Sample Size Determination Of The Nairobi Mombasa Road Accident Black Spots Within Makueni County

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.

In a normal distribution with mu = 34 and sigma = 5 what number corresponds to z = -4?

Slide 15: Here’s the same curve demonstrating the z-score values for 2.5% at each tail.

Decision Rule: The calculated test statistic of -3.024 does fall in the rejection region of z<-1.645, therefore I would reject the null and say there is sufficient evidence to indicate mu<50.

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.

The center for disease control wants to know the average life span of an ebola virus. Studies of similar types of viruses suggest the standard deviation will be 2.55 days (population parameter: use z-table), but they want a 99% confidence interval for the true average lifespan, and they want that confidence interval to have a width of 0.5 days. How many ebola viruses do they need to sample?

There are only 25 numbers in the sample collected. For purposes of analysis, it will be considered that the process itself is normally distributed, and its standard deviation is unknown. Under these conditions, the formula that gives the confidence interval is:

“t” is the trial’s number and C is another constant. The C value we use is the average C value of each date point C=0.1925. So z=0.1925*e^(1.0844*t)

Step 4: The sample’s score on the comparison distribution is determined which in his case is Z=

• Alternative Hypothesis Ha: µ ≥ 50,000, this means the test is a one tailed z test.

The 95% confidence interval for the population mean is 66,438 to 80,241. This means that there is a 95% confidence that this interval has the population mean.

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.