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11111nata11111 [884]
2 years ago
8

1. A waste management service attempts to design routes so that each of their trucks pick-up on average four tons of garbage or

less. A garbage collector believes, however, that he averages picking up more than four tons of garbage per day and decides to perform a hypothesis test. If the hypothesis test is performed at a 5% significance level and the resulting p-value is 0.04. Your conclusion should be:
2. It has been determined with 95% confidence that the proportion of on-line students at NYU who live in Brooklyn is between 0.73 and 0.77. Determine the sample proportion of on-line NYU students who live in Brooklyn.
3. Assume a normal distribution and use a hypothesis test to test the given claim.

According to city reports, it was found that the mean age of the prison population in the city was 26 years. Marc wants to test the claim that the mean age of the prison population in his city is less than 26 years. He obtains a random sample of 25 prisoners, and finds a mean age of 24.4 years and a standard deviation of 9.2 years. At a significance level of 0.05, what should his conclusion be?
Mathematics
1 answer:
IRINA_888 [86]2 years ago
6 0

Answer:

1.-Then we p-value indicates that we are in the rejection region we reject H₀

We support the claim of the garbage collector the average picking up more than 4 tn of garbage

2.- p  = 75 %

3.-3.-t(s) is in the rejection region we accept H₀ we have not evidence to support Marc´s claim

Step-by-step explanation:

1.- If p-value is 0,04   and significance level α = 5 %  or  α = 0,05 then p-value < α

Test hypothesis should be    ( x the average of garbage)      

Null hypothesis          H₀           x = 4 Tn

Alternative Hypothesis     Hₐ    x  > 4 Tn

Then  alternative hypothesis suggests a one tail-test to the right  and

p-value <  0,05

Then we p-value indicates that we are in the rejection region we reject H₀

We support the claim of the garbage collector the average picking up more than 4 tn of garbage

2.- As we are dealing with a normal distribution the CI  95 % is symmetrical with respect to the mean, therefore the proportion of student living in Brooklyn is:  

(0,73 + 0,77) /2

p = 0,75     p  = 75 %

Test hypothesis:

Null Hypothesis            H₀            μ    =  26

Alternative Hypothesis      Hₐ      μ   <  26

Alternative hypothesis tells us the test is a one-tail test to the left

Sample size   n = 25

Sample mean   μ  =  24,4

Sample Standard deviation  =  9,2

We assume normal distribution, and as n < 30 we use t-student table

with  24 degree of freedom

Significance level is 0,05  and df = 24  we find t (c)  in t- student table

t(c) = 1,7109      test to the left   t(c) = -1,7109

To calculate  t(s)

t(s)  =  (  24,4 - 26 ) / 9,2 / √25

t(s) =  - 1,6 * 5 / 9,2

t(s) = - 0,87

Comparing t(s)  and t(c) we have

t(s) > t(c)

t(s) is in the rejection region we accept H₀ we have not evidence to support Marc´s claim

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Based on a sample of 100 employees a 95% confidence interval is calculated for the mean age of all employees at a large firm. Th
babymother [125]

Answer:

a

\= x  =  40.85

b

E = 5.85

Ca  

   t_c =  2.08

Cb

   t_c  =   1.282

Explanation:

From the question we are told that

     The sample size is n  =  100

     The upper limit of the 95% confidence interval is  b =  47.2 years

     The lower limit of the 95% confidence interval is   a =  34.5 years

Generally the sample mean is mathematically represented as

         \= x  =  \frac{a + b }{2}

=>    \= x = \frac{47.2 + 34.5 }{2}

=>    \= x  =  40.85

Generally the margin of error is mathematically represented as

         E =  \frac{b- a }{ 2}

=>      E =  \frac{47.2- 34.5 }{ 2}

=>      E = 5.85

Considering question C a  

From the question we are told the confidence level is  90% , hence the level of significance is    

      \alpha = (100 - 90 ) \%

=>   \alpha = 0.10

The  sample size is  n =  22

Given that the sample size is not sufficient enough i.e n < 30 we will make use of the student t distribution table  

Generally the degree of freedom is mathematically represented as

           df =  n- 1

=>        df =  22 - 1

=>        df =  21

Generally from the student  t  distribution table the critical value  of  \frac{\alpha }{2} at a degree of freedom  of  21 is  

   t_c =t_{\frac{\alpha }{2} ,  21  } =  2.08

Considering question C b

From the question we are told the confidence level is  80% , hence the level of significance is    

      \alpha = (100 - 80 ) \%

=>   \alpha = 0.20

Generally from the normal distribution table the critical value  of  \frac{\alpha }{2} is  

   t_c  =Z_{\frac{\alpha }{2} } =  1.282

4 0
2 years ago
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