BNAD Exam 2
Suppose you are performing a hypothesis test on μ and the value of sigma is known. The the 5% significance level, the critical values for a two tailed test are:
-z.025 and z.025
In a binomial experiment there are 2 possible outcomes (k=2) and the probability of success (p) and the probability of failure (1-p) equals 1(P1+p2=1)
A multinomial experiment results in one of k possible outcomes (k≥3) and the probabilities added together equals 1 (p1+p2+...+Pk= 1)
With monster df
Always round down for df
Hypothesis Testing
Enables us to determine if the collected sample data is inconsistent with what is stated in the null hypothesis resolves conflicts between two competing opinions
Type II Error
Fail to reject the null hypothesis when the null hypothesis is false Denote the probability with β Miss
Null Hypothesis
Ho the presumed default state of nature or status quo equal sign goes in the null
The Goodness of Fit Test for Normality
Ho: Follows a normal distribution Ha: does not follow a normal distribution
What are the competing hypotheses if the thought that the proportion in group 1 watches TV more than in group 2?
Ho: p1-p2≤.20 HA: p1-p2>.20
Null Hypothesis for a Multinomial Experiment
Ho: p1=p2=p3=p4=0.25 Ha: not all population proportions are equal to .25 or Ho: p1=0.40, p2=0.30, p3=0.20, p4=0.10 HA: at least one of the proportions is different from its hypothesized value
You want to determine if proportion is more than .15
Ho: p≤ .15 HA: p>.15
The null hypothesis for a two sided test for a population mean would be noted as
Ho: μ=μ0
Hypothesis Tests for μD (Mean Difference)
Ho: μD=0 (there is no difference) HA: μD≠μ0 (there is a difference) reject the null if the range of the confidence interval does not include zero if the confidence interval includes zero we do not reject the null hypothesis and conclude that the mean does not differ from zero
Right Tailed Hypothesis Test
Ho: μ≤μ0 HA: μ>μ0 Alternative symbol points to the right
Left Tailed Hypothesis Test
Ho: μ≥μ0 HA: μ<μ0 Alternative symbol points to the left
Steps to Formulate the Null and Alternative
Identify the population parameter of interest Determine whether it is a one tailed or two tailed test Include some form of the equality sign in the null hypothesis and use the alternative hypothesis to establish a claim
Left Tail Probability of Alternative Hypothesis
P(Z≤z)
Right Tail Probability of Alternative Hypothesis
P(Z≥z)
Two Tail Probability of Alternative Hypothesis
P(Z≥z) if z>0 or P(Z≤z) if z<0
Confidence Intervals and Two-Tailed Hypothesis Tests
Reject H0 if the confidence interval does not contain the hypothesized value
Type I Error
Reject the null hypothesis when the null hypothesis is true denote the probability with α False alarm
P-Value
When testing μ, the probability of obtaining a sample mean at least as large or at least as small as the one derived from a given sample, assuming the null hypothesis is true
Multinomial Experiment
a generalization of a binomial experiment consists of a series of n independent and identical trials of a random experiment such that for each trial: -there are k possible outcomes or categories called cells; k≥2 -each time we repeat the trial the probability pi that one outcome falls into a particular cell remains the same -the sum of the cell cell probabilities is one, that is p1+p2+...+Pk= 1 Ho: p1+p2+...+Pk= 1
Matched-Pairs Sampling
a specific type of dependent sampling when the samples are paired in some way often a before and after study but does not necessarily need to be on the same individual parameter of interest is the mean difference we recognize by looking for a natural pairing between one observation in the first sample and one observation in the second sample
Test of Independence (Chi Square Test of a Contingency Table)
analyzes the relationship between two qualitative independent variables Ho: The two classifications are independent HA: The two classifications are dependent use a contingency table to conduct a hypothesis test that determines whether the classifications depend upon one another implemented as a right-tailed test and is valid when the expected frequencies in each cell are five or more
The basic principle for hypothesis testing is to first assume that the null hypothesis is true
and then determine if the sample data contradicts this assumption
α can only be reduced
at the expense of increasing β
We like to report the p-value
because people can make their own decisions about what α they want and whether or not they want to reject it
Binomial Distribution
can be approximated by a normal distribution for large sample sizes
Final conclusion to a statistical test
clearly interpret the results in terms of the initial claim
Two Tailed Hypothesis Test
defined for when the null hypothesis states a specific value for the population parameter of interest Ho: μ=μ0 HA: μ≠μ0 we can reject the null on either side of the hypothesized value of the population parameter
Goodness of Fit Test for a Multinomial Experiment
determines whether two or more population proportions equal each other or any predetermined set of values For example, are four candidates in an election equally favored by voters? or Do people rate food quality in a restaurant comparably to last year?
Degrees of Freedom for a Contingency Table
df=(r-1)(c-1)
Goodness-of-fit test
examines a single qualitative variable the chi-square test statistic will be at least zero
A required condition of Chi Test
expected frequency in each cell must be at least 5 One way to correct this potential problem is to combine categories
We can only do confidence intervals
for two tailed hypothesis tests
Both α (Type I Error) and β (Type II Error) will decrease
if n increases it is always in the best interest to have a sample size as large as you can possibly afford
two (or more) random samples are considered independent
if the process that generates one sample is completely separate from the process that generates the other sample
In the case when we construct a confidence interval for μ1-μ2 where σ²1 and σ²2 are unknown but assumed equal we calculate a pooled estimate of the common variance s²p
in the case where the variances are not assumed equal, we cannot calculate a pooled estimate of the population variance because of different variabilities in the two populations
One Tailed Hypothesis Test
involves a hypothesis that can only be rejected on one side of the hypothesized value (you're guessing that the mean is greater than (left tailed) or less than (right tailed) a certain value) α=.05 Z=1.645 α=.01 Z=2.33
d0
is a hypothesized mean difference if an interval does not include d0 we reject the null
The point estimate for the difference between two population means
is represented by the difference between two sample means
In the case when we construct a confidence interval for the difference between two proportions p1-p2 follows the general format of a point estimate ±
margin of error
Null and Alternative
mutually exclusive
Hypothesis testing for the difference between two sample means is only valid when the sapling distribution of X1-X2 is
normally distributed
The normal distribution approximation for a binomial distribution is valid when
np ≥ 5 and n(1-p) ≥ 5
The p-value is calculated assuming the
null hypothesis is true
When performing a hypothesis test on μ, the p-value is defined as the
observed probability of making a Type I error
Equivalent methods to test a one sided hypothesis
p-value approach critical value approach the conclusions are always the same; only the decision rules defer between the two approaches
When comparing two population proportions the parameter of interest is
p1-p2
A pooled estimate of proportions can be used when the underlying sample proportions are essentially the same estimates of the unknown
population proportion
Alternative Hypothesis
represented with HA the results that you want a contradiction of the default state of nature or status quo
The optimal values of Type I and Type II errors
require a compromise in balancing the costs of each type of error
The hypothesized value of the mean
resides in the null
Independent Random Samples
samples that are completely unrelated to one another; the process that generates one sample is completely independent of the process that generates the other sample
Contingency Table
shows the frequencies for two qualitative variables, x and y, where each cell of the table represents a mutually exclusive combination of the pair of x and y values
The power of a test
the probability of rejecting the null hypothesis when the null hypothesis is false the probability of being correct 1-β or 1-(the probability of not rejecting the null when it is false)
Statistical inference concerning the mean difference based on matched-pairs sampling requires one of two conditions
the sample size n≥30 The paired difference D=X1-X2 is normally distributed
In a test of independence
the test statistic follows the X^2df distribution
Given a right tailed hypothesis test, if the value of the test statistic is 1.82 and the critical value is 1.645
then we reject the null hypothesis
Inferential Statistics
use sample information to make decisions about an unknown population parameter
We calculate pooled estimate of the common variance by
using weighted averages of the sample variances
In order to conduct a test of independence
we calculate each cell's probability in a contingency table by applying the multiplication rule for independent events
When p<α
we reject the null
If a test statistic falls into the rejection region
we reject the null hypothesis
For most applications the hypothesized difference between two means is
zero
Significance Level
α the allowed probability of making a type I error choose the value before conducting a test
The parameter of interest for matched pairs sampling
μD
Mean Difference for Matched Pairs Sampling
μD= X1-X2 X1 and X2 are matched in a pair
We do not reject the null when the p value is
≥α