Textbook Notes (368,448)
BIOL499A (11)
Chapter 23

# Chapter 23final.pdf

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School
Department
Biology (Biological Sciences)
Course
BIOL499A
Professor
Blaine Mullins
Semester
Winter

Description
Chapter 23: Statistical Inference – Confidence Interval and Testing Hypothesis for One Population Mean In this chapter, we will talk about:  Estimation of a population mean,  Confidence interval for a population mean whnis unknown,  How to choose the sample size for a study about a population mean,  The necessary steps in the general hypothesis-testing procedure,  One-sample hypothesis test for a population mean when is unknown, Theorem: Suppose Y1, 2 Y n is a random sample from a normal population with mean of  and standard deviation of  respectively. Let 1 n 1 n Y   i and Y S Y   ( i 2 n 1 n 1 1 Y  Then T  has a standard normal distribution. In addition, the  / n Y  distribution ofT  is called Student's t distribution with n 1 degrees Sn/ of freedom. Notations:  Suppose Z has a standard normal distribution. Then, for given  (0  1), zz(or for simplisity ) is defined as: PZ()  z *    Suppose T has at -distribution with degrees of freedom of. Then, t *(or for simplisity t ) for given (0  1), df, is defined as: * P(T) t df,  Definition: a100(1)% confidence interval (C.I.) for a paramtis an interval (L, U) such that 1 PL( )    L and U are called lower and upper bound of the confidence interval, respectively. The percentage100(1)% is called confidence level. Twenty-five students are selected at random, find a 95% confidence interval for the average weight of students if we found0 and 10  Confidence Interval for One Population Mean: When population standard deviatin, , is unknown, we use sample standard deviation, S, to estimat. Therefore, a100(1)% confidence interval for (population mean) is given * s yt n * s The quantityME  t is called the margin of error. n Example 23.1: District court records provided data on sentencing for 19 criminals convicted of negligent homicide. The mean and standard deviation of the sentences were found to be 72.7 and 10.2 months, respectively. Determine a 95% confidence interval for the mean sentence for this crime. Solution: Top-Hat Question (Review 11-1): Example 23.2: The level of various substances in the blood of kidney dialysis patients is of concern because kidney failure and dialysis can led to nutritional problems. A researcher performed blood test on several dialysis patients on six consecutive clinic visits. One variable measured was the level of phosphate in the blood. Phosphate levels for a single person tend to vary normally over time. The data on one pa tient, in milligrams of phosphate per deciliter (mg/dl) of blood, are given below. 5.6 5.1 4.6 4.8 5.7 6.4 Find a 90% confidence interval for this patient's mean phosphate level. What is the lower limit of the confidence interval? Note: y s 5.3667 and 0.6653 Solution: 1  0.90    0.10 df  n 1 A 90% confidence interval for  is given by: s yt * n Conclusion: It is estimated with 90% confidence that the mean phosphate level for this patient is between mg/dl to mg/dl. Example 23.3: In a study on the nutritional qualities of fast foods, the amount of fat was measured for a random 75 hamburgers of a particular restaurant chain. The sample mean and standard deviation were found to be 30.2 and 3.8 grams, respectively. Use these data to construct a 90% confidence interval for the mean fat content in hamburgers served in these restaurants. Solution: Example 23.4: An entomologist sprayed 100 adult Melon flies with a specific low concentration of malathion and observed their survival times. The mean and standard deviation were found to be 18.9 and 4.2 days, respectively. (a) Use these data to construct a 95% confidence interval for the mean survival time. Solution: 1  0.95    0.05 df  n 1 A 95% confidence interval for is given by: s yt* n Conclusion: It is estimated with 95% confidence that the mean survival time is between days to days. (b) Find a 98% confidence interval for the true mean survival time. Solution: 1  0.98    0.02 df  n 1 A 98% confidence interval for is given by: * s yt n Conclusion: It is estimated with 98% confidence that the mean survival time is between days to days. Sample Size for a Study about a Population Mean: To determine how large a sample is needed for estimating aopulation mean, we must specify: ME = the desired margin of error 1 = the probability associated with the error margin. Then, the formula for calc
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