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Komok [63]
3 years ago
15

5. The superintendent of the local school district claims that the children in her district are brighter, on average, than the g

eneral population. To determine the aptitude of her district's children, a study was conducted. The results of her district's test scores were: 105, 109, 115, 112, 124, 115, 103, 110, 125, 99. If the mean of the general population of school children is 106, what could be said about her claim? Use alpha = .05
Mathematics
1 answer:
anygoal [31]3 years ago
4 0

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:

\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}  

where x_i are data points, \bar{x} is the mean and n is the number of observations.  

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{1117}{10} = 111.7

Sum of squares of differences = 642.1

S.D = \sqrt{\frac{642.1}{49}} = 8.44

We are given the following in the question:  

Population mean, μ = 106

Sample mean, \bar{x} = 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

H_{0}: \mu = 106\\H_A: \mu > 106

We use one-tailed(right) t test to perform this hypothesis.

Formula:

t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }

Putting all the values, we have

t_{stat} = \displaystyle\frac{111.7 - 106}{\frac{8.44}{\sqrt{10}} } = 2.135

Now,

t_{critical} \text{ at 0.05 level of significance, 9 degree of freedom } = 1.833

Since,                  

t_{stat} > t_{critical}

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.

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