Sample Size

this study is the use of convenience sampling, as previously mentioned. Black et al. (2000) make

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.

(b) The value of Z with an area of 5% in the right tail, but not the sample mean.

12. _____ For a given population, confidence intervals constructed from larger samples tend to be narrower than those constructed from smaller samples. Which statement below best describes why this is true? (A) The variability of the sample mean is less for larger samples. (B) The z-value for larger samples tends to be more accurate. (C) The population variance is larger for large populations. (D) As the sample size increases, the z-value (or t-value) becomes smaller. A machine dispenses potato chips into bags that are advertised as containing one pound of product. To be on the safe side, the machine is supposed to be calibrated to dispense 16.07 ounces per bag, and from long time observation, the distribution of the fill-weights is known to be approximately normal and the process is known to have a standard deviation of 0.15 ounces.

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:

standard deviation standardized value rescaling z-score normal model parameter statistic standard Normal model 68-95-99.7 Rule normal probability plot

I found this information by taking the Z score and looking at the Z table under the number curve, located in appendix of my textbook.

Consider the following scenario in answering questions 5 through 7. In an article appearing in Today’s Health a writer states that the average number of calories in a serving of popcorn is 75. To determine if the average number of calories in a serving of popcorn is different from 75, a nutritionist selected a random sample of 20 servings of popcorn and computed the sample mean number of calories per serving to be 78 with a sample standard deviation of 7.

The critical value for significance level α=0.05 for an upper-tailed z-test is given as 1.645.

guidelines that allow for bias and additional issues in ensuring a proper count of the

9. Assume you want to estimate with the proportion of students who commute less than 5 miles to work within 2%, what sample size would you need?

In order for you to understand how I did this, I will explain that a z-score gives you a way to compare different sets of data using their standard deviations and averages. In statistics, standard deviation is a measure that is used to quantify the amount of variation or dispersion of a set of data values. A standard deviation close to zero tells you the data points are close to the mean or average. Having a high standard deviation shows data points are spread out over a wider range of values. If x is a data point in a normal distribution, then the equation for the z-score of x is x equals x minus the average all divided by the standard deviation. Normal distribution is just a function that represents the distribution of many random variables as a symmetrical bell-shaped graph.

A field researcher is gathering data on the trunk diameters of mature pine and spruce trees in a certain area. The following are the results of his random sampling. Can he conclude, at the .10 level of significance, that

5) From calculations, computed z value is more than -1.65 and falls within Ho not rejected region. Ho is not rejected at α = 0.05 & α = 0.01 significance levels.

As a production manager, Michelle York was disappointed with this conclusion and decided to consult a professor on the best solution to this dilemma, after futile efforts to establish a sampling technique that would yield the desired improvement and results. Previously, the samples were carried under different settings since the company had to freeze on capital expenditures of a significant amount, and still, the replacement would have cost many times that amount. The professor’s advice was that the company should have used smaller sample sizes and taken more samples. After conferring with the professor, the company took twenty seven samples of five observations each. In their second sampling, the company got results with a sample mean range that indicated less deviation.