The proportions of blood types O, A, B, and AB in the general population of a particular country are known to be in the ratio 49:38:9:4, respectively. A research team, investigating a small isolated community in the country, obtained the following frequencies (i. e. observed number of cases) of blood type. Blood type O A B AB
Frequency 87 59 20 4
Test the hypothesis that the proportions in this community do not differ significantly from those in the general population.

The proportions differ significantly from those in the general population.

Step-by-step explanation:

We state our null and alternative hypotheses as

H0: p1= 49/100, p2= 38/100, p3= 9/100 and p4= 4/100 for a multinomial distribution involving four categories and n= 170

against

Ha: pi≠ pi0 for at least one value of i= 1,2,3,4

The significance level is set at ∝= 0.05

The test statistic under H0 is

χ² = ∑ (Oi- ei)²/ei

which has approximate chi square distribution with 3d.f  (n-1)

Computations: Under H0 the expected frequencies are

np10= Blood type O = 170 * 49/100= 83.3

np20 = Blood type A = 170 * 38/100= 64.6

np30 = Blood type B = 170 * 20/100= 11.76

np40 = Blood type AB = 170 * 4/100= 2.35

The value of χ² is computed as follows

Cell       Observed         Estimated        (Oi-ei)            (Oi-ei)²   (Oi-ei)²/ei

             Frequency      Frequency  

               Oi                    ei

1             87                        83.3               3.7               13.69        0.157

2            59                      64.6              -5.6                 31.36       0.531

3             20                     34                  -14                   196           9.8

4               4                      6.8                -2.8                  7.84         1.96      

∑             170                    188.7                                                  χ²= 12.43  

The critical region is χ²≥ χ²(0.05,3) = 7.82 for alpha = 0.05 and

χ²≥ χ²(0.01,3) = 11.34  for alpha = 0.01

As the calculated value lies in the critical region for both value of alpha we reject our null hypothesis and accept our alternate hypothesis. The proportions differ significantly from those in the general population.

hello,

b) 62.47% of the students are between 14 and 18 years old.

step-by-step explanation:

mean = 15 and standard deviation = 2

student between 14 to 18

formula used z-score

for age 14 , z score

using z-table , p(z> -0.5)=0.3085

for age 18 , z score

using z-table , p(z< 1.5)=0.9332

p(-0.5< z< 1.5)=0.9332-0.3085 = 0.6247

thus, 62.47% of the students are between 14 and 18 years old.

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