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riadik2000 [5.3K]
2 years ago
5

The table below shows the results of a screening program organized by Level 300 students of the department physiotherapy of the

College Health Sciences of the University of Ghana. Complete the table and answer the questions below it using your understanding of probability and its applications to biomedical data.
True Diagnosis for presence of E. Coli

Test results Disease No Disease Total
Positive 35 15
Negative 10 60
Total

a. What is the efficiency of the test? 3 marks
b. What is the sensitivity of the screening kit? 3 marks
c. What is the specificity of the screening kit? 3 marks
d. What is the predictive value (PPV) of the test? 3 marks
e. What is the negative predictive value (NPV) of the test? 3 marks
(15 marks)
2. In double blinded randomized control trial for hypertensive patients attending Cocoa Clinic, thirty (30) 50-59-year-old were admitted into an intervention program for 6 weeks. During the trial, the average improvement in their systolic blood pressure was 15. The average improvement in systolic blood pressure in the general population of hypertensive patients is 20 with a standard deviation of 2.

i. What are the null and alternative hypotheses in this RCT? (2 marks)
ii. What tail is required in this test? (2 marks)
iii. What is the most appropriate statistical test for this study? (2 marks)
iv. State the assumptions of the test. (2 marks)
v. Test the above hypothesis using the appropriate statistical tool (7 marks)
Critical value =3.6
Mathematics
1 answer:
Vladimir [108]2 years ago
7 0
The correct answer is C because they are asking for the specific screening kit marks
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Suppose the expected tensile strength of type-A steel is 103 ksi and the standard deviation of tensile strength is 7 ksi. For ty
ExtremeBDS [4]

Answer:

a

i So  the approximate distribution of \= X is \mu_{\= X} =103 and  \sigma_{\= X} = 0.783

ii So the approximate distribution of \= Y is \mu_{\= Y} =105 and  \sigma_{\= Y} = 0.645

b

 the approximate distribution of  \=X  - \= Y is E (\= X - \= Y)  = -2 and  \sigma_{\= X  - \=Y}=1.029

Here we can see that the mean of the approximate distribution is negative which tell us that this negative value of the  data for  \=X  - \= Y sample   are more and their frequency occurrence is higher than the positive values  

c

the value of  P(-1 \le \=X - \= Y  \le 1) is = -0.1639    

Step-by-step explanation:

From the question we are given that

       The expected tensile strength of the type A steel is  \mu_A = 103 ksi

        The standard deviation of type A steel is  \sigma_A = 7ksi

         The expected tensile strength of the type B steel is \mu_B = 105\ ksi

            The standard deviation of type B steel is  \sigma_B = 5 \ ksi

Also the assumptions are

       Let \= X be the sample average tensile strength of a random sample of 80 type-A specimens

Here n_a =80

      Let \= Y be  the sample average tensile strength of a random sample of 60 type-B specimens.

  Here n_b = 60

Let the sampling distribution of the mean be

             \mu _ {\= X} = \mu

                   =103

 Let the sampling distribution of the standard deviation be

               \sigma _{\= X} = \frac{\sigma }{\sqrt{n_a} }

                     = \frac{7}{\sqrt{80} }

                    =0.783

So What this mean is that the approximate distribution of \= X is \mu_{\= X} =103 and  \sigma_{\= X} = 0.783

For \= Y

 The sampling distribution of the sample mean is

               \mu_{\= Y} = \mu

                    = 105

  The sampling distribution of the standard deviation is

               \sigma _{\= Y} = \frac{\sigma }{\sqrt{n_b} }

                    = \frac{5}{\sqrt{60} }

                    = 0.645

So What this mean is that the approximate distribution of \= Y is \mu_{\= Y} =105 and  \sigma_{\= Y} = 0.645                      

Now to obtain the approximate distribution for \=X  - \= Y

               E (\= X - \= Y) = E (\= X) - E(\= Y)

                                =  \mu_{\= X} - \mu_{\= Y}

                                = 103 -105

                                = -2

The standard deviation of \=X  - \= Y is

               \sigma_{\= X  - \=Y} = \sqrt{\sigma_{\= X}^2 - \sigma_{\= Y}^2}

                         = \sqrt{(0.783)^2 + (0.645)^2}

                         =1.029

Now to find the value of  P(-1 \le \=X - \= Y  \le 1)

  Let us assume that F = \= X - \= Y

    P(-1 \le F \le 1) = P [\frac{-1 -E (F)}{\sigma_F} \le Z \le  \frac{1-E(F)}{\sigma_F} ]

                             = P[\frac{-1-(-2)}{1.029}  \le  Z \le  \frac{1-(-2)}{1.029} ]

                             =  P[0.972 \le Z \le 2.95]

                             = P(Z \le 0.972) - P(Z \le 2.95)

Using the z-table to obtain their z-score

                             = 0.8345 - 0.9984

                             = -0.1639

                   

3 0
3 years ago
Leah and 15 of her friends want to share 11 cookies equally. What decimal is equivalent to the fraction of a cookie each will ge
mote1985 [20]

Answer:

0.688

Step-by-step explanation:

There are 16 people total and 11 cookies.

This can be represented as:

\frac{16}{11},

which converted into a decimal is:

0.688

8 0
2 years ago
This is the last question need answered so please help me
shusha [124]
Assuming that is a square pyramid

that is the square base+4 triangles
that is
side^2+4(1/2bh)
where b=side=14in
and h=height=13in

surface area=14^2+4(1/2)(14)(13)
SA=196+(2)(182)
SA=196+364
SA=560 square inches
6 0
3 years ago
In a survey of 100 people, 10 preferred onions on their hot dogs. What percent preferred onions?
Aleks [24]
1/10 if you divide 10 by 100 it would be 1/10
4 0
3 years ago
Which is the graph of<br><img src="https://tex.z-dn.net/?f=y%20%3D%20%5Csqrt%5B3%5D%7Bx%20%2B%201%7D%20-%202" id="TexFormula1" t
Anna11 [10]

ANSWER

See attachment.

EXPLANATION

The given function is

y =  \sqrt[3]{x + 1}  - 2

The parent function is

y =  \sqrt[3]{x}

When we shift the parent graph to the left one unit, and down 2 units, we obtain the graph of the given function.

The graph of this function is shown in the attachment.

5 0
3 years ago
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