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totalpackagehypothesistesting-minitab.docHypothesis Testing
Hypothesis
The formal testing of hypothesis is a critical step not to be ignored. One of the jobs of statistics is to reduce the uncertainty in our decision making. It does so by helping to identify assumptions and check the validity, in essence, what is really there. It is my personal preference to evaluate stocks or mutual funds to see if they actually exceed the average market return of 10%. The 10% return is the historical return on stocks over the past 75 years, however, each year has its own unique return. When doing so, I am interested in did this mutual fund exceed the 10% annual return. When you look at hypothesis testing there are three ways you can set them up. There are as follows:
HO: = H1: not equal Word In story problem: Is Different
Ho: < or = H1: is greater than Word In story problem: more than, increased
HO: > or = H1: Is less than Word In story problem: decreased, dropped
The hypothesis come in “pre-packaged sets” so you can’t mix and match. So when you are setting them up, look at the alternative (H1) in setting them up. Many times this is the decision that interests you the most. In this case, I wanted returns that exceeded ten percent, so I looked at the alternative hypothesis of greater than when setting up the process. What is helpful in thinking about the hypothesis and alternative hypothesis are the unique qualities of the null or HO, hypotheses. The HO hypothesis has unique characteristics. The unique characteristics is that you assume there to either be “no difference” or status quo. So, I set the hypothesis up as follows:
HO: return is less than or equal to ten percent
H1: return is greater than ten percent
For the hypothesis test, I could have chosen to have it set up two other ways and those are:
HO: the return is equal to ten percent
H1; the return is not equal to ten percent
HO: The return is greater than or equal to ten percent
H1: The return is less than ten percent.
What is helpful in thinking about the hypothesis and alternative hypothesis are the unique qualities of the null or HO, hypotheses. The HO hypothesis has unique characteristics. The unique characteristics are that you assume there to either “no difference” or status quo. In the stating of HO there is always an implied equal. To help you understand this concept lets take a trip to the local vending machine.
Hypothesis Test Thinking and a Vending Machine
Imagine yourself walking up to a vending machine. You want to purchase a bottle of Pepsi for a $1.00. You walk up to the vending machine and assume you are going to put your dollar in the machine, press the Pepsi button, and the machine will dispense the bottle of Pepsi. Do you walk up to the machine, knowing that it is not going to give you your bottle of Pepsi after you insert your dollar? If so, would you proceed with the transaction. No you wouldn’t make the transaction and willingly lose your dollar. You assume that as you approach the machine and put your money in that you will receive your Pepsi because that is what has happened time after time, this is the “status quo” segment of the hypothesis. Now, when the machine doesn’t give you your Pepsi but keeps your dollar, you REJECT the hypothesis that this machine is fair and ACCEPT the alternative hypotheses that this machine has stolen my money and is a crooked machine. Now that you know about the formal hypothesis, let’s look at the rest of the procedure to separate fact from fiction.
Remaining Hypothesis Testing Steps
Decision Rule:
The remaining steps are to set up the rules and check our results. We set our rules up ahead of time so as not to be unduly influenced by the results. So, we decide on a level of significance. This is a significant level because it determines the point at which we either accept or reject our null hypothesis. The levels are normally.10, .05 or .01. These are similar to purchasing a Pepsi in a small, medium or large, sorry; we don’t have a super size in statistics. When choosing the levels of significance, one can look at industry norm or you can choose any one of the three. Normally, the .05 level of significance is chosen because this is middle ground. So we chose the .05 level of significance.
Test Statistic
The next step is to calculate our “test statistic”. This is the statistic that our computer software generates for us. This is most easily expressed in probability or what is normally called the “P” value. The “P” stands for probability and can be thought of this way. If the “true average”, the average that is really there, not our sample results, the probability of getting a sample average which we found, is some value. Lets say that we were testing the hypothesis that our mutual fund return is less than or equal to ten percent and we input our data, test the hypothesis at the .05 level of significance. The computer software calculates a “P value” of .15. A .15 value is greater than the .05 level of significance and so we don’t reject our hypothesis and conclude the stock we are looking at has a return less than or equal to 10%. However, if the “ P value” calculated from the computer program was .04; we would have rejected the null hypothesis and concluded that the average return of this mutual fund was greater than ten percent.
Side Note: if the “P or probability value calculated “is less than the level of significance, we reject the null hypothesis and accept the alternative. We do this because we are saying that if the null hypothesis is true, the probability of getting a sample average with this result is less than the decision rule statistic.
Decision:
The decision is where you compare the probability value calculated against the critical probability value set in your decision rule. If your test probability value is less than the critical, reject the null hypothesis and accept the alternative hypothesis. The P value testing allows for you to be slightly confident, if is just on the border line of accept or reject to very confident if the probability value is very small such as .001 (1/1000)
You may need to double check your values and if unsure, you can ask yourself do I want a 5% raise or do I want a .1% raise. This is a simple way of keeping your decision correct and not confusing the decision because you misinterpreted the p values size.
Traditional Approach
In 2004, a small dealership leased 21 Chevrolet Impalas on two year leases. When the cars were returned in 2006, the mileage recorded was (see table). Is the dealers mean significantly greater than the national average of 30,000 miles for two year leased vehicles at the 10% level of significance?
T value = 21-1 = 20 , one tailed test, .10 level of significance = 1.325
HO: The average miles is less than or equal to 30,000
H1: The average miles are greater than 30,000
Decision Rule: If the T value calculated is greater than 1.325, reject HO
Test Statistic: T= ( 33,950-30,000)/ (11,866/√21)
= ( 33,950-30,000)/ (11,866/4.58)
= 3,950/ 2,590
T value =1.52
Decision: Since the Test statistic of 1.52 is greater than the T critical value of 1.325, we reject HO and conclude the cars have more mileage.
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P value Approach
HO: The average miles is less than or equal to 30,000
H1: The average miles are greater than 30,000
Decision Rule: If the P value calculated is less than .10, reject HO
Test Statistic: .071
Decision: Since the Test statistic of .071 is less than the P critical value of .10, we reject HO and have weak evidence to conclude the cars have more mileage. For if the HO was true that the cars had less than or equal to 30,000 miles, the probability that you would have randomly found a sample average of 33,950 is 7.1%, therefore, it is possible but not probable.
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Box Plot Explanation: The red HO circle is the null hypothesis value. The blue line is the confidence interval and the xbar is the sample average from which a 90% confidence interval was created. Since the HO value is contained in the confidence interval, it is possible that it could be the “real mean”.
Data
|
40,060 |
24,960 |
14,310 |
17,370 |
44,740 |
44,550 |
20,250 |
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33,380 |
24,270 |
41,740 |
58,630 |
35,830 |
25,750 |
28,910 |
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25,090 |
43,380 |
23,940 |
43,510 |
53,680 |
31,810 |
36,780 |
Traditional Method
Investment Company of America
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One Mutual Fund Scenario
.05 Level of Significance
HO: The average return on the Investment Company of America is equal to 10%
H1: The average return on the Investment Company of America is not equal to 10%
Decision Rule: If the Z value is less than -1.96 or greater than 1.96, reject HO
Test Statistic: Z = (14.52 – 10)/(18.668/√71)
Z = 4.52/2.21
Z = 2.04
Decision: Since the the Z value calculated of 2.04 is greater than the Z critical of 1.96 we reject HO and conclude the fund annual return is different than 10%
Since this was a two tailed test, the .05 reject region was divided equally on both sides, thus is gives a .025 rejection zone on each side. Since 50% probability is on each side of the bell curve, we calculate the z value by 50% - 2.5% = 47.5%
.4750 = Z value of 1.96
NOTES:
2.21 is the standard error of the mean, when the standard deviation is adjusted to reflect sample size based upon the theory of the law of large numbers that as sample size increases, variation decreases.
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Suppose we are interested now in does the fund beat the historical average of 10%. It will take the following steps.
Note: This is a one tailed test and you need to remember which is the positive side and negative side of the bell curve with Left is negative, right is positive
HO: The average return on the Investment Company of America is less than or equal to 10%
H1: The average return on the Investment Company of America is greater than 10%
Decision Rule: If the Z value is greater than Z 1.65, reject HO
Test Statistic: Z = (14.52 – 10)/(18.668/√71)
Z = 4.52/2.21
Z = 2.04
Decision: Since the the Z value calculated of 2.04 is greater than the Z critical of 1.65,we reject HO and conclude the fund annual return is greater than 10%
Since this was a one tailed test, the .05 reject region is located on one side of the bell curve, thus to gain a Z value we subtract our level of significance, in this case .05 thus 50% - 5% = 455%
.45 = Z value of 1.645
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Suppose we are interested now in does the fund less the historical average of 10%. It will take the following steps.
HO: The average return on the Investment Company of America is greater than or equal to 10%
H1: The average return on the Investment Company of America is less than 10%
Decision Rule: If the Z value is less than Z -1.65, reject HO
Test Statistic: Z = (14.52 – 10)/(18.668/√71)
Z = 4.52/2.21
Z = 2.04
Decision: Since the the Z value calculated of 2.04 is greater than the Z critical of
- 1.65,we fail reject HO and conclude the fund annual return is greater than or equal to 10%
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P Value Approach
The formal steps look like this:
One Mutual Fund Scenario
One-Sample Z: Investment Company of America
Test of mu = 10 vs not = 10
The assumed standard deviation = 18.668
Variable N Mean StDev SE Mean 95% CI Z
Investment Company of Am 71 14.52 18.67 2.22 (10.18, 18.86) 2.04
Variable P
Investment Company of Am 0.041
HO: The average return on the Investment Company of America is equal to 10%
H1; The average return of the Investment Company of America of is not equal to 10%
Decision Rule: If the P value is less than .05 reject the HO
Test Statistic: .041
Decision: Since the probability statistics calculated of .041 is less than the critical probability value of .05, we reject the null hypothesis and are somewhat confident that the return on this fund is not equal to 10%. For if the HO was true that the funds real average is10%, the probability that you would have randomly found a sample average of 14.51 is 4.1%, therefore, it is possible but not probable
NOTES
One-Sample Z: Investment Company of America
Test of mu = 10 vs > 10
The assumed standard deviation = 18.668
95% Lower
Variable N Mean StDev SE Mean Bound Z P
Investment Company of Am 71 14.52 18.67 2.22 10.88 2.04 0.021
Suppose we are interested now in does the fund beat the historical average of 10%. It will take the following steps.
HO: The average return on the Investment Company of America is less than or equal to 10%
H1: The average return on the Investment Company of America is greater than 10%
Decision Rule: If the P value calculate is less than .05, reject HO
Test Statistic: .021
Decision: Since the probability statistics calculated of .021 is less than the critical probability value of .05, we reject the null hypothesis and are somewhat confident that the return on this fund is greater than 10%. For if the HO was true that the funds real average is less than or equal to10%, the probability that you would have randomly found a sample average of 14.51 is 2.1%, therefore, it is possible but not probable
NOTES
One-Sample Z: Investment Company of America
Test of mu = 10 vs < 10
The assumed standard deviation = 18.668
95% Upper
Variable N Mean StDev SE Mean Bound Z P
Investment Company of Am 71 14.52 18.67 2.22 18.16 2.04 0.979
Suppose we are interested now in does the fund earns less than the historical average of 10%. It will take the following steps.
HO: The average return on the Investment Company of America is greater than or equal to 10%
H1: The average return on the Investment Company of America is less than 10%
Decision Rule: If the P value calculate is less than .05, reject HO
Test Statistic: .979
Decision: Since the probability statistics calculated of .979 is greater than the critical probability value of .05, we fail to reject the null hypothesis and are very confident that the return on this fund is greater than or equal to 10%. For if the HO was true that the funds real average is greater than or equal 10%, the probability that you would have randomly found a sample average of 14.51 is 97.9%, therefore, it is possible and probable.
NOTES:
Hypothesis Testing Notes
E-Tickets
Most air traffic passenger use e-tickets. Electronic ticketing allows passengers not worry about paper tickets and it reduces the airlines costs. However in recent times airlines have begun to receive complaints from passengers regarding e-tickets, especially when having to switch planes. To investigate this problem an independent watchdog agency contacted a random sample of 21 airports and collected data on the number of complaints due to e-tickets.
At the .05 level of significance is there evidence to conclude that the ten year average of 15 complaints per month has now changed and actually increased due to electronic ticketing?
Problem Statement:
HO:
H1
Decision Rule
Test Statistic:
Decision
NOTES
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E-Tickets
Most air traffic passenger use e-tickets. Electronic ticketing allows passengers not worry about paper tickets and it reduces the airlines costs. However in recent times airlines have begun to receive complaints from passengers regarding e-tickets, especially when having to switch planes. To investigate this problem an independent watchdog agency contacted a random sample of 21 airports and collected data on the number of complaints due to e-tickets.
At the .05 level of significance is there evidence to conclude that the ten year average of 15 complaints per month has now changed and actually increased due to electronic ticketing?
Problem Statement:
The airline industry has undergone significant changes including the movement to e-tickets to reduce costs of paper tickets as well as allow for internet booking and reservation. However, given the tight markets and strong competition, the airline knows it needs to minimize consumer complaints as this translates into customer defection and loss of revenue.
T test is one tailed (.05) with sample size 21 – 1 = 20 degrees of freedom = 1.725
HO: Complaints are less than or equal to 15
H1: Complaints are greater than 15
Decision Rule If T is > 1.725, reject HO
Test Statistic: Test Statistic: T = (13.751 – 15)/(1.502/√21)
T= -1.429/.3277
T = -4.36
Decision: Since T calculated of -4.36 is less than T critical of 1.725, we fail to reject HO.
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E-Tickets
Most air traffic passenger use e-tickets. Electronic ticketing allows passengers not worry about paper tickets and it reduces the airlines costs. However in recent times airlines have begun to receive complaints from passengers regarding e-tickets, especially when having to switch planes. To investigate this problem an independent watchdog agency contacted a random sample of 21 airports and collected data on the number of complaints due to e-tickets.
At the .05 level of significance is there evidence to conclude that the ten year average of 15 complaints per month has now changed and actually increased due to electronic ticketing?
Problem Statement:
The airline industry has undergone significant changes including the movement to e-tickets to reduce costs of paper tickets as well as allow for internet booking and reservation. However, given the tight markets and strong competition, the airline knows it needs to minimize consumer complaints as this translates into customer defection and loss of revenue.
HO: Complaints are less than or equal to 15
H1: Complaints are greater than 15
Decision Rule: If P calculated is less than .05, reject HO
Test Statistic: 1.00 actually is .9999 but was rounded up to one
Decision: Since the probability value calculated of .9999 is greater than .05, we fail to reject HO and conclude there is strong evidence to support that the number of complaints has not increased. For if the HO was true that the funds real average of complaints are less than or equal to 15, the probability that you would have randomly found a sample average of 13.571 is 99.9%, therefore, it is possible and very probable.
Test of mu = 15 vs > 15
95% Lower
Variable N Mean StDev SE Mean Bound T P
E-Ticket 21 13.571 1.502 0.328 13.006 -4.36 1.000
Chicken Feed
Raising chickens in commercial chicken farms is a growing industry, as there is a shift from red meat to more white meat. New Jersey Red Chickens are a favorite chicken and the Feed is Us company has a new chicken food that they claim is excellent for increasing weight and naturally it costs more. You have heard this line before and therefore are skeptical but the potential pay off is huge. You buy a small amount and feed it to ten chickens, which you choose at random.
At the .01 level of significance is there evidence that the chicken weight exceeds the average chicken weight of 4.35 pounds.
Problem Statement:
HO:
H1
Decision Rule
Test Statistic:
Decision
Notes:
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Chicken Feed
Raising chickens in commercial chicken farms is a growing industry, as there is a shift from red meat to more white meat. New Jersey Red Chickens are a favorite chicken and the Feed is Us company has a new chicken food that they claim is excellent for increasing weight and naturally it costs more. You have heard this line before and therefore are skeptical but the potential pay off is huge. You buy a small amount and feed it to ten chickens, which you choose at random.
At the .01 level of significance is there evidence that the chicken weight exceeds the average chicken weight of 4.35 pounds .
Problem Statement:
HO: The chicken weight is less than or equal to 4.35 pounds
H1: The chicken weight is greater than 4.35 pounds
Decision Rule If T is > 2.821, reject HO
T test is one tailed (.01) with sample size 10 – 1 = 9 degrees of freedom = 2.821
Test Statistic: Test Statistic: T = (4.368 – 4.35)/(.0339/√10)
T= .018/.0107
T = 1.68
Decision: Since T calculated of 1.68 is less than T critical of 2.821, we do not reject HO.
Notes:
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Chicken Feed
Raising chickens in commercial chicken farms is a growing industry, as there is a shift from red meat to more white meat. New Jersey Red Chickens are a favorite chicken and the Feed is Us company has a new chicken food that they claim is excellent for increasing weight and naturally it costs more. You have heard this line before and therefore are skeptical but the potential pay off is huge. You buy a small amount and feed it to ten chickens, which you choose at random.
At the .01 level of significance is there evidence that the chicken weight exceeds the average chicken weight of 4.35 pounds .
Problem Statement:
HO: The chicken weight is less than or equal to 4.35 pounds
H1: The chicken weight is greater than 4.35 pounds
Decision Rule: If P calculated is less than .01, reject HO
Test Statistic: .064
Decision: Since the P value calculated of .064 is greater than .01, we fail to reject HO and are only somewhat confident the real chicken weight is less than or equal to 4.35 pounds For if the HO was true that the real average chicken weigh is less than or equal to 4.35, the probability that you would have randomly found a sample average of 4.368 is 6.4%, therefore, it is possible and probable.
Issue: small sample size, law of large numbers as sample size increases variation decreases.
Test of mu = 4.35 vs > 4.35
Small sample law of large number
99% Lower
Variable N Mean StDev SE Mean Bound T P
Chicken Weights 10 4.3680 0.0339 0.0107 4.3377 1.68 0.064
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Confidence interval contains the null hypothesis, so it is possible it contains the real average.
Notes:
AARP Work Survey
The American Association of Retired Persons (AARP) reports that 60% of retired persons under the age of 65 would be willing to return to work on a full time basis if a suitable job were available. A sample of 500 retired persons under the age of 65 revealed 315 would return to work. At the .05 level of significance can we conclude that more than 60% of retired people in the age group would return to work?
Problem Statement:
In today’s economy, each person is responsible for their own retirement investment decisions. Many retirees are forced to return to work to pay bills since their retirement savings are inadequate and there are also those who return to work for socialization and wanting to continue to make contributions to society. The question is today are more senior citizens returning to work than in previous generations? This has profound social and economic ramifications.
HO:
H1
Decision Rule
Test Statistic:
Decision
Notes:
AARP Work Survey
The American Association of Retired Persons (AARP) reports that 60% of retired persons under the age of 65 would be willing to return to work on a full time basis if a suitable job were available. A sample of 500 retired persons under the age of 65 revealed 315 would return to work. At the .05 level of significance can we conclude that more than 60% of retired people in the age group would return to work?
Problem Statement:
In today’s economy, each person is responsible for their own retirement investment decisions. Many retirees are forced to return to work to pay bills since their retirement savings are inadequate and there are also those who return to work for socialization and wanting to continue to make contributions to society. The question is today are more senior citizens returning to work than in previous generations? This has profound social and economic ramifications.
Z value is used since it is a proportion
HO: The proportion of retired persons returning to work is less than or equal to 60%
H1: The proportion of retired persons returning to work is greater than 60%
Decision Rule: If z calculated is greater than 1.645, reject HO
Test Statistic: Z = .63-.60/ (√.63*.37/500)
Z = 1.38
315/500= .63
1-.63 = .37
Decision Since the Z calculated of 1.38 is less than the critical value of 1.645, we fail to reject HO and conclude the average is still less than or equal to 60%.
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AARP Work Survey
The American Association of Retired Persons (AARP) reports that 60% of retired persons under the age of 65 would be willing to return to work on a full time basis if a suitable job were available. A sample of 500 retired persons under the age of 65 revealed 315 would return to work. . At the .05 level of significance can we conclude that more than 60% of retired people in the age group would return to work?
Problem Statement:
In today’s economy, each person is responsible for their own retirement investment decisions. Many retirees are forced to return to work to pay bills since their retirement savings are inadequate and there are also those who return to work for socialization and wanting to continue to make contributions to society. The question is today are more senior citizens returning to work than in previous generations? This has profound social and economic ramifications.
HO: The proportion of retired persons returning to work is less than or equal to 60%
H1: The proportion of retired persons returning to work is greater than 60%
Decision Rule: If P calculated is less than .05, reject HO
Test Statistic: .085
Decision: Since the P value calculated of .085 is greater than .05, we fail to reject HO and are only somewhat confident the real percentage of senior citizens returning to work is less than or equal to 60%. For if the HO was true that the number of people willing to return to work is less than or equal to 60%, the probability that you would have randomly found a sample average of 63% is 8.5%, therefore, it is possible and probable
AARP Work Survey
Test of p = 0.6 vs p > 0.6
95%
Lower
Sample X N Sample p Bound Z-Value P-Value
1 315 500 0.630000 0.594485 1.37 0.085
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Hypothesis Test Directions for Minitab
Step One Sample Z Hypothesis Test
Begin by clicking on stat, basic stat then on One Sample Z
Step Two for Z Hypothesis Test
1. Click on the column of data to be used: Investment Company of America
2. Enter the standard deviation value calculated from the graphical summary of the data which is 18.66
3. Click on perform hypothesis test
4. Enter the value for the test 10
5. Click on the Options Box
Step Three for Z Hypothesis Test
· Click on options
· Enter the alternative hypothesis test which in this example is greater than
· Enter the confidence interval value that corresponds with the level of significance in this example it is 95%
Step Four for Z Hypothesis Test
· Click on graphs
· Check the boxplot of data or other graph of your choosing.
· Click OK
Step Five for Z Hypothesis Test
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Click OK to run the test
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Step Six for Z Hypothesis Test
Results of the test
Results of the test are now showing in the session window and you can copy and paste them into a graph or word document.
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Directions for a One Sample T Test
Step 1
Enter your data and give them column headings for ease of use.
……..
Click on the menu and if you have a sample size smaller than 30 click on the one sample T or if your standard deviation is calculated from your sample data, you will use a one sample T test. One sample means you are only comparing one average calculated from the of data against the historical or status quo mean.
Step 2
Now that you have chosen the test, you will need to indicate where the data is found by clicking on the column where the data is located, unless you have what is called summarized data. Summarized data is when you have already calculated or been given the sample size, sample average and sample standard deviation.
If your data is in a column such as the example, click the data column heading to move it into the box.
If summarized data, where the mean, standard deviation and sample size are given, then click on the “summarized data” dot and enter the information into the boxes.
Step 3
· In this step you will need to check the hypothesis test box and then
· Enter the value tested in this case it is 4.35 pounds.
· Click Options
Step 4
· Enter the confidence interval level that corresponds with the level of significance. In this case our level of significance is .05, so the corresponding level of significance is 95%
· Select the correct H1
· Click on OK
Step 5
· In this step you will choose the graph options that would be most helpful to you. You can check all three and then look at the options as they appear to decide which is most helpful. In this case we choose a boxplot
· Click on boxplot
Step 6
Session Window with Results
· The hypothesis test listed with the alternative hypothesis being referred to as the “vs not”.
· The P value associated with the hypothesis test and other values are listed in the “”Variable output” section of this view.
Step 7
Graphical Analysis
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In graphical analysis you are looking to see if the HO is contained in the confidence interval as if the confidence interval is contained, you do not reject HO because it is possible that it could be the “real average”.
Summarized Data
Step One ( if given, skip this step and go to step two)
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Prior to using summarized data you can find the information in a graphical summary but most of the time it is given.Step Two
Enter the sample size, mean and standard deviation along with the hypothesized value.
Step Three
· Enter the confidence interval that corresponds with the level of significance
· Enter the H1
Click ok, since our data was summarized, we can’t do graph as we lack the individual data points.
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E-ticket |
Chicken Weights |
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Year ICA |
Investment Company of America |
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14 |
4.41 |
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2004 |
19.8 |
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14 |
4.37 |
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2003 |
26.3 |
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16 |
4.33 |
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2002 |
-14.47 |
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12 |
4.35 |
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2001 |
-4.05 |
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12 |
4.30 |
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2000 |
3.84 |
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14 |
4.39 |
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1999 |
16.21 |
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13 |
4.36 |
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1998 |
22.93 |
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15 |
4.38 |
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1997 |
29.81 |
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16 |
4.40 |
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1996 |
19.35 |
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15 |
4.39 |
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1995 |
30.62 |
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14 |
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1994 |
0.2 |
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12 |
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1993 |
11.6 |
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15 |
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1992 |
7 |
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15 |
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1991 |
26.5 |
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14 |
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1990 |
0.7 |
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13 |
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1989 |
29.4 |
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13 |
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1988 |
13.3 |
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12 |
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1987 |
5.4 |
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13 |
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1986 |
21.7 |
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10 |
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1985 |
33.4 |
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13 |
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1984 |
6.7 |
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1983 |
20.2 |
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1982 |
33.8 |
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1981 |
0.9 |
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|
|
1980 |
21.2 |
|
|
|
|
1979 |
19.2 |
|
|
|
|
1978 |
14.7 |
|
|
|
|
1977 |
-2.6 |
|
|
|
|
1976 |
29.6 |
|
|
|
|
1975 |
35.4 |
|
|
|
|
1974 |
-17.9 |
|
|
|
|
1973 |
-16.8 |
|
|
|
|
1972 |
15.9 |
|
|
|
|
1971 |
17 |
|
|
|
|
1970 |
2.6 |
|
|
|
|
1969 |
-10.7 |
|
|
|
|
1968 |
17 |
|
|
|
|
1967 |
28.9 |
|
|
|
|
1966 |
1 |
|
|
|
|
1965 |
26.9 |
|
|
|
|
1964 |
16.3 |
|
|
|
|
1963 |
22.9 |
|
|
|
|
1962 |
-13.2 |
|
|
|
|
1961 |
23.1 |
|
|
|
|
1960 |
4.5 |
|
|
|
|
1959 |
14.2 |
|
|
|
|
1958 |
44.8 |
|
|
|
|
1957 |
-11.9 |
|
|
|
|
1956 |
10.8 |
|
|
|
|
1955 |
25.4 |
|
|
|
|
1954 |
56.1 |
|
|
|
|
1953 |
0.4 |
|
|
|
|
1952 |
12.2 |
|
|
|
|
1951 |
17.8 |
|
|
|
|
1950 |
19.8 |
|
|
|
|
1949 |
9.4 |
|
|
|
|
1948 |
0.4 |
|
|
|
|
1947 |
0.9 |
|
|
|
|
1946 |
-2.4 |
|
|
|
|
1945 |
36.8 |
|
|
|
|
1944 |
23.3 |
|
|
|
|
1943 |
32.8 |
|
|
|
|
1942 |
16.8 |
|
|
|
|
1941 |
-7.4 |
|
|
|
|
1940 |
-2.4 |
|
|
|
|
1939 |
0.8 |
|
|
|
|
1938 |
27.6 |
|
|
|
|
1937 |
-38.5 |
|
|
|
|
1936 |
45.8 |
|
|
|
|
1935 |
83.1 |
|
|
|
|
1934 |
18.2 |
|
|
|
|
|
|
|
|
|
|
|
|
Fail to Reject
Test Stat 1.52
Reject > 1.325
Key Pieces for Formula
Negative side
Positive side
Key Data
HO value status quo
Standard deviation and sample size
Sample average
Fail to Reject
Test Stat 2.04
< - 1.96 Reject
Reject >1.96
HO value status quo
Standard deviation and sample size
Sample average
Fail to Reject
Test Stat 2.04
Reject >1.645
HO value status quo
Standard deviation and sample size
Sample average
Fail to Reject
Test Stat 2.04
Reject < -1.645
Lower Bound
P value
H1
Sample average
Reject >1.725
Fail to Reject
Test Stat -4.36
H1
Sample average
Test Stat 1.68
>2.821 Reject
Fail to Reject
Lower end of confidence interval
Sample average
Null
Test Stat 1.38
Fail to Reject
>1.645 Reject
H1
Stat
basic stat
Click on one sample Z
HO value
Standard deviation from graphical summary
Column of data
H1
Level of significance
Graph choice
Click OK
One sample T
Check circle if summarized data
Summarized data entry
Click on correct column of data
Options
HO value
Check box
Choose the H1
Enter confidence interval level
Graph options
Alternative Hypothesis listed
P value calculated
3rd quartile
1st quartile
Median
Sample average
Lower end of confidence interval
Null hypothesis
Sample size
Mean, standard deviation
Enter summarized data
HO
Choose H1
Choose level of significance
