degree of freedom:

df=n-1=8-1=7

 From t table, at\alpha = 0.05, reject null hypothesis if t.

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t=\frac{\bar{d}-\mu_d }{\frac{s_d}{\sqrt{d}}}=\frac{ -4.125- (-3)}{\frac{1.246}{ \sqrt{8}}} =-2.55

because the t=-2.553, removing the null assumption. Data promotes a food product manufacturer's assertion with a likelihood of Type 1 error of 0.05.

For point b:

From t table, at \alpha =0.01, removing the null hypothesis if t.

because t=-2.553 >-2.908, fail to removing the null hypothesis.  

The data do not help the foodstuff producer's point with the likelihood of a .01-type mistake.

For point c:

Hypotheses are:

H_0: \mu_d \geq -5\\\\H_a: \mu_d < -5

Degree of freedom:

df=n-1=8-1=7

From t table, at \alpha =0.05, removing the null hypothesis if t.

test statistic:  t=\frac{\bar{d}-\mu_d}{\frac{s_d}{\sqrt{n}}} =\frac{-4.125-(-5)}{\frac{1.246}{\sqrt{8}}}=1.986

Since t-1.986 >-1.895, The null hypothesis fails to reject. The results do not support the packaged food producer's claim with a Type 1 error probability of 0,05.

From t table, at\alpha= 0.01, reject null hypothesis ift.

Since t=1.986>-2.998 , fail to reject null hypothesis.  

Data do not support the claim of the producer of the dietary product with the probability of Type 1 error of .01.

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