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In a study on the effect of aluminum intake on the mental development of infants, a group of 54 infants 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 was 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 test to determine if infants who received the aluminum-depleted intravenous-feeding solution had a significantly different average BMDI score than in the general population. Be sure to include all the steps for full credit.

Given: Critical value = 2.00

Answer :

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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