I'm studying for my test and am reviewing the solution to some example problems. The problem is:

You are told that a new standardized test is given to 100 randomly selected third grade students in New Jersey. The sample average score is $\overline Y$ on the test is 58 and the sample standard deviation is $s_y$ = 8. The first part of the question asks you to construct a 95% confidence interval for the mean score of all New Jersey third graders.

First I found $SE_y$ = $\frac{s_y}{\sqrt{n}} = \frac{8}{\sqrt{10}} = 0.8.$

Then I did

$58 \pm (0.8)(1.64)$

However, instead of using $1.64$ they used $1.96$. Can someone explain why?


$1.96$ is used because the $95\%$ confidence interval has only $2.5\%$ on each side. The probability for a $z$ score below $-1.96$ is $2.5\%$, and similarly for a $z$ score above $+1.96$; added together this is $5\%$. $1.64$ would be correct for a $90\%$ confidence interval, as the two sides ($5\%$ each) add up to $10\%$.


To Find a critical value for a 90% confidence level.

Step 1: Subtract the confidence level from 100% to find the α level: 100% – 90% = 10%.

Step 2: Convert Step 1 to a decimal: 10% = 0.10.

Step 3: Divide Step 2 by 2 (this is called “α/2”). 0.10 = 0.05. This is the area in each tail.

Step 4: Subtract Step 3 from 1 (because we want the area in the middle, not the area in the tail): 1 – 0.05 = .95.

Step 5: Look up the area from Step in the z-table. The area is at z=1.645. This is your critical value for a confidence level of 90%.


hope this helps


Two reasons: 1) Students are in New Jersey.

2) the $z$-value associated with 95% is z=1.96; this can be seen as part of the 1-2-3, 65-95-99 rule that tells you that values within $ p.m 1-, 2- or 3$- deviations from the mean comprise 65- 95- or 99% of all values in the distribution . Or look it up in a table.


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