To: sbaker@odu.edu
From: Thomas Meyer
tmeyer@ph.vccs.edu
276 656-0283
Patrick Henry Community College
Subject: Statistics - w/ Dr.Spencer Baker - Homework Assignment #4, Ch 12 - Problem 09, page 306.
Date: March 25, 2004
Question:
For Each of the following examples, find the standard error of the mean,
critical value(s), and the rejection region(s) for the given null hypothesis
tested at the given a level.
| Population Mean | Population StandardDeviation | Sample Size | Null Hypothesis | alpha |
| a. 100 | 15 | 9 | H =100 | .05 |
| b. 500 | 100 | 25 | 500 | .01 |
| c. 16 | 1.33 | 30 | 16 | .01 |
| d. 18.8 | 1.75 | 20 | 18.8 | .05 |
| e. 63.2 | 9.73 | 80 | 63.2 | .001 |
| f. 100 | 15 | 36 | 100 | .05 |
Answer:
To find the standard error of the mean "sigma sub x-bar," use text page 263 definition and formula:The standard error of the mean is the standard deviation of the sampling distribution of the mean. It is the standard deviation of the sample means around Mu (u).
To computer sigma "sub X-bar" divide " sigma sub X" (the standard deviation of individual observations in the population) by the square root of the sample size.
A. 15 / square root of 9 = 15 / 3 = 5;
B. 100 / square root of 25 = 100 / 5 = 20
C. 1.33 / square root of 30 =1.33 / 5.4772256 = .2428237
D. 1.75 / square root of 20 = 1.75 / 4.472136 = .3913119
E. 9.73 / square root of 80 = 9.73 / 8.9442719 = 1.0878471
F. 15 / square root of 36 = 15 / 6 = 2.5
These standard errors are reported in the first column of the following
table.
Alpha and the null hypothesis are given.
Critical values of Z should be looked up in Appendix A beginning on text page
445.
Calculate = (Z) x (std. error of the mean) +
mu =
Critical Value
A. (5) x (+1.96) +
100
= 109.8
(5) x (-1.96) +
100
= 90.2
B. (20) x (+2.326) + 500 = 46.52 +500 = +546.52
C. (.2428237) X ( 2.576 ) +
16 =
.6233139 + 16 = 16.6233139
(.2428237) X ( -2.576 ) +
16 =
- .6233139 + 16 = 15.376686
D. (.3913119) x (-1.645) + 18.8 = -.6437081 + 18.8 = 18.156292
E. (1.0878471) x (+3.30 ) +
63.2
= 3.58898954 + 63.2 = 66.789895
(1.0878471) x (-3.30
) + 63.2
= -3.58898954 + 63.2 = 59.611011
F. (2.5) x (1.645 ) + 100 = 4.1125 + 100 = 104.1125
Important notes:
1. If the null hypothesis has an equal sign in it only, this will imply a
two-tail test, followed by to rejection regions.
2. If the null hypothesis has a less than (or greater than) sign, this
implies a one-tail test, and the rejection region reverses the less than to a
greater than symbol, or reverses a greater than symbol to a less than symbol.
| Std. Error of Mean | Alpha | Null Hypothesis | Critical Value from text page 444 |
Rejection Region(s) |
| 5 | .05 | Ho: mu = 100 (equal implies a 2-tail test) |
- or + 1.96 | >109.8 <90.2 |
| 200 | .01 | Ho: mu <= 500 | +2.326 | >546.52 |
| .2428237 | .01 | Ho: mu =16 (equal implies a 2-tail test) |
- or + 2.576 | >= 16.62 <=15.38 |
| .3913119 | .05 | Ho: mu >=18.8 | +1.645 | <= 18.16 |
| 1.0878471 | .001 | Ho: mu = 63.2 (equal implies a 2-tail test) |
- or + 3.30 | >=66.79 <=59.61 |
| 2.5 | .05 | Ho: mu <=100 | +1.645 | >=104.11 |
filename: StatHW3Ch12Prob09page306TomMeyer.doc
Tom Meyer
Thomas Meyer