Question

An SRS of 350 high school seniors gained an average of ¯ x = 21 points in their second attempt at the SAT Mathematics exam. Assume that the change in score has a Normal distribution with standard deviation σ = 52 . (a) Find a 99 % confidence interval for the mean change in score μ in the population of all high school seniors. (Enter your answers rounded to two decimal places.) lower bound of confidence interval: upper bound of confidence interval:

Answer 1

Answer:

Upper bound: 26.45

Lower bound: 15.55            

Step-by-step explanation:

We are given the following in the question:

Sample mean, $\bar{x}$ = 21

Sample size, n = 350

Alpha, α = 0.05

Population standard deviation, σ = 52

95% Confidence interval:

$\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}$

Putting the values, we get,

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

$21 \pm 1.96(\dfrac{52}{\sqrt{350}} ) = 21 \pm 5.45 = (15.55,26.45)$

Upper bound: 26.45

Lower bound: 15.55

Answer 2

The 99% confidence interval for the mean change in score is between 13.84 points to 28.16 points

How to calculate confidence interval

The z score of  99% confidence interval is 2.576

The margin of error (E) is:

$E = Z_\frac{\alpha }{2} *\frac{standard\ deviation}{\sqrt{sample\ size} } =2.576*\frac{52}{\sqrt{350} } =7.16$

The confidence interval = mean ± E = 21 ± 7.16 = (13.84, 28.16)

The 99% confidence interval for the mean change in score is between 13.84 points to 28.16 points

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