A manufacturer of muesli bars needs to describe the average fat content of the bars (the mean of the hypothetical population of fat contents that would be produced using the recipe). Several bars are analysed and their fat contents are measured.

The sample mean is a point estimate of the population mean, and a 95% confidence interval can also be found.

A particular brand of meusli bar is claimed by the manufacturer to have a fat content of 3.4g per bar. A consumer group suspects that the manufacturer is understating the fat content, so a random sample of bars is analysed.

The consumer group must assess whether the data are consistent with the statement (hypothesis) that the underlying population mean is 3.4g.

English Premier Soccer League, 2008/09

In the English Premier Soccer league, each of the 20 teams plays every other team twice (home and away) during the season. Three points are awarded for a win and one point for a draw. The table below shows the wins, draws, losses and total points for all teams at the end of the 2008/09 season.

We observed in an earlier simulation that there is considerable variability in the points, even when all teams are evenly matched. However, ...

A simulation

To assess whether there is any difference in skill levels, we can therefore run a simulation of the league, assuming evenly matched teams and generating random results with probabilities 0.372, 0.372 and 0.255 for wins, losses and draws. (A proportion 0.255 of games in the actual league resulted in draws.)

Click Simulate to simulate the 380 games in a season. The standard deviation of the final points is shown below the table. Click Accumulate then run the simulation about 100 times. (Hold down the Simulate button to speed up the process.)

The standard deviation of the points in the actual league table was 18.2. Since most simulated standard deviations are between 5 and 12, we conclude that such a high spread would be extremely unlikely if the teams were evenly matched.

Does Australia Post deliver on time?

The Herald-Sun newspaper published the following article on November 25 1992.

Campbell Fuller, Herald-Sun, 25 November 1992.

Is the author justified in disputing Australia Post's claim that 96% of letters are delivered on time?

A simulation

If Australia Post's claim is correct, and every letter independently has probability 0.96 of being delivered on time, we know that the number delivered on time out of 59 letters will be a random quantity. From the information in the article, we can deduce that 52 out of the Herald-Sun's 59 letters arrived on time (a proportion 52/59 = 0.881).

A simulation helps to answer this question.

Click Simulate to randomly 'deliver' 59 letters, with each independently having probability 0.96 of arriving on time. Click Accumulate then run the simulation between 100 and 200 times. (Hold down the Simulate button to speed up the process.)

Observe the distribution of the number of letters arriving on time. The proportion of simulations with 52 or fewer letters arriving on time is shown to the right of the dot plot. Observe that this rarely happens.

We will return to this example later.

Quality control for cornflake packets

In a factory producing packets of cornflakes, the weight of cornflakes that a filling machine places in each packet varies from packet to packet. From extensive previous monitoring of the operation of the machine, it is known that the net weight of '500 gm' packets is approximately normal with standard deviation σ = 10 gm.

The mean net weight of cornflakes in the packets is controlled by a single knob. The target is for a mean of µ = 520 gm to ensure that few packets will contain less than 500 gm. Samples are regularly taken to assess whether the machine needs to be adjusted. A sample of 10 packets was weighed and contained an average of 529 gm. Does this indicate that the underlying mean has drifted from µ = 520 and that the machine needs to be adjusted?

A simulation

If the filling machine is working to specifications, each packet would contain a weight that is sampled from a normal distribution with µ = 520 and σ = 10.

A simulation helps to answer this question.

Click Simulate to randomly generate the weights of 10 packets from a normal (µ = 520, σ = 10) distribution. Click Accumulate then run the simulation between 100 and 200 times. (Hold down the Simulate button to speed up the process.)

Observe that although many of the individual cornflake packets weigh more than 529 gm, it is rare for the mean weight to be as far from the target as 529 gm (i.e. either ≥529 gm or ≤511 gm).

We will return to this example later.

Characteristics of failed companies

A study in Greece compared characteristics of 68 healthy companies with those of another 33 that had recently failed. The jittered dot plots on the left below show the ratio of current assets to current liabilities for each of the 101 companies.

The mean asset-to-liabilities ratio for the sample of failed companies is 0.902 lower than that for the healthy companies, but the distributions overlap. Might this difference be simply a result of randomness, or can we conclude that there is a difference in the underlying populations?

Click Randomise to randomly pick 33 of the the 101 values for the failed group. If the underlying distribution of asset-to-liabilities ratios was the same for healthy and failed companies, each such randomised allocation would be as likely as the observed data.

Click Accumulate and repeat the randomisation several more times. Observe that the difference in means would rarely be as far from zero as -0.902 when we assume the same distribution for both groups. This strongly suggests that the distributions must be different.

Since the actual difference is so unusually large, ...

English Premier Soccer League, 2007/08 and 2008/09

We saw earlier that the distribution of points in the 2008/09 English Premier Soccer League Table was not consistent with all teams being evenly matched — the spread of points was too high. We will now investigate this further.

If some teams are better than others, the positions of teams in the league in successive years will tend to be similar. The table below shows the points for the teams in two seasons. (Note that the bottom three teams are relegated each year and three teams are promoted from the lower league, so we cannot compare the positions of six of the teams.)

Manchester United, Chelsea, Arsenal and Liverpool were the top four teams in both years. However, ...

Randomisation

If all other teams have equal probabilities of winning against any opponent, the 2008/09 points of 45 (which was actually obtained by Wigan) would have been equally likely to have been obtained by any of the teams in that year. Indeed, any allocation of the points (63, 62, 41, ..., 53) to the teams (Everton, Aston Villa, Blackburn, ..., Fulham) would be equally likely.

The diagram below performs this randomisation of the results in 2008/09.

Click Randomise to shuffle the 2008/09 points between the teams (excluding the top four teams and those that were only in the league for one of the seasons). If the teams were of equal ability, these points would have been as likely as the actual ones.

The correlation coefficient between the points in the two seasons gives an indication of how closely they are related. Click Accumulate and repeat the randomisation several more times. Observe that the correlation for the randomised values is only as far from zero as the actual correlation (r = 0.537) in about 5% of randomisations. Since a correlation as high as 0.537 is fairly unusual for equally-matched teams, ...

Use the pop-up menu below to check how the earlier examples in this section fit into the hypothesis testing framework.

Soccer league in one season Australia Post deliveries Net weight of corn flake packets Characteristics of failed companies Soccer leagues in two seasons

Australia Post example

A journalist trying to assess Australia Post's assertion that 96 percent of letters arrive 'on time' posted 59 letters and observed that only 52 arrived on time.

We model delivery of these letters as a random sample of 59 categorical values from a population with probability π of success (arrival on time). The null hypothesis of interest is therefore...

H₀:   π = 0.96

The alternative hypothesis is

H_A:   π < 0.96

Design of CD case

A company intends to manufacture a case that will hold 20 CDs. The design team are particularly keen on a design that is more expensive to manufacture than two competing designs. The manager wants to be sure that customers will prefer the more expensive case before starting production — the price is determined by competitors' CD cases so the more expensive one will have a reduced profit margin and can only be justified if sales are considerably higher.

To assess whether customers prefer the more expensive case, a limited number of each of the three designs is manufactured and placed for sale at the same price in a CD store. Out of the first 90 cases sold, the more expensive case was bought 36 times.

This situation can be modelled as random sampling of 90 values (the three case designs) from a categorical population in which the probability of picking the expensive case is π. The null hypothesis of interest is therefore...

H₀:   π = ¹/₃       (no preference)

The alternative hypothesis is

H_A:   π > ¹/₃       (preference for the expensive case)

Australia Post example

A journalist trying to assess Australia Post's assertion that 96 percent of letters arrive 'on time' posted 59 letters and observed that only 52 arrived on time.

H₀:   π = 0.96

H_A:   π < 0.96

In the diagram below, click Accumulate then hold down Simulate until about 100 samples of 59 letters have been generated. The proportion of these simulated samples in which 52 or fewer letters arrived on time is an approximation to the p-value for the test.

Since we know that the number arriving on time has a binomial (52, 0.96) distribution when the null hypothesis holds, the simulation is unnecessary. Select Binomial distribution from the pop-up menu. This binomial distribution is displayed, and the probability of 52 or fewer letters being delivered on time is shown to be 0.009 — the p-value for the test.

Since the p-value is so small, there would have been very little chance of the observed data arising if Australia Post's assertion had been correct. We can therefore conclude that there is strong evidence against their assertion. Note that this can be done without any simulations.

Design of CD case

In the trial of three CD cases that was described at the start of this section, all three were offered for sale at the same price in a CD store. Out of the first 90 cases that were purchased, 36 chose the design that was more expensive to manufacture than the other two. Since more than a third chose this case design, is there strong evidence that customers prefer it?

The null and alternative hypotheses are...

H₀:  π = ¹/₃       (no preference)

H_A:  π > ¹/₃       (preference for the expensive case)

The p-value is the probability of 36 or more expensive cases being purchased when π = ¹/₃. This can be obtained directly from a binomial distribution with π = ¹/₃ and n = 90.

Use the slider below to obtain the p-value for this test.

The p-value for the test is 0.1103, meaning that there is a probability of 0.1103 of the expensive case being purchased 36 of more times even if there is no real preference for it. We therefore conclude that there is no evidence of any preference from the data.

Ethics codes in companies

In 1999, The Conference Board surveyed 124 companies and found that 97 had their own ethics codes ("Business Bulletin", Wall Street Journal, Aug 19, 1999). In 1997, it was believed that 72% of companies had ethics codes, so is there any evidence that the proportion has changed?

This question is equivalent to asking whether a sample proportion of 97 out of 124 is consistent with sampling from a population with π = 0.72. This can be expressed as the hypotheses

We would expect about (0.72 x 124) = 89 of the companies to have ethics codes. A sample count that is either much greater than 89 or much less than 89 would suggest that the probability had changed. Use the slider below to obtain the p-value.

The probability of getting as many as 97 is 0.0718. Since this is a 2-tailed test, we must also take account of the probability of getting a count that is as unusually low, so the p-value is twice this, 0.1436. Getting 97 companies with ethics codes is therefore not unlikely, so we conclude that there is no evidence from these data of a change in the proportion of companies with ethics codes since 1997.

Home-based businesses owned by women

A recent study that was reported in the Wall Street Journal sampled 899 home-based businesses and found that 369 were owned by women.

Are home-based businesses less likely to be owned by females than by males? This question can be expressed as a hypothesis test. If the population proportion of home-based businesses owned by females is denoted by π, the hypotheses can be written as...

If the null hypothesis is true, the sample number owned by females will have a binomial distribution with parameters n = 899 and π = 0.5. The p-value for the test is therefore the sum of binomial probabilities,

A lot of probabilities must be evaluated and summed! And all are close to zero.

Home-based businesses owned by women

In this example, the sample size, n = 899 is large, so we can use a normal approximation to obtain the probability of 369 or fewer businesses owned by females if the underlying population probability was 0.5 (the null hypothesis).

Click Accumulate then simulate sampling of 899 businesses about 300 times. (Hold down the button Simulate.) From the simulation, it is clear that the probability of obtaining 369 or fewer businesses owned by females is extremely small — there is strong evidence against the null hypothesis.

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The same conclusion can be reached without a simulation.

Select Bar chart from the pop-up menu, then select Normal approximation. From the normal approximation, we can determine that the p-value for the test (the tail area below 369) is extremely close to zero.

Home-based businesses owned by women

The diagram below repeats the simulation that we used earlier to test whether the proportion of home-based businesses owned by women was less than 0.5:

The proportion owned by women in a sample of n = 899 businesses was ³⁶⁹/₈₉₉ = 0.410.

Again click Accumulate and hold down the Simulate button until about 100 samples of 899 businesses have been generated with a population probability of being owned by women of 0.5.

Select Statistical distance from 0.5 from the top pop-up menu to translate the proportions of female owners in the simulated samples into z-scores. Observe that most of these 'statistical distances from 0.5' are between -1 and +1.

The observed proportion owned by females was 0.410, corresponding to a statistical distance of z = -5.37, an unlikely value if the population proportion was 0.5.

Select Normal distribution from the lower pop-up menu to show the theoretical distribution of the z-scores. The p-value for the test is the tail area of this normal(0, 1) distribution below -5.37 and is virtually zero, so we again conclude that:

Blood pressure of executives

The medical director of a large company looks at the medical records of 72 male executives aged between 35 and 44 and observes that their mean blood pressure is  = 126.07. We model these 72 blood pressures as a random sample from an underlying population with mean µ (blood pressures of similar executives) .

Published national health statistics report that in the general population for males aged 35-44, blood pressures have mean 128 and standard deviation 15. Do the executives conform to this population? Focusing on the mean of the blood pressure distribution, this can be expressed as the hypotheses,

Filling milk containers

In a bottling plant, plastic containers are filled with a nominal 2 litres of milk. However the containers are filled so quickly that it is impossible to ensure that each contains exactly 2 litres. The volume of milk in a container is approximately normally distributed with standard deviation 0.005 litres, and the machinery is adjusted to give a mean volume of 2.012 litres. (Using the normal distribution, you can check that only 1% of containers should contain less than the nominal 2 litres of milk.)

At regular intervals, ten containers are sampled and the volume of milk in each is measured accurately to assess whether the machinery needs adjustment. (Overfilling wastes milk, but underfilling is illegal.) One sample is shown below.

Are the data consistent with the target mean volume of 2.012 litres? This can be expressed as a hypothesis test comparing...

Blood pressure of executives

From published information, the national distribution of blood pressure in males aged 35-44 is known to have a standard deviation σ = 15.

Filling milk containers

The variataion caused by the filling machinery is well understood and the standard deviation of the volume of milk is known to be σ = 0.005 litres.

Quality control for cornflake packets

The diagram below repeats the simulation that we used earlier to test whether a sample mean weight of 10 cornflake packets of 529 gm is consistent with a packing machine that is set to give normally distributed weights with µ = 520 gm and σ = 10 gm.

Again click Accumulate and hold down the Simulate button until about 100 samples of 10 packets have been selected and weighed. The p-value is the probability of getting a sample mean further from 520 gm than 529 gm — either below 511 gm or above 529 gm and the simulation provides an estimate. However a simulation is unnecessary since we can evaluate the p-value exactly.

Select Normal distribution from the pop-up menu on the bottom right to replace the simulation with the normal distribution of the mean,

From its tail area, we can calculate (without a simulation) that the probability of getting a sample mean as far from 520 as 529 is exactly 0.0044. This is the exact p-value for the test.

P-value from statistical distance

Finally, consider the statistical distance of our estimate of µ, 529 gm, from the hypothesised value, 520 gm.

Select 'Statistical distance' from 520 from the middle pop-up menu to show how the p-value is found using this z-score.

Weights of courier packages

A courier company suspected that the weight of recently shipped packages had dropped. From past records, the mean weight of packages was 18.3 kg and their standard deviation was 7.1 kg. These figures were based on a very large number of packages and can be treated as exact.

Thirty packages were sampled from the previous week and their mean weight was found to be 16.8 kg. The data are displayed in the jittered dot plot below.

If the null hypothesis was true, the sample mean would have the normal distribution shown in pale blue. Although the sample mean weight is lower than 18.3 kg, it is not particularly unusual for this distribution, so we conclude that there is no evidence that the mean weight has reduced.

The right of the diagram shows how the p-value is calculated from a statistical distance (z-score).

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Choose Modified Data from the pop-up menu. The slider allows you to investigate how low the sample mean must become in order to give strong evidence that µ is less than 18.3.

Returns from Mutual Funds

Investing in the share market can be risky for small investors since the value of individual companies can fluctuate greatly, especially over short periods of time. These risks can be reduced by buying shares in a mutual fund that spreads the investment amoung a wide portfolio of companies.

Different mutual funds invest in companies of different types and with different inherent risks of losing and (hopefully) gaining value. Some funds have been categorised as 'high-risk' funds and a sample of 25 of these is shown in the table below. The percentage return paid by these funds over a 3-year period (April 1997 to March 2000) is also shown. (The stock market did particularly well over this period!)

The corresponding annualised return from Federal Constant Maturity Rate Bonds over this period was 5.64%. Did the high-risk funds do any better on average than this 'safe' investment?

The hypotheses of interest are similar to those in the initial pages of this section,

However we no longer know the population standard deviation, σ. The only information we have about σ comes from our sample.

The diagram below generates random samples from a normal distribution. Click Take sample a few times to see the variability in the samples.

Click Accumulate then take about 50 random samples. Observe that the stacked dot plot of the t statistic conforms reasonably with a standard normal distribution.

Now use the pop-up menu to reduce the sample size to 5 and take a further 50-100 samples. You will probably notice that there are more 'extreme' t-values (less than -3 or more than +3) than would be expected from a standard normal distribution.

Reduce the sample size to 3 and repeat. It should now be clearer that the distribution of the t-statistic has greater spread than a standard normal distribution. Click on the crosses for the most extreme t-values and observe that they correspond to samples in which the 3 data values happen to be close together, resulting in a small sample standard deviation, s.

The diagram below shows the shape of the t distribution for various different values of the degrees of freedom.

Drag the slider to see how the shape of the t distribution depends on the degrees of freedom. Note that

• The t distribution's shape becomes close to the standard normal distribution (the red curve) as the degrees of freedom increase.
   
• When the degrees of freedom are small, the t distribution has much longer tails than the standard normal distribution.
  
  
• When the degrees of freedom are around 10, the centre of the t distribution looks close to a standard normal distribution. However an important use of the t distribution is in hypothesis tests where the tails of the distribution are particularly important. The tails of the distributions are quite different.

Returns from Mutual Funds

The example on the previous page asked whether the average annualised return on high-risk mutual funds was higher than that from Federal Bonds (5.64%) over the period April 1997 to March 2000. The population standard deviation was unknown and the hypotheses of interest were,

The diagram below shows the calculations for obtaining the p-value for this test from the t distribution with (n - 1) = 24 degrees of freedom.

Since the probability of obtaining such a high mean return from 25 funds is 0.000 (to 3 decimal places) if the underlying population mean is 5.64, we conclude that there is extremely strong evidence that the mean return on high-risk funds was over 5.64 percent.

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Select Modified Data from the pop-up menu and use the slider to investigate the relationship between the sample mean and the p-value for the test.

Saturated fat content of cooking oil

It has been claimed that the saturated fat content of soybean cooking oil is no more than 15%. A clinician believes that the saturated fat content is greater than 15% and randomly samples 13 bottles of soybean cooking oil for testing.

The clinician is interested in the following hypotheses.

The p-value of 0.04 means that there is moderately strong evidence against H₀ — i.e. moderately strong evidence that the mean saturated fat content is greater than 15%.

Saturated fat content of cooking oil

The clinician who tested the saturated fat content of soybean cooking oil was interested in the hypotheses.

If H₀ is rejected, the clinician intends to report the high saturated fat content to the media. The two possible errors that could be made are described below.

Ideally the decision should be made in a way that keeps both probabilities low.

Illustration

The diagram below relates to a normal population whose standard deviation is known to be σ = 4. We will test the hypotheses

The test is based on the sample mean of n = 16 values from this distribution. The sample mean has a normal distribution,

This normal distributions can be used to calculate the probabilities of the two types of error. The diagram below illustrates how the probabilities of the two types of error depend on the critical value for the test, k.

Drag the slider at the top of the diagram to adjust k. Observe that making k large reduces the probability of a Type I error, but makes a Type II error more likely. It is impossible to simultaneously make both probabilities small with only n = 16 observations.

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Note also that there is not a single value for the probability of a Type II error — the probability depends on how far above 10 the mean µ lies. Drag the slider on the row for the alternative hypothesis to observe that:

This is as should be expected — the further above 10 the population mean, the more likely we are to detect that it is higher than 10 from the sample mean.

Illustration

The diagram below is identical to the one on the previous page.

With the top slider, adjust k to make the probability of a Type I error as close as possible to 5%. This is the decision rule for a test with significance level 5%.

Drag the top slider to reduce the significance level to 1% and note that the critical value for the test increases to about k = 12.3.

Illustration

The following diagram again investigates decision rules for testing the hypotheses

based on a sample of n = 16 values from a normal population with known standard deviation σ = 4.

In the diagram, the decision rule is based on the p-value for the test. Use the slider to adjust the critical p-value and observe that the significance level (probability of Type I error) is always equal to the p-value used in the decision rule. Adjust the critical p-value to 0.01.

Illustration

The following diagram again investigates decision rules for testing the hypotheses

based on a samples from a normal population with known standard deviation σ = 4. We will fix the significance level of the test at 5%.

The top half of the diagram shows the normal distribution of the mean for a sample of size n = 16. Use the slider to increase the sample size and observe that:

• The spread of the mean's distribution decreases.
• The critical value for rejecting H₀ decreases to keep the significance level at 5%.
• The probability of a Type II error decreases and the power of the test increases.

Shareholder meeting vote

Two candidates, Mike Smith and Sarah Brown, stand for election as chairperson of the board of directors of a large company. Just before the shareholders' meeting at which the election will be held, 56 randomly selected shareholders are asked about their voting intentions. If the proportion intending to vote for Mike Smith is denoted by π, the hypotheses of interest are

The diagram below illustrates how the poll results might weigh the evidence for each candidate winning.

Drag the slider to see how different sample numbers choosing Mike Smith affect the evidence. Unless either candidate receives (say) three quarters of the sample vote, we should admit that there is some doubt about who will win — the sample may not accurately reflect the population proportions.

Market share estimation through audits

The traditional retail store audit is a widely used marketing research tool among consumer packaged goods companies. The retail stort audit involves periodic audits of a sample of retail audits to monitor inventory and purchases of a particular product. Another auditing procedure, weekend selldown audits, has been proposed as a less expensive alternative.

The market shares of 10 brands of fruit juice were estimated using both of the store audit methods. Do the two methods result in the same estimates, on average? The data are paired, so we analyse the difference in estimates for each product (traditional minus weekend selldown) and test whether the underlying population mean of these values is zero.

The diagram below illustrates the evidence obtained from a set of sample data.

Drag the slider to see the conclusions that might be reached for data sets with different means. The further the sample mean is from zero (on either side), the stronger the evidence that µ is not zero. We can get very strong evidence that H₀ does not hold if the sample mean is far from zero.

If  = 0, µ could just as easily be 0.0001 or -0.0002 (which correspond to H_A). We cannot distinguish, so the best we can say is that the data are consistent with the null hypothesis — the data provide no information against the µ being zero.

In the context of this example, the conclusion from a sample mean of zero would be that the experiment gave no evidence of any difference between the mean estimates from the two auditing methods. The mean estimates might be different, but the data did not detect the effect.

Market share estimation through audits

In the example on the previous page, two different auditing methods were used to estimate market share. Is there any difference between the mean estimate of market share for the two methods?

The diagram below weighs the evidence using the p-value from a t-test of whether the mean difference, µ = 0.

To illustrate these properties, we use a test for whether a population mean is zero.

In the diagram below, you will take random samples from a normal population for which H₀ is true and, separately, from populations for which H_A is true.

When H₀ holds

Initially the population mean is zero, so H₀ holds. A single sample from this population is shown on the left and the p-value for testing whether the population mean is zero is shown as a cross on the jittered dot plot on the bottom right.

Click the button Take sample a few times to take other samples from this population and add their p-values to the display on the bottom right. After taking 50 or more samples, you should observe that the p-values are spread evenly between 0 and 1. This supports our assertion that the p-values have a rectangular distribution between 0 and 1 when H₀ holds.

When H_A holds

Now use the slider to change the true population mean to 2.0. We are still testing whether the mean is zero, so H_A now holds. Take 40 or 50 samples and observe that the p-values are usually closer to 0 than to 1.

Click on some of the larger p-values on the jittered dot plot to display the samples that gave rise to them. The sample means vary and, by chance, some samples have means that are near 0.0, even when the population mean is 2.0; these samples result in larger p-values.

Repeat this exercise with different population means (try at least 1.0, 2.0, 3.0 and -2.0). The further the population mean from the value targetted by H₀, 0.0, the more tightly clustered the p-values are around 0.0.

Again, we use the specific hypothesis test for

in order to demonstrate these general results.

Click the button Take sample 50 or more times to take samples from this population and add their p-values to the display on the right. From the diagram on the top right, we can read off the proportion of p-values that are less than any value. Approximately 50% of p-values are less than 0.5, 20% are less than 0.2, etc. when the null hypothesis is true.

Use the slider to change the true population mean to 1.5 and repeat. From the diagram on the top right, you should observe that more than 50% of p-values are less than 0.5, more than 20% are less than 0.2, etc. when the alternative hypothesis holds.

Another type of test

The normal distribution is often used as a hypothetical population from which a set of data are assumed to be sampled. But are the data consistent with an underlying normal population, or does the population distribution have a different shape?

One popular test for assessing whether a random sample come from a normal population is the Shapiro-Wilkes W test. The theory behind the test is advanced and the formula for the p-value cannot be readily evaluated by hand. However most statistical programs will perform the test.

A random sample of 40 values from a normal population is displayed in a jittered dot plot on the left of the diagram. The p-value for the Shapiro-Wilkes W test is shown under the dot plot and also graphically on the right.

Click Take sample a few times to take more samples and build the distribution of the p-values for the test. You should observe that the p-values have a rectangular distribution between 0 and 1 when the null hypothesis is true (i.e. if the samples are from a normal distribution).

Drag the slider on the top left of the diagram to change the shape of the population distribution. Repeat the exercise above and observe that when the null hypothesis does not hold, the p-values tend to be closer to 0.

Click on crosses on the display of p-values in the bottom right to display the sample that produced that p-value. P-values near zero usually correspond to samples that have very long tails to one or both sides, or have very short tails to one or both sides.

Returns from Mutual Funds

As a numerical example, the table below gives the annual returns for a sample of 137 mutual funds in 1999 (a period of rapid growth in the US economy).

A histogram of the data is shown below.

The best-fitting normal distribution (with mean and standard deviation equal to those of the data) has been superimposed on the histogram. There is a suggestion of skewness in the distribution of returns. Are the data really skew, or might this amount of skewness be possible in random samples from a normal distribution?

Applying the Shapiro-Wilkes W test to the data using the statistical program Minitab gives a p-value of "under 0.01". We conclude that there is strong evidence that the distribution is not normal. Even after deleting the 'outlier' — the First American Technology fund had a return of 191.8% — there is still strong evidence of skewness in the distribution of returns.