Question

# 1. Suppose we suspected an unusual distribution of blood groups in patients undergoing a certain surgical...

1. Suppose we suspected an unusual distribution of blood groups in patients undergoing a certain surgical procedure. The data below gives the results for 100 patient’s blood type:

O         A         B       AB

35        25        30        10

We want to test the hypothesis that the population served by the hospital that performs the surgery is 30% group O, 20% group A, 40% group B, and 10% group AB. Test at an alpha level of 0.05. Find the test statistic.

a 3.68

b 1.98

c 9.49

d 4.58

2. Use the following set of observed frequencies to test the independence of the two variables. Variable one has values of 'A' and 'B'; variable two has values of 'C', 'D', and 'E'.

 C D E A 12 10 8 B 20 24 26

Using a = 0.05, the estimates of the expected frequency in row 1 (A) column 1 (C) when the two variables are independent is _______.

a 9.6

b 12

c 16

d 10

3. For the following ANOVA table, the df Treatment value is ___________.

 Source of Variation SS df MS F Treatment 150 Error 40 20 Total 23

a 43

b 3

c 1.15

d 460

4. A researcher is interested in using a chi-square test of independence to determine if age is independent of minutes spent reading. Age is divided into four categories while minutes spent reading is classified as high, medium, low. The number of degrees of freedom for this test is 12

a true

b false

5. Each person in a random sample of 54 students living in the dorms at a large university was asked whether he or she preferred to study in the dorm room, in the dorm’s study lounge, or at the university library. The resulting data is summarized in the table below.

 Dorm Room Study Lounge Library 14 17 23

In a hypothesis test to determine if there is evidence that the three locations are not equally preferred by students at the university who live in the dorms, what is the expected count for the library category?

a. 54

b. 18

c. 20

d. 12

e. 10

Que.1

Observation table:

 Oi Pi Ei=N*Pi (Oi-Ei)2/Ei 35 0.3 30 0.8333333 25 0.2 20 1.25 30 0.4 40 2.5 10 0.1 10 0 N = 100 4.5833333

Test statistic, Que.2

Observation table:

 C D E Total A 12 10 8 30 B 20 24 26 70 Total 32 34 34 100 Que.3

df (treatment) = df(Total) - df(error) = 23 - 20 = 3

Que.4

False.

Degrees of freedom = (r-1) (c-1) = (4-1) (3-1) = 6

Que.5

Expected count = N *P(library) = 54 *(1/3) = 18

Where N = 14 +17 +23 = 54

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