i need help calculating levels of confidence:?

Question

i need 89% confidence.

how do i come up with the z-score? i know, 10.5 on one side and 10.5 on the other. ...but what is the corrosponding z-score that i need and how do i calculate it?

i know that pascal. ...but for example at 95% confidence you would use 1.96.

so the 95% is between -1.96 and 1.96. ....1.96 being z scores.


i think thats all the info i can give this is confusing for me.

wow, pascal youre a machine!

no wonder no one could give me a good answer that looks pretty complicated! thanks for the table.

...ill try and remember to mark you as a best answer!

Answer 1

The z score is the number of standard deviations away from the mean your score is. So for a score of x, the corresponding z-score is given by (x-μ)/σ, where μ is the mean and σ the standard deviation.

Edit: Ah, I see, you mean that you want to find out what z-score corresponds to a given confidence interval. My mistake.

The standard normal distribution (with μ=0 and σ=1) is given by e^(-x²/2)/√(2π). The probability that the z-score falls between a and b is given by the integral of this function from a to b (the √(2π) factor ensures that the integral of this function from -∞ to ∞ is equal to 1). Thus, to compute the 89% confidence interval, you would need to find a such that [-a, a]∫e^(-x²/2)/√(2π) dx = 0.89. Unfortunately, this can only be done numerically, so in most cases one refers to a precomputed table such as this one (http://www.digitalreview.com.ar/normaldistribution/ ). If you have 11% outside your confidence interval, then 5.5% of it will appear to the left of the interval and 5.5% to the right. All told, you are looking for the z-score such that the cumulative distribution is 94.5%, or 0.945. As you can see from the table, that happens at a z-score of 1.6, so your 89% confidence interval would be from -1.6 to 1.6

Edit 2: Thanks for the vote of confidence. That said, if I was a machine I wouldn't make mistakes, but the first time I typed this I accidentally gave the normal distribution as e^(x²/2)/√(2π) instead of the correct e^(-x²/2)/√(2π) (I missed the minus in front of the x²). I've corrected it above. Good luck with the rest of your studies.

Answer 2

Pascal is on the right track. There are tables available that give you the z-score corresponding to a "level of confidence". You need to be careful about what you are talking about. Apparently you are talking about a normal distribution curve. The z-score tells you the amount of the distribution within a range of z-values. For instance, a z=1 would include the center 68 percent of the distribution, z=2, the center 95 percent, and so-on. Thus, your 10.5 on one side etc. is really the bounds that include 79 per-cent of the distribution, and not an 89 % confidence. For an 89 percent inclusion, two-sided, you would have to find the z corresponding to 5.5 % of the distribution.