1. For a and b, conduct every step of the Hypothesis testing process (1-5).The distribution of SAT scores is normal with M = 500, with a standard deviation alpha = 100. Mable believes that she scored significantly higher than average, at x = 643. Did Mable scores significantly higher than the average SAT score?A sample of n = 30 adults is taken from the above population and their average SAT score is measured to be M = 507. Does this sample differ from the population?

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Description: 1. For a and b, conduct every step of the Hypothesis testing process (1-5).The distribution of SAT scores is normal with M = 500, with a standard deviation alpha = 100. Mable believes that she scored significantly higher than average, at x = 643. Did

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MATLAB: An Introduction with Applications

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Author:Amos Gilat

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1. For a and b, conduct every step of the Hypothesis testing process (1-5).

The distribution of SAT scores is normal with M = 500, with a standard deviation alpha = 100. Mable believes that she scored significantly higher than average, at x = 643. Did Mable scores significantly higher than the average SAT score?
A sample of n = 30 adults is taken from the above population and their average SAT score is measured to be M = 507. Does this sample differ from the population?

Expert Solution

Step 1

1. The null and alternative hypothesis is given by;

$H_{0} : μ = 500 H_{a} : μ > 500 L e t t h e l e v e l o f s i g n i f i c a n c e b e α = 0.05 G i v e n; x = 643 n = 1 σ = 100 T h e t e s t s t a t i s t i c i s g i v e n b y; z = \frac{\sqrt{n} (x - μ)}{σ} = \frac{\sqrt{1} (643 - 500)}{100} = 1.43 T h e c r i t i c a l v a l u e i s g i v e n b y; z_{α} = z_{0.05} W e u s e E x c e l t o f i n d t h e c r i t i c a l v a l u e;$

Hence the critical value is;

z_0.05 = 1.645 [Since Normal Distribution is symmetric]

Here;

z < z_0.05

We accept the null hypothesis. Hence Mable did not score significantly higher than the average SAT score.

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