1. Zoology: Hummingbirds
Allen’s hummingbird has been studied by zoologist Bill Alther. A small group of 15 Allen’s hummingbirds has been under study in Arizona. The average weight for these birds is x-bar=3.15 grams. Based on previous studies, we can assume that the weights of Allen’s hummingbirds have a normal distribution, with σ = 0.33 gram.
(a) Find an 80% confidence interval for the average weights of Allen’s humingbirds in the study region. What is the margin of error?
Error = z (s) / sq. root of 15
= 1.28 * .33/ 3.87
Confidence Interval 3.04 gm < µ < 3.26 gm; Margin of Error = 0.11 gm.
(b) What conditions are necessary for your calculations?
(c) Give a brief interpretation of your results in the context of this problem (Interpret the confidence interval)
If the weights of 100 hummingbirds was taken, 80 of them would cross µ. This means that we are 80% confidence that the µ lies between 3.04 gm and 3.26. 20% of the cases will not cross µ.
2. Assume you’ve administered a worker satisfaction test to a random sample of 25 workers at your company. The test is purported to have a population standard deviation or σ = 4.50. The test results reveal a sample mean (x-bar) of 78. Based on that information, develop an estimate of the mean score for the entire population of workers, using a 95% confidence interval.
Answer: 76.25 < µ < 79.76
3. The NAEP survey includes a short test of quantitative skills, covering mainly basic arithmetic and the ability to apply it to realistic problems. Scores on the test range from 0 to 500. For example, a person who scores 233 can add the amounts of two checks appearing on a bank deposit slip; someone scoring 325 can determine the price of a meal from a menu; a person scoring 375 can transform a price in cents per ounce into dollars per pound. In a recent year, 840 men 21 to 25 years of age were in the NAEP sample. Their mean quantitative score was x-bar = 272. These 840 men are a random sample from the population of all young men. On the basis of this sample, what can we say about the mean score µ in the population of all 9.5 million young men of these
ages? Suppose σ = 60.
1. Establish what is µ, σ, x-bar, n
2. Calculate the standard error : (σ/ sq. N)
3. Decide which confidence interval you’d like to use (which percentage)
4. Construct and interpret the confidence interval.
Central Limit Theorem
A certain strain of bacteria occurs in all raw milk. Let x be the bacteria count per milliliter of milk. The health department has found that if the milk is not contaminated, then x has a distribution that is more or less mound-shaped and symmetrical. The mean of the x distribution is µ = 2500, and the standard deviation is σ = 300. In a large commercial dairy, the health inspector takes 42 random samples of the mlik produced each day. At the end of the day, the bacteria count in each of the 42 samples is averaged to obtain the sample mean bacteria count x-bar.
(a) assuming the milk is not contaminated, what is the distribution of x-bar?
standard error = sigma/ sq.root of n
= 300/ sq. root of 42
(b) Assuming the milk is not contaminated, what is the probability that the average bacteria count x-bar for one day is between 2350 and 2650 bacteria per mililiter?
First step: We need to use the standard error because this a central limit theorem application. so standard error is 46.15
find z-score (x-x-bar/st. error) ****The trick with these questions is that that the z-score uses the corresponding information from the new Central Limit Theorem Chart. So,it actually amounts to x-bar – µ / st. error.
We need to find two z-scores.
z – 2350-2500/46.15
z = 150/ 46.15
z = 3.25
z = 2650 – 2500 / 46.15
z= 150 / 46.15
So we know the area between mean and z is 49.94. Therefore we can say that 49.94 X 2 = almost 100%, so almost 100 % of the average bacteria count will be between 2350-2650.
1. Know how to calculate a scatterplot using a set of data such as the following:
Consider: which value of r indicates a stronger correlation: r=0.731 or r=-0.845 Explain your reasoning.
2. Identify the explanatory variable and the response variable. A nutritionist wants to determine if the amount of water consumed each day by persons of the same weight and on the same diet can be used to predict individual weight loss.