Symmetric about the vertical axis through the mean μ
Approaches the horizontal axis asymptotically
Total area under the curve and above the horizontal is 1
Approximately 68% of observations fall within 1 σ from the mean
Approximately 95% of observations fall within 2 σ from the mean
Approximately 99.7% of observations fall within 3 σ from the mean
68.27% 95.45 % 99.73%
Standard Normal Distribution
Special type of normal distribution where μ =0
Used to avoid integral calculus to find the area under the curve
Standardizes raw data
Z = X - μ 0 σ
Given the normal distribution with μ = 49 and σ = 8, find the probability that X assumes a value:
Less than 45
More than 50
Example 2 The achievement sores for a college entrance examination are normally distributed with the mean 75 and standard deviation equal to 10. What fraction of the scores would one expect to lie between 70 and 90.
Distribution of all possible sample statistics
Population All Possible Samples Sample Means 1, 2, 3, 4 1, 2, 3 2.00 1, 2, 4 2.33 3, 4, 1 2.67 2, 3, 4 3.00 μ = 2.5; σ = 1.18 n = 3 μ xbar = 2.5
Central Limit Theorem
Given a distribution with a mean μ and variance σ² , the sampling distribution of the mean approaches a normal distribution with a mean (μ) and a variance σ²/N as N, the sample size, increases.
The mean of the population and the mean of the sampling distribution of means will always have the same value .
The sampling distribution of the mean will be normal regardless of the shape of the population distribution.
As the sample size increases , the distribution of the sample average becomes less and less variable.
Hence the sample average X bar approaches the value of the population mean μ .
Example 3 An electrical firm manufactures light bulbs that have a length of life normally distributed with mean and standard deviation equal to 500 and 50 hours respectively. Find the probability that a random sample of 15 bulbs will have an average life ofless than 475 hours.
Normal Distribution and Hypothesis Testing
A hypothesis is a conjecture or assertion about a parameter
Null v. Alternative hypothesis
Proof by contradiction
Null hypothesis is the hypothesis being tested
Alternative hypothesis is the operational statement of the experiment that is believed to be true
Alternative hypothesis specifies a one-directional difference for parameter
H 0 : μ = 10 v. H a : μ < 10
H 0 : μ = 10 v. H a : μ > 10
H 0 : μ 1 - μ 2 = 0 v. H a : μ 1 - μ 2 > 0
H 0 : μ 1 - μ 2 = 0 v. H a : μ 1 - μ 2 < 0
Alternative hypothesis does not specify a directional difference for the parameter of interest
H 0 : μ = 10 v. H a : μ ≠ 10
H 0 : μ 1 - μ 2 = 0 v. H a : μ 1 - μ 2 ≠ 0
Also known as the “rejection region”
Critical region contains values of the test statistic for which the null hypothesis will be rejected
Acceptance and rejection regions are separated by the critical value, Z .
Type I error
Error made by rejecting the null hypothesis when it is true .
Denoted by the level of significance, α
Level of significance suggests the highest probability of committing a type I error
Type II error
Error made by not rejecting (accepting) the null hypothesis when it is false .
Probability denoted by β
Notes on errors
Type I ( α ) and type II errors ( β ) are related . A decrease in the probability of one, increases the probability in the other.
As α increases , the size of the critical region also increases
Consequently, if H 0 is rejected at a low α , H 0 will also be rejected at a higher α .
Testing a Hypothesis on the Population Mean Z = X - μ 0 σ /√n t = X - μ 0 S /√n υ = n - 1 H 0 Test Statistic H a Critical Region σ known μ = μ 0 μ < μ 0 μ > μ 0 μ ≠ μ 0 z < -z α z > z α |z| > z α /2 σ unknown μ = μ 0 μ < μ 0 μ > μ 0 μ ≠ μ 0 t < -t α t > t α |t| > t α /2
critical value test statistic Reject H 0
critical value test statistic Do not reject H 0
Example 4 It is claimed that an automobile is driven on the average of less than 25,000 km per year. To test this claim, a random sample of 100 automobile owners are asked to keep a record of the kilometers they travel. Would you agree with this claim if the random sample showed an average of 23,500 km and a standard deviation of 3,900 km? Use 0.01 level of significance.