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jok3333 [9.3K]
3 years ago
9

he labor force participation rate is the number of people in the labor force divided by the number of people in the country who

are of working age and not institutionalized. The BLS reported in February 2012 that the labor force participation rate in the United States was 63.7% (Calculatedrisk). A marketing company asks 120 working-age people if they either have a job or are looking for a job, or, in other words, whether they are in the labor force. What is the probability that fewer than 60% of those surveyed are members of the labor force?
Mathematics
1 answer:
tangare [24]3 years ago
8 0

Answer:

z= \frac{0.6-0.637}{0.0439} =-0.843

So then we can find the probability like this:

P(p

And using the normal standard table or excel we got:

P(p

Step-by-step explanation:

For this case we can check if we can use the normal approximation for the proportion and we have this:

np = 120*0.637 =76.44 >10

n(1-p) = 120*(1-0.637) = 43.56>10

Then we can conclude that we can use the normal approximation. And we have this:

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

So the mean is given by:

\mu_p = 0.637

And the deviation is given by:

\sigma_p = \sqrt{\frac{0.637*(1-0.637)}{120}}= 0.0439

And for this case we want to find this probability:

P( p

And we can use the z score given by:

z = \frac{p -\mu}{\sigma_p}

And for this case the z score is:

z= \frac{0.6-0.637}{0.0439} =-0.843

So then we can find the probability like this:

P(p

And using the normal standard table or excel we got:

P(p

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Step-by-step explanation:

17.33 h

2438.1/17.33 = 140.68

+140.68 m/h

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Create a set of five lengths so that:
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1, 2, 3, 7, 10

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A researcher claims that the variation in the salaries of elementary school teachers is greater than the variation in the salari
guajiro [1.7K]

Answer:

a

The null hypothesis is  H_o :  \sigma^2_1 = \sigma^2 _2

The alternative hypothesis is H_a :  \sigma_1 ^2 > \sigma^2_2

b

F_{critical} = 1.8608

c

F = 2.9085

d

    The decision rule is  

Reject the null hypothesis

e

There is sufficient evidence to support the researchers claim

Step-by-step explanation:

From the question we are told that

 The first sample size is  n_1 = 30

 The sample variance for elementary school is  s^2_1 = 8324

 The second sample size is  n_2 = 30

  The sample variance for the secondary school is  s^2_2 = 2862

   The significance level is  \alpha = 0.05

The null hypothesis is  H_o :  \sigma^2_1 = \sigma^2 _2

The alternative hypothesis is H_a :  \sigma_1 ^2 > \sigma^2_2

Generally from the F statistics table  the critical value of \alpha = 0.05 at first and  second degree of freedom df_1 = n_1 - 1 = 30 - 1 = 29 and  df_2 = n_2 - 1 = 30 - 1 = 29 is  

         F_{critical} = 1.8608

Generally the test statistics is mathematically represented as

       F = \frac{s_1^2 }{s_2^2}

=>   F = \frac{8324 }{2862}

=>   F = 2.9085

Generally from the value obtained we see that  F >  F_{critical } Hence

   The decision rule is  

Reject the null hypothesis

    The conclusion is  

  There is sufficient evidence to support the researchers claim

   

4 0
3 years ago
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