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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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The table gives the relative frequencies of recipes that contains sugar and salt, contain at least one of those ingredients, or
tigry1 [53]

<u>Answer-</u>

<em>The probability that a randomly selected recipe does not contain sugar, given that it contains salt is 22.4%</em>

<u>Solution-</u>

The given table in the link shows the relative frequencies of recipes that contains sugar and salt, or contains at least one of those ingredients, or contains neither of those ingredients.

We have to find the conditional probability that the recipe doesn't contain sugar, given that it contains salt.

We know that, the conditional probability of occurrence of A given that B occurs is,

P(A|B)=\frac{P(A\ and\ B)}{P(B)} =\frac{P(A\bigcap B)}{P(B)}

P(\text{Doesn't contain sugar}\ |\ \text{Contains salt})=\frac{P(\text{Doesn't contain sugar}\ and\ \text{Contains salt})}{P(\text{Contains salt})}


P(\text{Doesn't contain sugar}\ and\ \text{Contains salt})=\frac{0.15}{1} =0.15\\P(\text{Contains salt})=\frac{0.67}{1}=0.67

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