Question

A study claims that girls and boys do not do equally well on math tests taken from the 2nd to 11th grades (Chicago Tribune, July 25, 2008). Suppose in a representative sample, 344 of 430 girls and 369 of 450 boys score at proficient or advanced levels on a standardized math test. (You may find it useful to reference the appropriate table: z table or t table)

Let p₁ represent the population proportion of girls and p₂ the population proportion of boys.

a. Construct the 95% confidence interval for the difference between the population proportions of girls and boys who score at proficient or advanced levels. (Negative values should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places and final answers to 2 decimal places.)

b. Select the appropriate null and alternative hypotheses to test whether the proportion of girls who score at proficient or advanced levels differs from the proportion of boys.

• H₀: p₁ − p₂ = 0; H_A: p₁ − p₂ ≠ 0

• H₀: p₁ − p₂ ≤ 0; H_A: p₁ − p₂ > 0

• H₀: p₁ − p₂ ≥ 0; H_A: p₁ − p₂ < 0

c. At the 5% significance level, what is the conclusion to the test? Do the results support the study’s claim?

• Reject H₀; the study's claim is supported by the sample data.

• Reject H₀; the study's claim is not supported by the sample data.

• Do not reject H₀; the study's claim is supported by the sample data.

• Do not reject H₀; the study's claim is not supported by the sample data.

Answer 1

sample #1   -----> GIRL
first sample size,     n1=   430          
number of successes, sample 1 =     x1=   344          
proportion success of sample 1 , p̂1=   x1/n1=   0.8000          
                  
sample #2   ----->   BOY
second sample size,     n2 =    450          
number of successes, sample 2 =     x2 =    369          
proportion success of sample 1 , p̂ 2=   x2/n2 =    0.820          
                  
difference in sample proportions, p̂1 - p̂2 =     0.8000   -   0.8200   =   -0.0200

level of significance, α =   0.05              
Z critical value =   Z α/2 =    1.960   [excel function: =normsinv(α/2)      
                  
Std error , SE =    SQRT(p̂1 * (1 - p̂1)/n1 + p̂2 * (1-p̂2)/n2) =     0.0265          
margin of error , E = Z*SE =    1.960   *   0.0265   =   0.0519
                  
confidence interval is                   
lower limit = (p̂1 - p̂2) - E =    -0.020   -   0.0519   =   -0.0719
upper limit = (p̂1 - p̂2) + E =    -0.020   +   0.0519   =   0.0319
                  
so, confidence interval is (   -0.08 < p1 - p2 <   0.03 )  

.................

B)

H0: p1 − p2 = 0; HA: p1 − p2 ≠ 0

..............

C)

• Do not reject H₀; the study's claim is not supported by the sample data.

.............

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