In a study of the effect of aluminum intake on the mental development of infants, a group of 54infants who had been born prematurely were given a special aluminum depleted intravenous-feeding solution. At an adjusted age of 18 months,the neurologic development of the infants were measured using the Bayley Mental Development Index (BMDI), a unitless scale with higher scores indicating higher development. They had a sample mean BMDI of 97.4 and a sample standard deviation of 18.0.The (BMDI) is similar to an IQ score, with 100 being the average in the general population.

Assuming all relevant assumptions are met, carry out the appropriate testto determine if infants who receivedthealuminum depleted intravenous-feeding solution had a significantly differentaverage BMDI score thaninthe general population. Be sure to include all the stepsfor full credit.

Given:critical value= = 2.00

Answer:

The decision rule is  

  Fail to reject the null hypothesis

The conclusion is

   The no sufficient evidence to conclude  that the average  BMDI score is different from the general population

Step-by-step explanation:

From the question we are told that

    The  sample size is  n  =  54

    The sample mean is  [tex]\= x  =  97.4[/tex]

    The standard deviation is  [tex]s = 18.0[/tex]

     The population mean is  [tex]\mu  =  100[/tex]

     The critical value  is  [tex]t_{critical } =  2.0[/tex]

Generally the test statistics is mathematically represented as  

       [tex]t =  \frac{\= x - \mu}{\frac{s}{\sqrt{n} } }[/tex]

=>    [tex]t =  \frac{ 97.4  - 100}{\frac{ 18}{\sqrt{54} } }[/tex]

=>    [tex]t =-1.062[/tex]

Comparing the obtained t statistics with the critical value we see that

   The  critical value is not within the region of rejection (i.e -1.062 to 1.062  )

Hence the decision rule is  

  Fail to reject the null hypothesis

The conclusion is

   The no sufficient evidence to conclude  that the average  BMDI score is different from the general population

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