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snow_tiger [21]
3 years ago
4

State study on labor reported that one-third of full-time teachers in the state also worked part-time at another job. For those

teachers, the average number of hours worked per week at the part-time job was 13. After an increase in state teacher salaries, a random sample of 400 teachers who worked part-time at another job was selected. The average number of hours worked per week at the part-time job for the teachers in the sample was 12.5 with a standard deviation of 6.5 hours. Is there convincing statistical evidence, at the level of α=0.05, that the average number of hours worked per week at part-time jobs decreased after the salary increase?
(A) No. The p-value of the appropriate test is greater than 0.05.
(B) No. The p-value of the appropriate test is less than 0.05.
(C) Yes. The p-value of the appropriate test is greater than 0.05.
(D) Yes. The p-value of the appropriate test is less than 0.05.
(E) Not enough information is given to determine whether there is convincing statistical evidence
answer a show work, please
Mathematics
1 answer:
VLD [36.1K]3 years ago
7 0

Answer: a) no. The p value of the appropriate test is greater than 0.05

Step-by-step explanation:

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sergeinik [125]

Answer:

I think so

Step-by-step explanation:

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AB corresponds to EF

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3 years ago
Plz help me!!!!!i need to get this correct plz help!!!
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Answer:

Yes

Step-by-step explanation:

Yes they are because if you divide 18 by 3 you get six and if you divide 15 by 3 you get 5, so it simplifies to 6:3 which is the same as the ratio for rings

3 0
3 years ago
At NC State University, 16.4% of the undergraduate classes have more than 50 students. If a random sample of 200 undergraduate c
Lena [83]

Answer:

We need to check the conditions in order to use the normal approximation.

np=200*0.164=32.8 \geq 10

n(1-p)=200*(1-0.164)=167.2 \geq 10

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

The mean is given by:

p =0.164

And the deviation is given by:

\sigma_{p}= \sqrt{\frac{0.164*(1-0.164)}{200}}= 0.0262

So then the correct options for this case are:

The sampling distribution will be approximately normal.

The mean of the sampling distribution will be 16.4%.

The standard deviation of the sampling distribution will be 0.0262.

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=200, p=0.164)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

We need to check the conditions in order to use the normal approximation.

np=200*0.164=32.8 \geq 10

n(1-p)=200*(1-0.164)=167.2 \geq 10

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

The mean is given by:

p =0.164

And the deviation is given by:

\sigma_{p}= \sqrt{\frac{0.164*(1-0.164)}{200}}= 0.0262

So then the correct options for this case are:

The sampling distribution will be approximately normal.

The mean of the sampling distribution will be 16.4%.

The standard deviation of the sampling distribution will be 0.0262.

6 0
3 years ago
A school surveyed its
Alinara [238K]

Answer:

12.5%

Step-by-step explanation:

6 0
3 years ago
Gabriel has 268 miniture trains.he lines them up in 2 equal rows. how many trains are in each row
Gwar [14]

Answer:

134

Step-by-step explanation:

268/2=134

since its equal both rows should have the same number.

3 0
3 years ago
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