I am told to find the z value corresponding to the confidence interval level of 80.3% for the population mean where the sample size is n = 40. I am looking at the formula and it seems like I need the standard deviation. How would I go on about solving this without a standard deviation?
[Math] finding z value for confidence interval without standard deviation
statistics
Related Solutions
Hint: Let random variables $\bar{X}$ and $\bar{Y}$ be sample means from random samples drawn from our two distributions. Let $W=\bar{X}-\bar{Y}$. Then $W$ has mean $\mu_1-\mu_2$. So it is an unbiased estimator of $\mu_1-\mu_2$.
The random variable $W$ has normal distribution. The variance of $W$ is $\frac{25}{25}+\frac{9}{36}$.
The notion of confidence interval is somewhat intuitive, but that may be keeping you from understanding what it means in more depth.
Say I have multiple samples $x_i$ from a population, and I wish to estimate the population mean $\mu$. A CI of, say, 95\% represents an interval of possible values of $\mu$ such that given my samples, the "probability" that the $\mu$ lies in that interval is 95%.
We immediately see that there can be more than one such interval, since I could trade probability past the upper end for probability at the lower end of the interval, thus shifting the interval. Let's skirt that issue by demanding a symmetric interval about my sample mean.
But the "probability" is not well defined from the information I just presented!
In order to assign a probability, I have to make some assumptions about the population. The usual assumption is that the population variance is equal to the unbiased estimator of variance obtained from our sample. But we still have things backward: We can't honestly talk about the probability of the population mean being in some range, without any assumption about the a priori (before I saw my samples) probabilities of the mean being various values.
So we apply the usual sleight-of-mind logic employed by the frequentist point of view. We ask:
Given that the population variance is our unbiased sample variance estimate, What are the highest and lowest values of the population mean $\mu$ such that the chance of or sample being as far away from $\mu$ as it is, is lower than 100%-95% = 5%.
Now let's go back to your problem. Since the population is finite, as you draw more samples (without replacement) you actually do learn something about the population. If you had drawn all the objects but one, if you take your sample unbiased variance as the population variance, your 95% confidence interval for the value of that remaining one object would be roughly $2\sigma$ but your estimate of the population mean will have a variance of $\sigma/N$. This is quite a bit smaller than would be the case for an infinite population or a small sampling of a large population.
Now when you draw that last sample, you know everything about the distribution. In particular, you know the mean exactly. Therefore any interval that includes the actual mean is a 100% CI. If you then say that the real CI is the tightest such interval, then it has width zero.
Best Answer
The $z$-value is a normalized value and is unrelated to the sample size or the s.d.
I'm going to suppose you are using a 2-tail Z-test.
We know that $$2P(Z\ge z)=1-0.803$$ Since $Z\sim N(0,1)$,we have $$z\approx1.29015$$