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Umnica [9.8K]
2 years ago
15

ANSWER ASAP PLEASE!!!

Mathematics
1 answer:
Dmitrij [34]2 years ago
6 0
The answer is D. <span>No, it is not a valid inference because his classmates do not make up a random sample of the students in the school.</span>
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Clark Heter is an industrial engineer at Lyons Products. He would like to determine whether there are more units produced on the
larisa [96]

Answer:

No, at the 0.05 significance level, the number of units produced on the night shift is not larger.

Step-by-step explanation:

We are given that the mean number of units produced by a sample of 54 day-shift workers was 345. The mean number of units produced by a sample of 60 night-shift workers was 351.

Assume the population standard deviation of the number of units produced on the day shift is 21 and 28 on the night shift.

Let \mu_1 = population mean number of units produced on the day shift

      \mu_2 = population mean number of units produced on the night shift

So, <u>Null Hypothesis</u>, H_0 : \mu_1-\mu_2\geq0  or  \mu_1\geq\mu_2    {means that the mean number of units produced on the night shift is same or lesser on the day shift}

<u>Alternate Hypothesis,</u> H_A : \mu_1-\mu_2  or  \mu_1    {means that the mean number of units produced on the night shift is larger}

The test statistics that will be used here is <u>Two-sample z test statistics</u> as we know about population standard deviations;

              T.S.  =  \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma_1^{2} }{n_1}+\frac{\sigma_2^{2} }{n_2} } }  ~ N(0,1)

where, \bar X_1 = sample mean number of units produced by a sample of 54 day-shift workers = 345

        \bar X_2 = sample mean number of units produced by a sample of 60 night-shift workers = 351

       \sigma_1  = population standard deviation of the number of units produced on the day shift = 21

        \sigma_2 = population standard deviation of the number of units produced on the day shift = 28

        n_1 = sample of day-shift workers = 54

        n_2 = sample of night-shift workers = 60

So, <em><u>test statistics</u></em>  =  \frac{(345-351)-(0)}{\sqrt{\frac{21^{2} }{54}+\frac{28^{2} }{60} } }

                              =  -1.302

Now at 0.05 significance level, the z table gives critical value of -1.6449 for left-tailed test. Since our test statistics is more than the critical value of z as -1.302 > -1.6449 so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the mean number of units produced on the night shift is same or lesser than those produced on the day shift.

4 0
2 years ago
Sarah and her friends are going blueberry picking. Sarah picks $b$ blueberries. Michael picks $3$ more than half as many as Sara
yKpoI14uk [10]
The correct answer is b=3+1/2 divided by 5+3 after that +10-3*5*3 . add the answers up and thats the solution
6 0
2 years ago
Which congruence theorem can be used to prove △MLQ ≅ △NPQ?<br><br> AAS <br> SSS <br> ASA <br> SAS
storchak [24]
Answer: SSS

Proof:
In ΔMLQ and ΔNPQ,
MQ = NQ (given) S
Since Q is the midpoint of LP, by definition, LQ = QP (S)
LM = PN (given) S

∴ ΔMLQ ≡ ΔNPQ (SSS)
4 0
2 years ago
Read 2 more answers
Point M and N are midpoints of AB and DA respectively.
Fed [463]
<span>triangle CDN ≅ triangle CBM
</span><span>
Statement | Reason
line DC ≅ __BC___ | The sides of a square are congruent.
line BM ≅ __DN___ | They are both half the length of a side of the square.
angle __D___ ≅ angle B | All the angles of a square are congruent.
triangle CDN ≅ CBM | 
by
</span><span>SAS</span>
5 0
2 years ago
The quality control team of a company checked 800 digital cameras for defects. The team found that 20 cameras had lens defects,
dmitriy555 [2]

Answer:

0.025

Step-by-step explanation:

-This is a conditional probability problem.

-Let L denote lens defect and C charging defect.

#We first calculate the probability of a camera having a lens defect;

P(lens)=\frac{Lens}{Total}\\\\=\frac{20}{800}\\\\=0.025

#Calculate the probability of a camera having a charging defect:

P(Charging)=\frac{Charging}{Total}\\\\=\frac{25}{800}\\\\=0.03125

The  the probability that a camera has a lens defect given that it has a charging defect is calculated as:

P(L|C)=\frac{P(C)P(L)}{P(C)}\\\\=\frac{0.025\times 0.03125}{0.03125}\\\\=0.025

Hence,  the probability that a camera has a lens defect given that it has a charging defect is 0.025

6 0
2 years ago
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