Answer:
We conclude that children in district are brighter, on average, than the general population.
Step-by-step explanation:
We are given the following data set:
105, 109, 115, 112, 124, 115, 103, 110, 125, 99
Formula:
[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]
where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.
[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]
[tex]Mean =\displaystyle\frac{1117}{10} = 111.7[/tex]
Sum of squares of differences = 642.1
[tex]S.D = \sqrt{\frac{642.1}{49}} = 8.44[/tex]
We are given the following in the question:
Population mean, μ = 106
Sample mean, [tex]\bar{x}[/tex] = 111.7
Sample size, n = 10
Alpha, α = 0.05
Sample standard deviation, s = 8.44
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 106\\H_A: \mu > 106[/tex]
We use one-tailed(right) t test to perform this hypothesis.
Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]t_{stat} = \displaystyle\frac{111.7 - 106}{\frac{8.44}{\sqrt{10}} } = 2.135[/tex]
Now,
[tex]t_{critical} \text{ at 0.05 level of significance, 9 degree of freedom } = 1.833[/tex]
Since,
[tex]t_{stat} > t_{critical}[/tex]
We fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.
We conclude that children in district are brighter, on average, than the general population.
