Statistics Homework

WEEK 6 – HOMEWORK 6: LANE CHAPTERS, 11, 12, AND 13; ILLOWSKY CHAPTERS 9, 10

INTRODUCTION TO HYPOTHESIS TESTING

WHAT IS A HYPOTHESIS TEST?

Here we are testing claims about the TRUE POPULATION’S STATISTICS based on SAMPLES we have taken. The most common statistic of interest is of course the POPULATION MEAN (µ). But, we can also test its VARIANCE and its STANDARD DEVIATION. (We can also compare TWO or more means to see if there are significant differences.

We must have a basic hypothesis, referred to as the NULL Hypothesis (Ho) and an ALTERNATE Hypothesis (Ha).

Our NULL ( and ALTERNATE) Hypotheses can take three forms:

(1) Ho: µ < some number; Ha: µ > that number (< is “less than or equal to” and > is “greater than or equal to” ),

(2) Ho: µ > some number; Ha: µ < that number , or

(3) Ho: µ = some number; Ha µ ≠ that number ; (≠ means “not equal to”)

NOTE THAT Ho MUST HAVE THE “EQUALS” IN IT WHEREAS Ha NEVER DOES.

(1) Is referred to as a “ONE-TAILED TEST TO THE LEFT”

(2) Is a “ONE-TAILED TEST TO THE RIGHT”

(3) Is a “TWO-TAILED TEST”

NEXT, we need to decide what level of significance, i.e.(how sure we want to be about our hypothesis. This is where α comes in again. Do we want to test at the 10%, 5% or 1% level of significance? Another wrinkle is that for the TWO-TAILED test, since our value could be greater OR less than some number, we use α /2 for each extreme, so for 10% it’s 5% (0.050) at each end (tail of the curve), for 5% it’s 2.5% (0.0250) at each end, and for 1% it’s 0.5% (0.0050) at the ends. You have heard about this kind of split before with confidence intervals, but think about it. Here is a graphical display of all this:

As you can see, there is a CRITICAL z-VALUE for each of these test depending on the significance level alpha (α) or α/2.

In HW4 questions 1 and 2, you found the critical z-values for alpha’s of 1%, 5% and 10%, which would work for the one-tailed tests. For the two tailed tests we need to split these alphas (α/2) and find the critical z-values (at the positive and negative tails of the graph) So, for an α of 1% (0.0100) it would be α/2 or 0.005 in the left tail (negative z-value) = -2.575 and for the far right tail (0.005 in that tail) we would have to find the z-value for an area to the LEFT of 99.5% (0.9950) and this is +z = +2.575

Continuing on, for an α of 5% for a two-tailed test the z-values for α/2 would correspond to areas under the curve of 0.0250 at each end. The far left tail would have a negative z-value of -1.96 (see picture above) and the far right tail would have a positive z-value of +1.96 that in the Table represented an area of 97.5% (0.9750) to the LEFT.

Lastly, for an alpha of 10%, hence an α/2 at both ends of 5% (the two-tailed test), the negative z-value would be -1.645.

The positive z-value marking the upper 5% (Table value from 95% to the left) is +1.645.

SO, FOR YOUR USE IN ALL HYPOTHESIS TEST (AND WORKS FOR CONFIDENCE INTERVALS TOO) HERE IS A TABLE OF THE CRITICAL Z-VALUES FOR THE VARIOUS LEVELS OF SIGNIFICANCE (ALPHA’s) MOST COMMONLY USED.

ALPHA (α)	-Z-value (LEFT tail)	+Z-value (RIGHT tail)	ALPHA/2 (α/2)	-Z-value (LEFT tail)	+Z-value (RIGHT tail)
1% (0.0100)	-2.33	+2.33	0.5% (0.0050)	-2.575	+2.575
5% (0.0500)	-1.645	+1.645	2.5% (0.0250)	-1.96	+1.96
10% (0.1000)	-1.28	+1.28	5% (0.0500)	-1.645	+1.645

(For those in the NAVY the left and right colors are for PORT and STARBOARD and the history of these terms is interesting (“google” them).

HYPOTHESIS TESTING:

1. WRITE OUT YOUR NULL (Ho) AND ALTERNATE (Ha) HYPOTHESES, MAKING SURE THE NULL INCLUDES AN “EQUALS”.

2. DECIDE ON THE LEVEL OF SIGNIFICANCE (α) YOU WANT (OR ARE TOLD TO USE-BUT IF NOT TOLD ASSUME 5%)

3. SOMETIMES THE LEVEL IS REFERRED TO AS 95% (FOR 5%), 90% FOR (10%) AND 99% FOR (1%). THIS IS JUST TO CONFUSE US.

4. USING THE FORMULAS PROVIDED (IN THE TEXTS) CALCULATE THE TEST STATISTIC, FOR NOW A z-VALUE OR t-VALUE

5. COMPARE THE CALCULATED z OR t TEST STATISTIC TO THE CRITICAL VALUE (YOU ONLY HAVE THE CRITICAL z-VALUES IN THE TABLE ABOVE) AND IF THE TEST STATISTIC IS A SMALLER NUMBER THAN THE NEGATIVE CRITICAL VALUE IT IS IN THE “UNUSUAL” (RED) AREA, HENCE IT IS “UNLIKELY” THE NULL HYPOTHESIS (Ho) IS CORRECT, SO IN THIS CASE WE REJECT Ho AND ACCEPT Ha.

6. ALSO, IF THE CALCULATED TEST STATISTIC IS GREATER THAN THE POSITIVE CRITICAL VALUE WE AGAIN REJECT Ho.

FOR EXAMPLE: ONE-TAILED TEST TO THE LEFT WITH ALPHA = 5% : OUR CRITICAL LEFT TAIL NEGATIVE z-VALUE IS -1.645.

WE CALCUALTE OUR TEST STATISTIC AND GET A -1.650. DO WE ACCEPT OR REJECT Ho ? WE REJECT Ho BECAUSE -1.650 IS LESS THAN -1.645 (FURTHER LEFT IN THE RARE AREA TAIL)

SIMILARLY, FOR A ONE-TAILED TEST TO THE RIGHT WITH ALPHA = 5%: OUR CRITICAL RIGHT TAIL POSITIVE z-VALUE IS +1.645. IF OUR CALCULATE TEST STATISTIC IS 1.650 WE AGAIN REJECT Ho SINCE +1.650 IS FURTHER RIGHT IN THE (RED) “RARE” AREA THAN THE THRESHOLD z-VALUE OF +1.645.

HYPOTHESIS TESTING USING CRITICAL t-VALUES AND t-TEST STATISTICS: IF WE ARE NOT GIVEN THE POPULATION STANDARD DEVIATON WE HAVE TO USE THE t-VALUES (t-TEST) RATHER THAN THE z-VALUES (z-TEST). IT WOULD BE DIFFICULT TO SET UP A TABLE OF THE CRITICAL t-VALUES (LIKE WE DID FOR THE CRITICAL z-VALUES ABOVE) SINCE THE t-VALUES VARY WITH THE SAMPLE SIZE (N) THAT DETERMINES THE DEGREES OF FREEDOM (df = N-1) WE ENTER THE t-TABLE WITH.

NOW, ON TO THE HOMEWORK FOR WEEK 6:

REVIEW: Using the t-Table (NOT the z-Table). The BODY of the t-Table actually contains the t-values, NOT the areas or probabilities like the z-Tables. As review, we locate our desired probability along the TOP row of the t-Table based on the alpha selected. NOTE that these are the UPPER-TAIL probabilities – area shown in blue. We then go down that column (under the probability) until we reach the row with the calculated degrees of freedom (df = N-1). The number at that intersection is our desired t-Value.

#1. WHAT are the critical t-Values with 20 df (NOT 14 as the example circles show) at the 1%, %5 and 10% levels of significance (one-tailed test – right tail). What would the Ho and the Ha be in this problem (THIS IS QUESTION #1)

THIS WEEK’S CONCEPT: HYPOTHESIS TESTING DEALING WITH POPULATION MEANS AND THE SAMPLE MEANS AND STANDARD DEVIATIONS THAT ARE USED TO TEST THOSE POPULATION MEANS.

HOWEVER, WE MUST ADJUST THE SAMPLE’S SD FOR “SAMPLING ERROR” and this is done by dividing the sample SD (abbreviated “s”) by the square root of its sample size N. So, the “adjusted” s = original sample s/ √n. This can also be calculated as s2 / n. NOTE that the s2 is actually the VARIANCE. Remember too from Week 1 that the variance is just the square of the distance of each data point from the true mean. The “squaring” gets rid of negative values as you should recall. IT’S ALL ABOUT THESE DATA POINT DISTANCES FROM THE TRUE MEAN.

FOR THESE HYPOTHESIS TESTS, If we KNOW the POPULATION’S STANDARD DEVIATION, we use the z-TEST.

LOOK AT ILLOWSKY’S SAMPLE PROBLEM 9.9 ON A TEXT PAGE AROUND 380 AS YOU READ ON

ILLOWSKY USES THE P-VALUE TO TEST HYPOTHESES AND COMPARES THEIR CALCULATED P-VALUE TO THE ALPHA LEVEL OF SIGNIFICANCE WE CHOSE. IF THE P-VALUE IS LESS THAN THE ALPHA WE REJECT Ho. THIS IS FINE, BUT IT REQUIRES SOFTWARE TO GET THE PRECISE PROBABILITY (P-VALUE).

WE WILL BE USING THE TEST STATISTIC APPROACH AND ACTUALLY READING THE TABLES. (YOU WILL SEE THAT YOU CAN STILL GET A CLOSE APPROXIMATION OF THE P-VALUE FROM THE TABLE AS WELL.). ILLOWSKY HAS AN EXAMPLE USING TEST STATISTICS ON ABOUT PAGE 486 (EXAMPLE 9.11)

BOTH OF THESE HYPOTHESIS TESTING APPROACHES: P-VALUE AND TEST STATISTIC WILL SUPPORT THE SAME CONCLUSION (ACCEPT OR REJECT Ho). CONFIDENCE INTERVALS CAN ALSO BE USED TO TEST HYPOTHESIS, BUT THIS APPROACH CAN (BUT NOT ALWAYS) SUPPORT DIFFERENT CONCLUSIONS.

AN HYPOTHESIS TEST EXAMPLE: ILLOWSKY 9.9 (around page 480)

A baker claims his bread height is at least 15 cm. Customers say it isn’t that high. This is a one-tailed test and we will use the customer’s statement as our NULL Hypothesis

Ho: mean bread height is less than or equal to 15 cm so Ha: mean bread height is greater than 15 cm OR

Ho: µ < 15 cm and Ha: µ > 15 cm (NOTE that the Ho has the “equals” in it as it must)

This phrasing makes it a one-tailed test to the RIGHT. If the customers had said the bread simply was NOT 15 cm, we could have ended up with a two tailed test with Ho: mean equals 15 cm and Ha: mean is simply not equal to 15 cm (height is higher or lower than 15cm). But, let’s keep it simple.

He bakes 10 loaves that have an average height of 17cm (this is a SAMPLE)

The Baker has produced enough loaves of bread (his POPULATION of bread) over his career to have calculated a standard deviation = 0.5 and we adjust this for “error” as well so it is SD/√N and N = 10 so the adjusted SD = 0.5/√10 = 0.16

NOW, since we have the POPULATION standard deviation we can use the z-Test statistic. The 10 loaf sample mean is 17 cm, this is our x-value we are using to test the accuracy of the Population mean of 15 cm. We simply standardize that x-value: z-test statistic = (x – mean) /adjusted SD = (17 – 15) / 0.16 = 12.5

Since this is the z-TEST value we now go to our +z-Table with. You will quickly see that a z-value of +12.5 is WAY off the chart with more than 99.999% to the LEFT ( and less than 0.000000001 to the right as the PROBABILITY. Normally you would compare this calculated z-Test statistic to our critical values, but you can see that at whatever level of significance we chose (1%, 5% or 10%) we are MUCH further to the right (in the “reject Ho) area than any of them.

So, we REJECT Ho and accept Ha: This Baker’s bread has an average height of greater than 15 cm (we don’t say equal to or greater than since the Ho included the equals and we rejected all of it).

NO POPULATION STANDARD DEVIATION

MORE OFTEN we do NOT know the POPULATION’S SD, so we substitute the SAMPLE’S STANDARD DEVIATION and use the t-TEST.

LOOK AT ILLOWSKY EXAMPLE 9.16 ON ABOUT PAGE 488 THAT USES THE t-TEST STATISTIC. THEY STILL USE THE P-VALUE BUT WE WILL CALCULATE THE TEST STATISTIC

This example deals with scores on a Statistics test. Student believe the mean score is 65 and the instructor believes it is higher. So, let’s go with Ho: mean < 65 and Ha: µ > 65 We don’t have POPULATION mean of SD. So we use a sample.

This is a “single population mean” one-tailed “t” test to the right (since the Ha: has the “>”)

The chosen level of significance is 5%, but we can’t use the critical z-values we tabulated earlier. We need to find the critical t-Value. Ten student tests were sampled so N= 10 and the df = N – 1 = 10 – 1 = 9

The sample of the 10 tests had a mean (X-bar as it’s referred to) of 67. And we need to calculate the sample’s standard deviation (s) which ends up s = 3.197 and then correct it for sampling error σ = s/√10 = 3.197/3.162 = 1.011

Now, our t-Test statistic = (X-bar – mean)/ adjusted SD = (67 – 65)/1.01 = 1.978

WHAT IS THE CRITICAL t-VALUE WITH AN ALPHA OF 5% AND df = 9 ?

Go to the T-Table and find 5% (0.05) along the TOP row. Go down that COLUMN till you come to the row for a DF of 9 and read: 1.833. THIS IS OUR CRITICAL t-VALUE FOR THIS PROBLEM.

COMPARE THE CALCULATED t-TEST STATISTIC 1.978 TO THE CRITICAL VALUE 1.833 AND WE SEE THAT THE TEST STATISTIC IS GREATER THAN (FURTHER TO THE RIGHT) OF THE CRITICAL VALUE, HENCE WE REJECT Ho. THE MEAN IS GREATER THAN 65.

ILLOWSKY USED SOFTWARE BUT LET’S GO INTO THE t-TABLE WITH OUR t-TEST STATISTIC CALUCLATED VALUE OF 1.978 AT A df OF 9. FIND IT? NOW, LOOK UP TO THE TOP ROW OF THE TABLE AND WHAT PROBABILITIES ARE WE BETWEEN?

WE ARE LESS THAN 0.0500 BUT MORE THAN 0.025. PER ILLOWSKY, THE ACTUAL P-VALUE IS 0.0396. SEE HOW YOU CAN GET AN APPROXIMATE P-VALUE JUST FROM THE TABLE? HOWEVER, WE WILL USE THE TEST STATISTIC APPROACH.

YOU CAN CERTAINLY PRACTICE USING THE P-VALUE SOFTWARE, BUT SUBMIT ANSWERS USING TEST STATISTICS.

WE CAN ALSO TEST PROPORTIONS RATHER THAN MEANS. THERE IS NO MEAN PROVIDED. WE GET WHAT WE NEED FROM A SAMPLE: SAMPLE SIZE AND SAMPLE PROPORTION. OUT ASSUMED PROPORTION IS WHAT WE SPECIFY IN THE NULL HYPOTHESIS. WE USE THESE VALUES TO CALCULATE THE STANDARD DEVIATION (OF THE PROPORTIONS) AND THEN USE THAT VALUE TO CALCULATE THE Z-TEST STATISTIC. CHECK ILLOWSKY EXAMPLE 9.17 AROUND PAGE 489)

HERE Ho: PROPORTION (P) IS 50% (0.50) MEANING THAT 50% OF FIRST TIME BRIDES ARE THE SAME AGE AS THEIR GROOMS. AND, Ha: PROPORTION IS ≠ 50% HENCE IT’S A TWO-TAILED TEST.

THE LEVEL OF SIGNIFICANCE CHOSEN IS 1% (0.01) BUT SINCE IT’S TWO-TAILED IT’S α/2 SO IT’S 0.005 AT EACH END. THE CRITICAL Z-VALUE OF THIS α/2 OF 0.005 (LEFT TAIL) OR 0.995 (RIGHT TAIL) = +2.575

THE SAMPLE OF 100 MARRIAGES (N = 100) SHOWED A PROPORTION OF 53% (p’ = 0.53)

THE STANDARD DEVIATION (σ) = SQRT [P * (1 – P)/N] = SQRT [ 0.5 * (1 – 0.5)/100] = SQRT [0.0025] = 0.05

THE Z-TEST STATISTIC = (p’ – P) / σ = (0.53 – 0.50)/0.05 = 0.6 AND SINCE 0.6 IS NOT A LARGE AS THE CRITICAL VALUE OF 2.575 WE FAIL TO REJECT Ho.

CHECKING THE PROBABILITIES: A Z-VALUE OF 0.6 CORRESPONDS TO A PROBABILITY IN EACH TAIL (TWO-TAILED TEST) OF 0.2743 (FROM THE TABLE) AND WE MUST DOUBLE THIS TO ACCOUNT FOR THE TWO-TAIL TOTAL PROBABILITY: 2 * 0.2743 = 0.5486 = 55%

SINCE α = 0.01 (1%) IN TOTAL AND OUR CALCULATED PROBABILITY VALUE = 0.5486 WE ARE NO WHERE NEAR THE TAILS AT EITHER END, HENCE CAN NOT REJECT Ho.

ONE LAST HYPOTHESIS INVOLVES COMPARING THE MEANS OF TWO POPULATIONS. THESE POPULATIONS CAN BE INDEPENDENT OR “MATCHED PAIRS” (NOT INDEPENDENT). IF THE POPULATION STD DEVIATIONS ARE KNOWN WE USE THE Z-TEST, BUT IF THEY ARE NOT IT’S A t-TEST PROBLEM

THE Ho: would be that the means are equal expressed as Ho: M1 –M2 = 0 and the Ha: could be a < or > or ≠ depending on how it’s stated.

Illowsky around pages 530-531 have the formulas for t-test statistic calculation and the degrees of freedom (df) calculation FOR THESE MEAN COMPARISON HYPOTHESIS TESTS. Both are a little involved, but you NEED TO USE THEM (NOT software) to calculate these necessary values. Round off your calculated df to the nearest whole number and then YOU CAN AND NEED TO USE A t-TABLE TO FIND THESE VALUES.

HYPOTHESIS TESTING CAN MAKE MISTAKES. THESE ARE REFERRED TO AS TYPE I AND TYPE II ERRORS. WE COULD REJECT Ho WHEN IT IS ACTUALLY TRUE OR ACCEPT Ho WHEN IT IS ACTUALLY FALSE. BIG MISTAKES LEGALLY.

ACTION	Ho IS ACTUALLY
	TRUE	FALSE
Do NOT reject Ho	Correct Outcome	Type II Error
Reject H0	Type I Error	Correct Outcome

CONTINUING ON WITH THE HW PROBLEMS (DON’T OVERLOOK QUESTION #1 WAY EARLIER)

( ILLOWSKY CHANGES PAGE NUMBERS, LANE HASN’T.)

#1 (way earlier in the discussion above)

#2. (LANE C-11) THE ALPHA LEVEL IS 0.05 WHEN YOU ANALYZE THE DATA IF THE NULL HYPOTHESIS (Ho) IS TRUE, WHAT IS THE PROBABILITY THAT YOU COULD MAKE A TYPE I ERROR.? b. IF THE NULL HYPOTHESIS IS FALSE, WHAT IS THE PROBABILITY THAT YOU COLD MAKE A TYPE I ERROR.

#3. IS STUDYING FOR A COURSE IN ONE LONG PERIOD MORE USEFUL THAN STUDYING IN INTERVALS?

TWO GROUPS: #1 12 STUDENTS TOOK A TEST AFTER STUDYING TWO HOURS CONTINUOUSLY; #2 12 DIFFERENT STUDENTS TOO THE SAME TEST AFTER STUDYING 4 ONE-HALF HOUR SESSIONS WITH 30 MINUTES BETWEEN THESE SESSIONS. (LANE PROVIDES A GOOD EXAMPLE OF THIS TYPE OF PROBLEM BEGINNING AROUND PAGE 408)

GROUP #1: MEAN TEST SCORE WAS 75 AND THE VARIANCE WAS 120 (WHAT WAS THE STD DEVIATION?)

GROUP #2: MEAN TEST SCORE WAS 86 AND THE VARIANCE WAS 100 (WHAT WAS THE SD?)

(a) ARE THE MEAN TEST SCORES OF THESE TWO GROUPS SIGNIFICANTLY DIFFERENT AT THE 5% (0.05) LEVEL ?

IS THIS A t-TEST OR z-TEST PROBLEM AND WHY? DETERMINE THE VALUE OF THE TEST STATISTIC AND FIND ITS CRITICAL VALUE IN THE TABLE AND THEN COMPARE TO MAKE YOU DECISION (ACCEPT OR REJECT Ho) . YOU CAN ALSO (BUT NOT IN PLACE OF) FIND THE PROBABILITY (P-VALUE) WITH SOFTWARE AND SEE IF YOU GET TO THE SAME CONCLUSION.

(b) USE THE SAME PROCEDURES TO DRAW YOU CONCLUSION IF ONLY 6 STUDENTS WERE IN EACH GROUP

#4. (LANE C-13) LIST 3 WAYS YOU COULD INCREASE THE POWER OF AN EXPERIMENT AND EXPLAIN WHAT THEY MEAN OR HOW THEY DO IT. [ PROBLEM #4 USED TO BE ASSIGNED, BUT IT IS VERY CONFUSING – CHECK IT OUT IF YOU HAVE TIME. THE RANK ORDER ANSWER IS C, A, B, D ]

#5 ASSUME THAT THE AVERAGE PERSON’S IQ IS 107 AND THAT OF A FISH IS 4. YOU HAVE TROUBLE CATCHING FISH SO YOU FEEL THAT THEIR IQ MUST BE HIGHER.

YOU CATCH 12 FISH (N = 12) AND A “FISH PSYCHOLOGIST” (NOT A RECOMMENDED MAJOR) TESTS THEIR IQ’s AND DETERMINES THOSE IQ’S TO BE: 5, 4, 7, 3, 6, 4, 5, 3, 6, 3, 8, 5

Ho: IQ ? 4 and Ha: IQ ? 4 (ONE-TAILED TEST) ASSUME α = 5% (0.05) [replace the “?” appropriately)

CALCULATE BY HAND THE TEST STATISTIC AND COMPARE IT TO THE CRITICAL VALUE TO DRAW YOUR CONCLUSION. THE SAME RESULTS. YOU CAN USE SOFTWARE TO GENERATE A P-VALUE FOR COMPARISON VALUE ONLY NO CREDIT FOR JUST A SOFTWARE GENERATED P-VALUE. ILLOWSKY HAS USEFUL EQUATIONS AROUND PAGES 530-531 FOR THE t-TEST STATISTIC AND THE DEGREES OF FREEDOM (THE LATTER TO GET THE CRITICAL t-VALUE). LANE ALSO HAS PERHAPS A SIMPLER VERSION OF THESE EQUATIONS (CERTAINLY FOR df)..

#6 CAREFUL ! THIS IS A “MATCHED PAIRS” PROBLEM (same group both experiments hence not independent): A GROUP OF 8 STUDENTS (N = 8) TIMED HOW LONG THEY COULD HOLD THEIR BREATH BEFORE AND AFTER EXERCISING. HERE ARE THEIR RESULTS:

BEFORE EXERCISE TIMES (MINUTES): 26, 21, 47, 40, 30, 28, 22, 21 AFTER EXERCISE: 23, 25, 45, 43, 37, 35, 29, 32

Ho: NO DIFFERENCE (MEAN OF TIME 2 – MEAN OF TIME 1 = 0) Ha: (M2 – M1) ≠ 0

ASSUME A SIGNIFICANCE LEVEL: α = 5% or (0.05) BUT SINCE IT’S TWO-TAILED IT’S α/2 = 0.025 IN EACH TAIL

#7 WHICH DEGREE OFFERS THE BEST POTENTIAL STARTING SALARY? A SURVEY OF 50 ENTRY LEVEL PSYCHOLOGY MAJORS WAS COMPARED TO A SURVEY OF 60 (NOT 50) ENTRY LEVEL ENVIRONMENTAL MGMT MAJORS.

LANE AND ILLOWSKY OFFER DIFFERENT FORMULAS FOR THIS PROBLEM (BUT THEY GIVE THE SAME RESULT) CHECK LANE PAGE 408 AND ILLOWSKY AROUND PAGE 530 SECTION 10.1 USE EITHER APPROACH.

PSYCHOLOGY MAJORS: SAMPLE MEAN = $46,100 AND THE SAMPLE STD DEVIATION: $3450

ENMT MAJORS: SAMPLE MEAN = $46,700 AND THE SAMPLE STD DEVIATION: $4210

WE DO NOT HAVE THE POPULATION STANDARD DEVIATIONS SO IS THIS A t OR Z TEST ? (answer)

Ho: MEAN P-MAJ $ ? MEAN ENMT MAJ $ Ha: MEAN P $ ? MEAN ENMT $ (replace “?” appropriately)

ASSUME A 5% (0.05) LEVEL OF SIGNIFICANCE (α=0.05) ONE –TAILED TEST

#8 LET’S TAKE ANOTHER SHOT AT THE COFFEE / NON-COFFEE DRINKING STUDENT AWAKE TIMES THAT WERE THE TOPIC OF OUR WEEK 5 DISCUSSION. IN THAT DISCUSSION WE USED THE CONFIDENCE INTERVAL APPROACH TO TEST OUR HYPOTHESIS THAT COFFEE DRINKING INCREASED AWAKE TIME DURING THE LECTURE.

Here are the times (in minutes) that each of these 15 coffee drinking students was able to stay awake:

33, 60, 45, 80, 120 , 90, 100, 90, 115, 120, 80, 100, 90, 95, 100

Here are the times these 15 non-coffee (in class) drinking students stayed awake:

20,  50, 90, 15, -5 (fell asleep before the lecture even started), 30, 45, 60, 55, 45, 45, 60, 75, 80, 115  

(a) WHAT ARE THE Ho AND Ha FOR THIS ANALYSIS ? ASSUME A SIGNIFICANCE LEVEL OF 5% (α = 0.05)

(b) IS THIS A t-TEST OR z-TEST AND WHY?

(c) WHAT ARE THE SAMPLE MEANS, VARIANCES, AND STANDARD DEVIATIONS ?

(c) WHAT IS THE VALUE OF THE CALCULATED TEST STATISTIC ( AND df IF IT’S A t-TEST)

(d) WHAT IS THE CRITICAL VALUE OF THE t OR z ?

(e) IF YOU CAN (USING SOFTWARE) WHAT IS THE P-VALUE

(f) COMPARING THE TEST STATISTIC TO ITS CRITICAL VALUE, WHAT CONCLUSION DID YOU DRAW?

AND, DID THE P-VALUE EXCEED THE ALPHA OR NOT, HENCE AGREE WITH THE TEST STATISTIC APPROACH?

(g) HOW DID THIS CONCLUSION COMPARE TO THE ONE MADE BASED ON THE CONFIDENCE INTERVALS ?

THAT’S ENOUGH FOR THIS WEEK (DON’T FORGET THE QUIZ !)