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Mashcka [7]
2 years ago
5

The business college computing center wants to determine the proportion of business students who have personal computers (PCʹs)

at home. If the proportion exceeds 25%, then the lab will scale back a proposed enlargement of its facilities. Suppose 200 business students were randomly sampled and 65 have PCʹs at home. Find the rejection region for this test using α = 0.01. (Note that this is a right-tailed test).
A) Reject H0 if 0 z > 2.33.
B) Reject H0 if 0 z < -2.33.
C) Reject H0 if 0 z > 2.575 or z < -2.575.
D) Reject H0 if 0 z = 2.
Find the test statistic, 0 t , to test the claim about the population mean μ < 6.7 given n = 20, x = 6.3, s = 2.0.
A) -1.233
B) -0.872
C) -1.265
D) -0.894
Determine the test statistic, 0 z , to test the claim about the population proportion p < 0.85 given n = 60 and x = 39.
A) -4.34
B) -1.96
C) -1.85
D) -1.76
Mathematics
1 answer:
mojhsa [17]2 years ago
5 0

Answer:

Question  5

    correct option is  A

Question 6

    correct option is  D

Question 7  

      correct option is  A  

Step-by-step explanation:

Considering question 5

From the question we are told that

     The population proportion considered is  p  =  0.25

      The sample size is  n  =  200

      The number that had a personal computer at home is  k  =  65

      The  level of significance is  \alpha  = 0.01

The null hypothesis is  H_o :  p =  0.25

The alternative hypothesis is  H_a :  p > 0.25

Generally from the z-table the critical value of  \alpha  = 0.01  to the right of the curve is

          z_{\alpha } =  2.33

Generally given that it is a right-tailed test , the rejection region is  

      z > 2.33

Considering question 6

  The sample size is  n = 20

  The standard deviation is  s = 2

   The sample mean is  \=x  =  6.3

   The  population mean   \mu =  6.7

Generally the test statistics is mathematically represented as

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

=>       t =  \frac{ 6.3 -  6.7  }{ \frac{ 2}{ \sqrt{20} }  }

=>       t  = -0.894

Considering question 7

  The sample size is  n = 60

   The sample mean is  x = 39

   The  population proportion    p = 0.85

Gnerally the sample proportion is mathematically represented as  

           \^ p  =  \frac{x}{n}

=>       \^ p  =  \frac{39}{60}  

=>       \^ p  =  0.65

Generally the standard error of this distribution is mathematically represented as

     SE =  \sqrt{ \frac{ p(1 - p ) }{ n } }

=> SE =  \sqrt{ \frac{  0.85 (1 - 0.85  ) }{ 60  } }

=> SE =  0.0461

Generally the test statistics is mathematically represented as

          z  =  \frac{ \^  p  -  p   }{SE}

=>       z =  \frac{ 0.65 - 0.85  }{0.0461}

=>       z  = -4.34

         

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