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AfilCa [17]
3 years ago
8

A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 439.0 gram setting.

It is believed that the machine is underfilling the bags. A 47 bag sample had a mean of 433.0 grams. A level of significance of 0.05 will be used. Determine the decision rule. Assume the standard deviation is known to be 21.0. Enter the decision rule.
Mathematics
1 answer:
Vlada [557]3 years ago
3 0
<h2>Answer with explanation:</h2>

Let \mu be the average weight of chocolate chips in each bag.

As per given , we have

H_0: \mu =439.0\\\\ H_a: \mu

Since H_a is left-tailed , so we perform a left-tailed test.

Also, the population standard deviation is known \sigm=21.0.

So we use z-test.

Test statistic : z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqt{n}}}

Substitute given value ,\overline{x}=433.0\ , n=47,\ \sigma=21 \ , \mu=439

z=\dfrac{433-439}{\dfrac{21}{\sqrt{47}}}\approx-1.96

Significance level = 0.05

Decision rule : Reject H_0 when p-value < 0.05.

By z-table , P- value for left-tailed test  = P(z<-1.96)= 1- P(z<1.96)

[∵ P(Z<-z)=1-P(Z<z)]

= 1- 0.9750

=0.025

Since , P-value(0.025) < 0.05 , so we reject the null hypothesis.

We support the claim that the machine is underfilling the bags at 0.05 significance level .

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In microbiology, colony-forming units (CFUs) are used to measure the number of microorganisms present in a sample. To determine
zimovet [89]

Answer:

(a)  0.2650

(b)  0.0111

(c)  0.0105

(d)  0.0006

Step-by-step explanation:

Given that:

In microbiology, colony-forming units (CFUs) are used to measure the number of microorganisms present in a sample.

Suppose that the number of CFUs that appear after incubation follows a Poisson distribution with mean μ = 15.       &;

If the area of the agar plate is 75cm²;

what is the probability  of observing fewer than 4 CFUs in a 25 cm² area of the plate.

We can determine the mean number of CFUs that appear on a 25cm² area of the plate as follows;

75cm²/25cm² = 3

Since;

mean  μ = 15  

mean number of CFUs that appear on a 25cm² = 15/3 = 5 CFUs

Thus ; the probability of observing fewer than 4 CFUs in a 25 cm² area of the plate is estimated as:

= P(X < 4)

Using the EXCEL FUNCTION ( = poisson.dist(3, 5, TRUE) )

we have ;

P(X < 4) = 0.2650

b) If you were to count the total number of CFUs in 5 plates, what is the probability you would observe more than 95 CFUs?

Given that the total number of CFUs = 5 plates; then the mean number of CFUs in 5 plates =  15×5 = 75 CFUs

The probability is therefore = P( X > 95 )

= 1 - P(X ≤ 95)

= 1 - poisson.dist(95,75,TRUE) ( by using the excel function)

= 0.0111

c) Repeat the probability calculation in part (b) but now use the normal approximation.

Let assume that the mean and the variance of the poisson distribution are equal

Then;

X \sim N (\mu = 75 , \sigma^2 = 75)

We are to repeated the probability calculation in part (b) from above;

So:

P( X > 95 )

use the normal approximation

From standard normal variable table:

P(Z > 2.3094)

Using normal table

P(Z > 2.3094) = 0.0105

(d)  Find the difference between this value and your answer in part (b).

So;

the difference between the value in part c and part b is;

=  0.0111 - 0.0105

= 6*10^{-4}

= 0.0006 to four decimal places

5 0
3 years ago
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