Statistics Cheat Sheet

From the above output, we can see that the p-value is 0.000186, which is smaller than 0.05 (if we select a 0.05 significance level).

The area under the curve to the left of the unknown quantity must be 0.7 (70%). So, we must first find the z value that cuts off an area of 0.7 in the left tail of standard normal distribution. Using the cumulative probability table, we see that z=0.53.

2. In order to determine the average amount spent in November on Amazon.com a random sample of 144 Amazon accounts were selected. The sample mean amount spent in November was $250 with a standard deviation of $25. Assuming that the population standard deviation is unknown, what is a 95% confidence interval for the population mean amount spent on Amazon.com in November?

(c) The mean of the sample and the value of Z with an area of 5% in the left tail.

of 175 pounds and a standard deviation of 10 pounds.. Find the probability, P(W < 190), that

The area under a normal curve with mu = 35 and sigma = 7 is 0, 1, or 2?

If the total sample size is over 15, two sample t tests are safe if there are no

An intelligence test for which the scores are normally distributed has a mean of 100 and a standard deviation of 15. Use this information to describe how the scores are distributed.

18. Suppose that the scores of architects on a particular creativity test are normally distributed. Using a normal curve table, what percentage of architects have Z scores:

3. In one elementary school, 200 students are tested on the subject of Math and English. The table below shows the mean and standard deviation for each subject.

2. For the following set of scores, fill in the cells. The mean is 74.13 and the standard deviation is 9.98.

H_a: μ 30 we will use a z-test for the mean to test the given hypothesis.

“Hypothesis testing is a decision-making process for evaluating claims about a population” (Bluman, 2013, p. 398). This process is used to determine if you will accept or reject the hypothesis. The claim is that the bottles contain less than 16 ounces. The null hypothesis is the soda bottles contain 16 ounces. The alternative hypothesis is the bottles contain less than 16 ounces. The significance level will be 0.05. The test method to be used is a t-score. The test statistic is calculated to be -11.24666539 and the P-value is 1.0.  The P-value is the probability of observing a sample statistic as extreme as the test statistic, assuming the null hypothesis is true. The T Crit value is 1.69912702. The calculations show there is enough evidence to support the claim that the soda bottles do

At the .01 significance level is there a difference in the mean amount purchased on an impulse at the two stores? Explain these results to a person who knows about the t test for a single sample but is unfamiliar with the t test for independent means.

s squared subscript p=(n1-1)s squared2 subscript1+(n2-1)s squared2 subscript2]/n1+n2-2. The numerator of the function, n1+n2-2, is the degrees of freedom.