Question

In: Statistics and Probability

Assume that we would like to test at significance level 0.01 whether there is enough evidence...

Assume that we would like to test at significance level 0.01 whether there is enough evidence to claim that average height of children by the end of age three in families with low-socioeconomic status is less than the general average height for this age, which is 94 cm. Assume that height measurements by the end of age three follow a normal distribution with standard deviation 6 cm.

(a) Write the null and alternative hypothesis for the hypothesis testing procedure that may potentially lead to proving the claim in the question described.

(b) For this study, assume that 9 children (from families with low socioeconomic status )will be chosen to test the null paired with the appropriate alternative hypothesis. Assume that this test is run twice each time with a different sample data set. For the first sample the mean was measured as 92.0, and the second sample mean was 94.2. State the definition of p-value, and using this definition compute the p-value for each outcome. ie. p-value of 92.0 and p-value of 94.2. Which of the two values provides stronger evidence in support of the claim about the height of children born to families with low socio-economic status?

Solutions

Expert Solution

Ho :   µ =   94
Ha :   µ <   94

Level of Significance ,    α =    0.01                  
population std dev ,    σ =    6.0000                  
Sample Size ,   n =    9                  
Sample Mean,    x̅ =   92.0000                  
                          
'   '   '                  
                          
Standard Error , SE = σ/√n =   6.0000   / √    9   =   2.0000      
Z-test statistic= (x̅ - µ )/SE = (   92.000   -   94   ) /    2.0000   =   -1.00
                                
                          
p-Value   =   0.1587   [ Excel formula =NORMSDIST(z) ]              
Decision:   p-value>α, Do not reject null hypothesis

.......

population std dev ,    σ =    6.0000                  
Sample Size ,   n =    9                  
Sample Mean,    x̅ =   94.2000                  
  
Standard Error , SE = σ/√n =   6.0000   / √    9   =   2.0000      
Z-test statistic= (x̅ - µ )/SE = (   94.200   -   94   ) /    2.0000   =   0.10

p-Value   =   0.5398

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

mean value with 92 provides stronger evidence in support of the claim about the height of children born to families with low socio-economic status

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

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