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maks197457 [2]
3 years ago
5

According to the Bureau of Labor Statistics, citizens remain unemployed for an average of 15.9 weeks before finding their next j

ob (June, 2008). Suppose you want to show that Louisiana has been effective in getting their unemployed back to work sooner. You take a random sample of 50 citizens who were unemployed six months earlier and ask them to report the duration. You find that the average time spent unemployed was 13.4 weeks. Which of the following statements is the correct alternative hypothesis?
a. -2.64
b. -2.32
c. -2.11
d. -1.28
e. none of these are correct
Mathematics
1 answer:
Ahat [919]3 years ago
7 0

Complete Question

According to the Bureau of Labor Statistics, citizens remain unemployed for an average of 15.9 weeks before finding their next job (June, 2008). Suppose you want to show that Louisiana has been effective in getting their unemployed back to work sooner. You take a random sample of 50 citizens who were unemployed six months earlier and ask them to report the duration. You find that the average time spent unemployed was 13.4 weeks with a sample standard deviation of the time unemployed is 6.7 weeks.

1 Which of the following statements is the correct alternative hypothesis?

2 The test statistic for testing the hypothesis is

a. -2.64

b. -2.32

c. -2.11

d. -1.28

e. none of these are correct

Answer:

1  

 The  alternative hypothesis  H_a  :  \mu < 15.9

2

The  test statistics z =  -2.64  

Step-by-step explanation:

From the question we are told that

     The population  mean value for time  citizens remain unemployed is  \mu  =  15.9 \  weeks

     The  sample size is  n =  50

     The  sample standard deviation is  6.7 weeks.

      The  sample mean value for time  citizens remain unemployed is  \mu  =  15.9 \  weeks

       

     The  null hypothesis is  H_o  :  \mu \ge 15.9

      The  alternative hypothesis  H_a  :  \mu < 15.9

Generally test statistics is mathematically represented as

       z =  \frac{ \= x  - \mu }{ \frac{s}{\sqrt{n} } }  

=>    z =  \frac{13.4 - 15.9}{\frac{6.7}{\sqrt{50}}}

=>     z =  -2.64  

   

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Answer:

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Step-by-step explanation:

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That is z with a pvalue of 1 - 0.025 = 0.975, so Z = 1.96.

Now, find the margin of error M as such

M = z\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

Assume that the standard deviation in the amount of caffeine in 8 ounces of decaf coffee is known to be 2 mg.

This means that \sigma = 2

If we wanted to estimate the true mean amount of caffeine in 8 ounce cups of decaf coffee to within /- 0.5 mg, how large a sample would we have to take to achieve this result?

We would need a sample of n.

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Dividing both sides by 0.5

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