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Salsk061 [2.6K]
2 years ago
7

A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 430 gram setting. It

is believed that the machine is underfilling the bags. A 21 bag sample had a mean of 421 grams with a standard deviation of 15. Assume the population is normally distributed. A level of significance of 0.1 will be used. Find the P-value of the test statistic. You may write the P-value as a range using interval notation,
Mathematics
1 answer:
Zanzabum2 years ago
6 0

Answer: We reject H₀  we find that the machine is underflling the bottles

P ( 421 ± 4,3376 )

Step-by-step explanation:

We assume a normal distribution

Population mean 430 grs

Unknown standard deviation

We have a one tail condition "underfilling"

And our test is:

Null hypothesis      H₀         X  =  μ₀

Alternative hypothesis    Hₐ     X  < μ₀

We must use t student distribution and find the interval

X ± t*(s/√n)

In that expession   X is the sample mean  421 grs, "s"  is sample standard deviation, n is sample size, then

421 ±  t * ( 15 / √21 )          (1)

We go to t table and look for t value for  α = 0,1 and df = 21 - 1   df = 20

we get t (remember it is a one tail test)  t = 1,325, plugging this value in equation (1) we get the interval

421 ± 1,325 * ( 15/√21 )    ⇒  421 ±  1,325 * ( 3,2737)

421 ± 4,3376  

421 + 4,3376  = 425,34

421 - 4,3376  = 416,66

As we can see the mean value of the population 430 grs is not inside the interval  [ 416,66 ;  425,34 ] then we can assure the machine is underfilling the bags, and not meeting the setting spec

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Answer:

The ballonist is at a height of 3579.91 ft above the ground at 3:30pm.

Step-by-step explanation:

Let's call:

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a_2 the horizontal distance between the second observer and the ballonist

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S=\frac{h^{2} }{\tan(\alpha _1)}+\frac{h^{2} }{\tan(\alpha _2)}\\S=h^{2}*(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})

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h=\frac{a}{(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})}\\h=\frac{7220 ft}{(\frac{1 }{\tan(35.6)}+\frac{1}{\tan(58.2)})}\\h=3579,91 ft

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