You cannot from the information provided.
A recent survey of 50 executives who were laid off from their previous position revealed it took a mean of 26 weeks for them to find anoher positio. the standard deviation of the sample was 6.2 weeks. construct a 95 % confidence interval for the population. Is it reasonable that the population mean is 28 weeks? Justify your answer
a) T or F The sampling distribution will be normal. Explain your answer. b) Find the mean and standard deviation of the sampling distribution. c) We pick one of our samples from the sampling distribution what is the probability that this sample has a mean that is greater than 109 ? Is this a usual or unusual event? these are the rest of the question.
The standard error of the mean and sampling error are two similar but still very different things. In order to find some statistical information about a group that is extremely large, you are often only able to look into a small group called a sample. In order to gain some insight into the reliability of your sample, you have to look at its standard deviation. Standard deviation in general tells you spread out or variable your data is. If you have a low standard deviation, that means your data is very close together with little variability. The standard deviation of the mean is calculated by dividing the standard deviation of the sample by the square root of the number of things in the sample. What this essentially tells you is how certain are that your sample accurately describes the entire group. A low standard error of the mean implies a very high accuracy. While the standard error of the mean just gives a sense for how far you are away from a true value, the sampling error gives you the exact value of the error by subtracting the value calculated for the sample from the value for the entire group. However, since it is often hard to find a value for an entire large group, this exact calculation is often impossible, while the standard error of the mean can always be found.
You are studying the sample because you want to find out information about the whole population. If the sample you have drawn from the population does not represent the population, you will find out about the sample but will not find out about the population.
You're an idiot. It's standard deviation. Google that for your answer.
Can someone help me find the answer for a sample n=36 with a population mean of of 76 and a mean of 79.4 with a standard deviation of 18?
You cannot from the information provided.
Information is not sufficient to find mean deviation and standard deviation.
A recent survey of 50 executives who were laid off from their previous position revealed it took a mean of 26 weeks for them to find anoher positio. the standard deviation of the sample was 6.2 weeks. construct a 95 % confidence interval for the population. Is it reasonable that the population mean is 28 weeks? Justify your answer
You calculate standard deviation the same way as always. You find the mean, and then you sum the squares of the deviations of the samples from the means, divide by N-1, and then take the square root. This has nothing to do with whether you have a normal distribution or not. This is how you calculate sample standard deviation, where the mean is determined along with the standard deviation, and the N-1 factor represents the loss of a degree of freedom in doing so. If you knew the mean a priori, you could calculate standard deviation of the sample, and only use N, instead of N-1.
we calculate standard deviation to find the avg of the difference of all values from mean.,
No, you have it backwards, the standard deviation is the square root of the variance, so the variance is the standard deviation squared. Usually you find the variance first, as it is the average sum of squares of the distribution, and then find the standard deviation by squaring it.
The proportion is approx 95%.
To calculate variance, (population standard deviation squared), usually written as σ2, first find the mean. In this case, ( 3 + 5 + 6 + 4 + 7 + 7 +8 ) / 7 = 40/7 ≈ 5.7143 Then find the difference between every number and the mean. 5.7143 - 3 = 2.7143 5.7143 - 5 = 0.7143 5.7143 - 6 = -0.2857 5.7143 - 4 = 1.7143 5.7143 - 7 = -1.2857 5.7143 - 7 = -1.2857 5.7143 - 8 = -2.2857 Then square all of the differences. 7.3674 0.5102 0.0816 2.9388 1.6530 1.6530 5.2244 Then add the products together. 19.4284 Then divide by the number of numbers in the set of data (7) for the variance. σ2 ≈ 2.78 Since you are using the entire population, you use the population standard deviation instead of the sample standard deviation. In statistics, you are more likely to see sample standard deviation, but since there are only seven numbers in the data set, we use the population standard deviation.
You cannot because the median of a distribution is not related to its standard deviation.
a) T or F The sampling distribution will be normal. Explain your answer. b) Find the mean and standard deviation of the sampling distribution. c) We pick one of our samples from the sampling distribution what is the probability that this sample has a mean that is greater than 109 ? Is this a usual or unusual event? these are the rest of the question.