Question

Human measurements provide a rich area of application for statistical methods. In a longitudinal study the development of elementary school children explored the relationship between IQ and private speech. Private speech is considered to be when children talk to themselves. It was thought that private speech would be related to IQ, because IQ is supposed to measure mental maturity, and it was known that private speech decreases as students progress through the primary grades. The study included 34 students whose first-grade IQ scores are given below. Use the R code to calculate the sample summary statistics, histogram and boxplot.

IQ.data = c(78, 91, 99, 102, 103, 103, 106, 107, 107, 107, 108, 108, 109, 109, 110, 110, 112, 112, 112, 113, 114, 114, 118, 118, 125, 125, 126, 127, 127, 132, 136, 140, 140, 151)

Answer 1

> summary(IQ.data)

And we got the following result

Min. 1st Qu. Median Mean 3rd Qu. Max.

78.0 107.0 112.0 114.7 125.0 151.0

For this case we have this:
`Min = 78, Q_1 = 107.0, Median =112, Q_3 = 125, Max= 151`

The interquartile range would be:

`IQR = Q_3 -Q_1 = 125-107 = 18`

For the histogram we can use this code:

> hist(IQ.data)

And the result is on the first figure attached and we can see that the distribution is not symmetrical and a little skewed to the left

And for the boxplot we can use this code:

boxplot(IQ.data)

And we see on this figure a presence od one outlier the 78 for this case.

Explanation:

For this case we have the following data:

IQ.data = c(78, 91, 99, 102, 103, 103, 106, 107, 107, 107, 108, 108, 109, 109, 110, 110, 112, 112, 112, 113, 114, 114, 118, 118, 125, 125, 126, 127, 127, 132, 136, 140, 140, 151)

We want to calculate the sample summary statistics and we can use the following code:

> summary(IQ.data)

And we got the following result

Min. 1st Qu. Median Mean 3rd Qu. Max.

78.0 107.0 112.0 114.7 125.0 151.0

For this case we have this:
`Min = 78, Q_1 = 107.0, Median =112, Q_3 = 125, Max= 151`

The interquartile range would be:

`IQR = Q_3 -Q_1 = 125-107 = 18`

For the histogram we can use this code:

> hist(IQ.data)

And the result is on the first figure attached and we can see that the distribution is not symmetrical and a little skewed to the left

And for the boxplot we can use this code:

boxplot(IQ.data)

And we see on this figure a presence od one outlier the 78 for this case.