Buec chapter 9

A phone industry manager thinks that customer monthly cell phone bills have increased, and now average over $52 per month. The company wishes to test this claim to a 90% confidence level with a sample of 64 customers where the mean value of $53.1 was found. (Assume o= $10 is known)

1) State H0 and Ha: H0: μ ≤ 52 Nothing new so continue to assume $52 per month Ha: μ > 52 The average is greater than $52 per month (i.e. there is sufficient evidence to support the manager's claim with the necessary level of confidence) 2) Specify Significance Level and Sample Size 90% confidence means a = 0.10 The sample size available was n = 64 3) Select appropriate test statistic Since σ is known and n ≥ 30, use Z statistic 4) Find Rejection Region a= 0.10 converts to Z = 1.28 from Table A.5 Reject H0 if Z > 1.28 5) Obtain sample and compute the test statistic Obtain sample and compute the test statistic A sample is taken with the following results: X = 53.1, n = 64 ( = 10 was assumed known) Then the test statistic is: Zstat=.88 6) Reach a decision and interpret the result: Do not reject H0 since Z = 0.88 ≤ 1.28 Conclusion: There is not sufficient evidence at the 90% confidence level to conclude that the mean bill is over $52

Challenge the historically based assertion that the true mean number of computers in Canadian homes is 3, with a 5% probability of making a Type 1 error. 0=.8 A survey of 100 homes finds X = 2.84

1. Formulate the appropriate null and alternative hypotheses H0: μ = 3 Ha: μ ≠ 3 (This is a two-tail test) 2. Specify the desired level of significance and sample size a = 0.05, n = 100 3. Select appropriate test statistic Since σ is known and n > 30, use Z statistic 4.Determine the critical values that define the rejection region Reject H0 if ZSTAT ≤ ZCRIT(low) or if ZSTAT ≥ ZCRIT(high) otherwise, do not reject H0 Z(crit)=+-1.96 a/2=.025 5. Zstat=-2 Make a Decision: identify where the test statistic lands to decide which hypothesis prevails? Since Zstat = -2.0 ≤ -1.96, we reject the null hypothesis and conclude: There is sufficient evidence at the 95% confidence level that the mean number of computers in Canadian homes is not equal to 3.

The 6 Steps of Critical Value Hypothesis Testing

1.Formulate and state the null (H0) and alternative (Ha) hypotheses 2.Specify the desired significance level (α) and sample size (n) 3.Select the appropriate test statistic distribution. (Z, t, or other) 4.Calculate the resulting rejection region using critical values derived from your choice of α in Step 2. (Zcrit, tcrit, etc.) 5.Take a random sample and use its data to calculate the test statistic. (Zstat, tstat, etc.) 6.Compare values to make the statistical decision and state conclusion in terms of the question. If the test statistic falls into the non-rejection region, do not reject the null hypothesis H0. If the test statistic falls into the rejection region, reject the null hypothesis. State the managerial conclusion in the context of the problem.

What is a Hypothesis?

A claim (assertion) about the value of a population parameter. It may or may not turn out to be true, but a hypothesis can be tested. Population Mean Example: Example: The mean monthly cell phone bill of Burnaby is μ = $42 Example: The proportion of adults in Burnaby with cell phones is π = 0.78

What is Hypothesis Testing

A procedure to evaluate a hypothesis about the population parameter with actual results obtained from a sample of data The purpose of hypothesis testing is to choose between exactly two contradictory alternatives of the value of a population parameter The two competing hypotheses are called the Null Hypothesis (H0 pronounced "H-naught") and the Alternative Hypothesis (Ha pronounced "H-one", or "H-a").

The Null Hypothesis, H0

Begin with the assumption that the null hypothesis is valid (the default hypothesis). Similar to the idea of "innocent until proven guilty" Refers to the value reached with the least amount of assumed knowledge. (most boring) This is often but not always the currently assumed best guess value. Sometimes called the "working hypothesis". Always contains equivalence in the form of one of "=" , "≤" or "" symbols. May or may not be rejected by sample data depending on whether sample data is consistent with the hypothesis or not. A statement about the numerical value of a population parameter to be tested Examples: The average number of TV sets in Canadian homes is still equal to three HO=u=3 The average number of TV sets in Canadian Homes is less than or equal to three HO=u<=3 The average number of TV sets in Canadian Homes is three or more HO>=3

p -Value Approach Solution

Calculate the p-value and compare to a (assuming that μ = 52.0) since Z=.88 then p-val=.5-.3106=.1894 .1894>a(.10) DO NOT REJECT

Hypothesis Testing Process

Conventional opinion: The population mean age of model railroaders is 60. HO: U=60 Now select a random sample Suppose the sample mean age is 20: XBAR = 20 Suffiently unlikely reject null hypothesis Suppose the sample mean age was X = 20. This is significantly lower than the H0 hypothesis that the mean population age is 60. If the null hypothesis were true, the probability of getting such a different sample mean would be very small, so you reject the null hypothesis . In other words, getting a sample mean of 20 is so unlikely if the population mean was 60, you conclude that the population mean must not be 60.

Connection Between Two Tail Tests and Confidence Intervals

For X = 2.84, σ = 0.8 and n = 100, the 95% confidence interval is: 2.6832 ≤ μ ≤ 2.9968 Since this interval does not contain the hypothesized mean (3.0), we reject the null hypothesis at a = 0.05

The average cost of a hotel room in Vancouver is believed to be $168 per night. This is challenged by a random sample of 25 hotel rooms resulting in: X = $172.50 S = $15.40 = 0.05 significance level (Assume the population distribution is roughly normal so n=25 is large enough for central limit theorem)

H0: μ = 168 H1: μ ¹ 168 a = 0.05 n = 25 3) o is unknown and population is normal, so use a t statistic 4) a = 0.05 𝒂/𝟐 = 0.025 (two tail) n = 25 d.f. = n-1 = 24 X = 172.5 S = 15.4 t crit= ± 2.0639 5) Tstat=1.461 6) Do not reject H0: There is not sufficient evidence at the 95% confidence level that the true average hotel room cost is different from $168.

Management at an auto factory assumes that, like years before, there are still an average of two defects per car. The workers union disagrees with the production manger and argues that the workers have made great strides in quality control so the average is now lower. State the null and alternative hypotheses associated with this issue.

H0: μ ≥ 2 ( at least two errors per car) Ha: μ < 2 ( fewer than 2 errors per car)

n = 2 1-α = .95 x = 14 μ=7 σ = .707

H0: μ≤7Ha: μ>7 n=2, α=.05 (by continuing with n=2, we are making the assumption that the population distribution is close to normal, otherwise this is much too small! The close to normal assumption is reasonable since this situation is technically a Poisson distribution.) Since we are given a value for σ we will be using the Z distribution Zcrit=1.645 (referenced with knowledge that α=.05) Zstat= 14.002 (calculated using formula) Comparing these values, we find |Zstat| > |Zcrit| putting it in the region of rejection. Conclusion: We are 95% certain that using an axe against an advancing zombie horde is more effective than a gun.

The Sample Statistic and Critical Values

If the sample mean is close to the stated population mean, the null hypothesis is not rejected. If the sample mean is far from the stated population mean, the null hypothesis is rejected. How far is far enough to reject H0? The critical value for a test statistic creates the dividing line for decision making -- it answers the question of how far is far enough before you reject H0! critical values Region of Rejection.Region ofNon-Rejection.Region of Rejection .=critical range

One-Tail Tests

In many cases, the alternative hypothesis focuses on a particular direction This is a lower-tail test since the alternative hypothesis is focused on the lower tail below the mean of 3 HO:μ>=3 HA:μ<3 This is an upper-tail test since the alternative hypothesis is focused on the upper tail above the mean of 3 H0: μ ≤ 3 Ha: μ > 3

Alternative Hypothesis (Ha)

Is the complement of the null hypothesis e.g. The average number of TV sets in Canadian homes is less than 3 ( Ha: µ < 3 ) Challenges the previous belief by suggesting the null hypothesis parameter is wrong. Never contains the "=" , "≤" or "" symbols, so can only contain "≠", "<" or ">". May or may not be accepted depending on whether null hypothesis is rejected from analyzing sample data. Is generally the hypothesis that the researcher is trying to test so it is sometimes called the "test hypothesis". Examples: The average number of TV sets in Canadian homes is not equal to three HA:U=/3 The average number of TV sets in Canadian Homes is more than three HA:U>3 The average number of TV sets in Canadian homes is less than three HA:U<3 Note: The H0 and Ha are always complements of each other, so when you can define one, you have defined the other automatically.

Do You Ever Truly Know σ?

Probably not! In virtually all real world business situations, σ is not known. If there is a situation where σ is known then µ is also known (since to calculate σ you need to know µ.) If you truly know µ there would be no need to gather a sample to estimate it. If the population standard deviation is unknown, you instead use the sample standard deviation S. Because of this change, you use the t distribution instead of the Z distribution to test the null hypothesis about the mean. When using the t distribution you must assume the population you are sampling from follows a normal distribution. All other steps, concepts, and conclusions are the same.

Burden of Proof

Statistical inference (sampling) can NEVER prove the null hypothesis, regardless of whether it is actually true or not. It can only REJECT IT or NOT-REJECT IT probabilistically, depending on where the test statistic lands. This places the burden of proof on the alternative hypothesis which is only accepted when it can be statistically supported with evidence. The only way to completely prove a null hypothesis would be to measure absolutely every point in the population (perform a census) and discover that the parameter matches the null hypothesis value, effectively eliminating every alternative possibility. e.g. It is essentially impossible to prove that there IS NOT a teapot orbiting amongst the objects in the asteroid belt, even though there is no objective reason to assume that there is one. The null assumption therefore must be that there is no space teapot and the burden of proof is correspondingly assigned to providing evidence that there IS a space teapot.

Level of Significance, a

The probability of rejecting the null hypothesis when it is actually true. (making a Type 1 error) Defines the area of rejection region of the sampling distribution Is designated by Typical values are 0.01, 0.05, or 0.10 Is selected by the researcher at the beginning as their accepted risk of making a Type 1 sampling error. Used to calculate the critical value(s) of the test.

p-value

The probability that the value of a given sample's statistic (x, p,...) would be at least as far as it is from the null hypothesis value (µ,π,...) if the null hypothesis were true. Mathematically: P(|Z| ≥ |Zstat|) if H0 is true. Found by referencing P(Zstat) on Z table and finding area of tail. Also called "observed level of significance" Obtain the p-value from a table or computer Compare the p-value with a If p-value ≤ a , reject H0 If p-value > a , do not reject H0

Lower-Tail Tests

There is only one critical value, since the rejection area is in only one tail Reject H0 if Z ≤ ZCRIT; otherwise do not reject H0

Upper-Tail Tests

There is only one critical value, since the rejection area is in only one tail Reject H0 if Z ≥ ZCRIT; otherwise do not reject H0

How did we find p-value for a t-distribution!?!

This is impractical to read from a series of t-tables. In the real world, you would use the Excel function: =T.DIST.RT(tstat,df) This returns the probability of the right tail (.0785), which you would then double. (This example is a two tailed test and you are looking for the total probability of finding a sample mean as far in either direction from the H0 population mean, so you add both tails.) You will not be asked to calculate a p-value for a t-distribution, but you might be asked to find the p-value for a Z=distribution on the final exam

Type I and II Error Relationship

Type I and Type II errors can not happen at the same time because: Type I errors can only occur if H0 is true Type II errors can only occur if H0 is false For a constant sample size (n), if Type I error probability ( a ) down , then Type II error probability ( β ) up automatically. go to slide 20 idk what it means

Possible Errors in Hypothesis Test Decision Making

Type l Error Rejecting the null hypothesis (H0) when it is true. Imagine a Type 1 error as a "False Alarm" Considered a serious type of error because it changes your decision away from the right way. The probability of making a Type 1 Error is a a is called the level of significance of the test a is chosen by the researcher in advance of data analysis Type II Error Not rejecting a null hypothesis when it is actually false. Imagine a Type II error as a "missed opportunity" to change your opinion to a more accurate one. The probability of making a Type II Error is β β is a calculated value, typically not selected by the researcher. Decision H0 True . H0 False do not reject(HO) NO ERROR (1-a) TYPE 2 B Reject(HO) TYPE 1 A NO ERROR(1-B) The confidence coefficient (1-α) is the probability of not rejecting H0 when H0 is true. The probability of avoiding a false positive (Type I Error) The confidence level of a hypothesis test is confidence coefficient expressed as a percentage(1-α) X 100%. If the confidence coefficient is .95, the confidence level is 95% The power of a statistical test (1-β) is the probability of rejecting H0 when H0 is false. The probability of avoiding a missed opportunity (Type II Error)

How likely is it to see a sample mean of 2.84 (or a value further from the mean in either direction) if the true mean is U = 3.0? (using σ=.8, n=100 as before)

Z(stat)=-2.0 p-value=.0228 .0228x2<.025x2 so REJECT