Choosing the correct analysis (and reading Minitab output)

(#1-2) One numerical variable

If you have one numerical variable:

you have one population that you are trying to describe
it would be natural to summarize the data using a mean. (Note that you could also summarize the data using a median, but we've not learned how to perform hypothesis tests and confidence intervals for a median.)
so, use 1 sample t procedures. Perform the hypothesis test if you want to test if the population mean equals something. Calculate the confidence interval if you want to estimate the value of the population mean m.

Example

A random sample of 32 Stat 250 students were asked: What is your current cumulative GPA? The data were entered into a column in Minitab. Two (2) of the students declined to answer the question, and so are denoted as missing ("*"). The data are:

gpa

      *   3.60   3.01   3.41   2.50   2.75   3.59   3.00   3.91   3.13
   2.40   3.94   3.25   3.00   3.50   3.98   2.50   3.45   2.69   2.67
   3.35   2.80   3.40   2.88   3.73   2.10   2.45   3.42   2.50   2.00
      *   2.74

The Minitab output looks like:

T Confidence Intervals

Variable     N      Mean    StDev  SE Mean       95.0 % CI
gpa         30    3.0550   0.5444   0.0994  (  2.8517,  3.2583)

T-Test of the Mean

Test of mu = 3.0000 vs mu > 3.0000

Variable     N      Mean    StDev   SE Mean        T          P
gpa         30    3.0550   0.5444    0.0994     0.55       0.29

The output tells us:

that we asked Minitab to test the null hypothesis H₀: m = 3.0 against the alternative hypothesis H_A: m > 3.0.
that N = 30 students responded
the sample mean g.p.a ("Mean") is 3.055. That is, the average g.p.a. of the 30 students in the sample is 3.055.
the sample standard deviation ("StDev") is 0.5444. That is the standard deviation of the g.p.a. of the 30 students is 0.5444
the standard error of the mean ("SE Mean") is 0.0994. Recall that the standard error of the mean is the sample standard deviation divided by the square root of the sample size. So, here SE Mean is 0.5444/sqrt(30) = 0.0994.
A 95% confidence interval for the population mean is 2.852 to 3.258. That is, we can be 95% confident that the mean g.p.a. of all students in the conceptual population of Stat 250 students is between 2.852 and 3.258. Recall that, if n > 30, the 95% confidence interval will be close to the sample mean ± (2 × standard errors). So, as you can see here, the 95% confidence interval is approximately 3.055 ± (2 × 0.0994).
The t-statistic ("T") for the requested hypothesis test is 0.55, which translates into a P-value of 0.29. (That is, the area under the t curve with N-1 = 29 degrees of freedom to the right of 0.55 is 0.29. That is, there is a 29% chance that we'd get a sample mean as large as 3.055 if the population mean is 3.00. The sample mean is not unlikely, so we don't reject the null hypothesis.)

(#3-4) One categorical (binary) variable

If you have one categorical (binary) variable:

you have one population that you are trying to describe
it would be natural to summarize the data using a proportion (percentage)
so, use 1 proportion Z procedures. Perform the hypothesis test if you want to test if the population proportion equals something. Calculate the confidence interval if you want to estimate the value of the population proportion p.

Example

A random sample of 87 students were asked: Have you ever dyed your hair? Yes or No. The data were entered into a column in Minitab -- the Yes and No responses were coded as a 1 and 0, respectively. The data are:

dyed

     0     0     0     1     1     1     1     0     1     0     0     1
     0     1     1     0     1     0     0     0     1     0     1     0
     1     0     0     1     0     0     1     1     0     0     1     1
     1     1     1     0     0     1     1     1     1     1     1     0
     0     1     0     0     0     1     0     1     1     1     1     0
     0     0     1     0     0     0     1     0     1     1     0     0
     0     1     1     0     0     0     0     0     0     0     0     0
     0     0     0

The Minitab output looks like:

Test and Confidence Interval for One Proportion

Test of p = 0.5 Vs p < 0.5

Success = 1

Variable          X      N  Sample p        95.0 % CI       Z-Value  P-Value
dyed             38     87  0.436782  (0.332560, 0.541004)    -1.18    0.119

The output tells us:

that we asked Minitab to test the null hypothesis H₀: p = 0.5 against the alternative hypothesis H_A: p < 0.5
that N = 87 students were surveyed
that X = 38 of the surveyed students said they had once dyed their hair ("Success = 1")
the sample proportion ("Sample p") is 0.436782. That is, 38/87, or 43.6782% of the surveyed students reported having dyed their hair once
A 95% confidence interval for the population proportion is 0.333 to 0.541. That is, we can be 95% confident that between 33.3% and 54.1% of the population of all students have dyed their hair once.
The Z-statistic ("Z-value") for the hypothesis test is -1.18, which translates into a P-value of 0.119. (That is, the area under the normal curve to the left of -1.18 is 0.119. That is, there is a 11.9% chance that we'd get a sample proportion as small as 0.437 if the population proportion is 0.50. The sample proportion is not unlikely, so we don't reject the null hypothesis.)

(#5-6) One categorical (binary) variable that forms independent groups and one numerical variable

If you have one categorical (binary) variable that forms independent groups and one numerical variable:

you have two populations that you are trying to compare. The populations are formed by each level of the binary variable.
If the binary variable forms independent groups, use these 2-sample (#5-6) procedures. If the binary variable forms paired groups, use the paired (#7-8) procedures instead. Hint on deciding: Ask yourself if the same (or similar) person (or object) was measured twice. If so, then use the paired (#7-8) procedures. If not, then use these 2-sample (#5-6) procedures.
it would be natural to summarize the numerical variable data calculating a mean for the first group and a mean for the second group
so, use 2 sample t procedures. Perform the hypothesis test if you want to test if the population means are the same (i.e. if the difference in the population means is 0). (Note that testing to see if the population means are the same is equivalent to testing whether or not there is a relationship between the binary variable and the numerical variable.) Calculate the confidence interval if you want to estimate the difference in the population means m₁ - m₂.

Note that here we are effectively treating the binary variable that forms the groups as the explanatory variable and the numerical variable by which we're comparing the groups as the response variable. If we switched the variables around and treated the numerical variable as the explanatory variable and treated the binary variable as the response variable, we'd need to do a "logistic regression analysis," which is beyond the scope of this course.

Example

A random sample of 31 students were asked two questions:

What is your gender? Male or Female.
How much money did you spend on textbooks this semester?

The data were entered into two columns in Minitab. One column contains the gender ("subscripts") of the student, while one column contains the amount of money spent on books ("samples"). Males were coded as a 1 and females as a 2. The data are:

 Row   Sex   Books



   1     1    400
   2     1    500
   3     2    210
   4     1    200
   5     2    240
   6     1    160
   7     1    400
   8     2    300
   9     2    250
  10     2    300
  11     1    465
  12     1    300
  13     2    145
  14     1    345
  15     2    350
  16     1    350
  17     2    300
  18     2    300
  19     1    300
  20     2    300
  21     1    350
  22     1    550
  23     1    245
  24     1    230
  25     2    340
  26     1    129
  27     2    300
  28     2    200
  29     2    159
  30     1    330
  31     1    270

The Minitab output looks like:

Two Sample T-Test and Confidence Interval

Two sample T for Books

Sex          N      Mean     StDev   SE Mean

1           17       325       116        28
2           14     263.9      64.4        17

95% CI for mu (1) - mu (2): ( -7,  129)
T-Test mu (1) = mu (2) (Vs not =): T = 1.85  P = 0.076  DF = 25

The output tells us:

that we asked Minitab to test the null hypothesis H₀: m₁ = m₂ against the alternative hypothesis H_A: m₁ m₂
that N = 17 males ("Sex = 1") and N = 14 females ("Sex = 2") responded
that the sample mean for the males is $325 and the sample mean for the females is $263.90.
that the sample standard deviation for the males is $116, and the sample standard deviation for the females is $64.40.
that the standard error of the mean for the males is $28, and the standard error of the mean for females is $17.
A 95% confidence interval for the difference in the population means, m₁ - m₂, is -7 to 129. That is, we can be 95% confident that the difference in the population means is between -$7 and $129. That is, the difference in the means could be anywhere from the males spending $7 less to $129 more than the females. Because the confidence interval contains 0, we can't conclude the mean amount spent by the males differs than the mean amount spent by the females.
The t-statistic ("T") for the hypothesis test is 1.85, which translates into a P-value of 0.076. (That is, 2 × the area under the t curve with 25 degrees of freedom ("DF") to the right of 1.85 is 0.076. That is, there is a 7.6% chance that we'd get a difference in the sample means as large as $61.10 ($325 minus $263.90) if the difference in the population means is 0. The difference in the sample means is not unlikely enough, so we don't reject the null hypothesis.)

(#7-8) One categorical (binary) variable that forms paired groups and one numerical variable

If you have one categorical (binary) variable that forms paired groups and one numerical variable:

you have two populations that you are trying to compare. The populations are formed by each level of the binary variable.
If the binary variable forms independent groups, use the 2-sample (#5-6) procedures. If the binary variable forms paired groups, use these paired (#7-8) procedures instead. Hint on deciding: Ask yourself if the same (or similar) person (or object) was measured twice. If so, then use these paired (#7-8) procedures. If not, then use the 2-sample (#5-6) procedures.
it would be natural to summarize the numerical variable data calculating the mean of the paired differences of the two groups.
so, use paired procedures. Perform the hypothesis test if you want to test if the mean of the paired differences is 0 (i.e. if the population means are the same). Calculate the confidence interval if you want to estimate the mean of the paired differences m_D.

Example

The pulse rates of a random sample of 10 students were measured. Then, the students were asked to march in place. Their pulse rates were measured again. The data were entered into two columns in Minitab. One column contains the students' pulse rates before marching, while one column contains the students' pulse rates after marching. The data are:

  Row   Bef    Aft

   1     60     72
   2     62     92
   3     80     84
   4     67     68
   5     70     80
   6     52     72
   7     56     80
   8     75     88
   9     56     64
  10     80    104

Note that we could calculate and enter the differences, and then use the 1-sample t procedures on the column of differences. Instead, we can let Minitab do the dirty work of calculating the differences by just using Mintab's paired t procedures. The Minitab output looks like:

Paired T-Test and Confidence Interval

Paired T for Bef - Aft

                  N      Mean     StDev   SE Mean

Bef              10     65.80     10.21      3.23
Aft              10     80.40     12.14      3.84
Difference       10    -14.60      9.51      3.01

95% CI for mean difference: (-21.40, -7.80)
T-Test of mean difference = 0 (vs < 0): T-Value = -4.85  P-Value = 0.000

The output tells us:

that we asked Minitab to consider the Bef - Aft differences. That is, subtract the after pulse rates from the before pulse rates.
that N = 10 students were measured
the sample mean of the differences, Bef - Aft, is -14.60 beats per minute.
the sample standard deviation of the differences, Bef - Aft, is 9.51 beats per minute.
the standard error of the mean of the differences, Bef - Aft, is 3.01. Recall that the standard error of the mean is the sample standard deviation divided by the square root of the sample size. So, here SE Mean is 9.51/sqrt(10) = 3.01.
A 95% confidence interval for the mean of the Bef-Aft differences, m_D, is -21.40 to -7.80. That is, we can be 95% confident that the mean of the Bef-Aft differences is between -21.40 to -7.80 beats per minute. That is, we can be 95% confident that the mean pulse rate before exercise is between 21.4 and 7.8 beats lower than the mean pulse rate after exercise.
we asked Minitab to test the null hypothesis H₀: m_D = 0 against the alternative hypothesis H_A: m_D<0. That is, we want to test that the pulse rates have increased on average.
The t-statistic ("T") for the hypothesis test is -4.85, which translates into a P-value of 0.000 (to 3 decimal places). (That is, the area under the t curve with N-1 = 9 degrees of freedom to the left of -4.85 is 0.000... That is, there is a very small chance (P < 0.001) that we'd get a sample mean difference as large as -14.60 if the mean difference in the population is 0. The sample mean difference is very unlikely, so we reject the null hypothesis without hesitation.)

Note that if we had asked Minitab to consider the Aft-Bef differences, our conclusions would not change. The output would look like:

Paired T-Test and Confidence Interval

Paired T for Aft - Bef

                  N      Mean     StDev   SE Mean

Aft              10     80.40     12.14      3.84
Bef              10     65.80     10.21      3.23
Difference       10     14.60      9.51      3.01

95% CI for mean difference: (7.80, 21.40)
T-Test of mean difference = 0 (vs > 0): T-Value = 4.85  P-Value = 0.000

Note that:

we'd have to change our alternative hypothesis to H_A: m_D > 0, since we are still interested in testing that the pulse rates increased on average. It's just now that a positive number suggests an increase.
The sample mean difference (14.60) and the 95% confidence interval (7.80 to 21.40) just changed signs. The confidence interval tells us that we can be 95% confident that the mean pulse rate after exercise is between 7.8 and 21.4 beats higher than the mean pulse rate before exercise.
The P-value is still 0.000...., so we still reject the null hypothesis, and conclude that the mean difference is greater than 0, which is consistent with concluding that the mean pulse rate after exercise is greater than the mean pulse rate before exercise.

(#9-11) One categorical (binary) variable that forms independent groups and one categorical (binary) variable

Here, one categorical variable forms two or more independent groups. It would be natural to summarize the second categorical variable by calculating a proportion for the first group and a proportion for the second group and a proportion for the third group and so on.

The chi-square test (#11) and the 2-proportions Z procedures (#9) are used to see:

if there is a relationship between the two categorical variables
if the proportions for each group are equal

That is, the above two goals are equivalent.

How to decide what to use:

If the categorical variable that defines the groups forms three or more groups (you have "greater than" a 2 × 2 table), you must use the chi-square test (#11) to compare the proportions. (In this case, you cannot calculate a confidence interval for the three or more groups simultaneously. You can only calculate a confidence interval for the difference in two of the proportions at a time. If you do this, use the 2-proportions Z-interval.)
If the categorical variable that defines the groups forms two groups (i.e. you have a 2 × 2 table), you can either use the chi-square test (#11) or the 2 proportions Z-test (#10) to compare the proportions. The two different P-values that you get should be nearly the same. The advantage of using the 2 proportions Z-test is that Minitab will also automatically calculate and report a confidence interval for the difference in the proportions, too.
If you want to estimate the difference in two population proportions, calculate a 2-proportions Z-interval (#9).

Example

A random sample of 70 Stat 250 students were asked two questions:

What is your gender? Male or female.
Do you read your horoscope regularly? Yes or No.

The data were entered into two columns in Minitab. One column contains the students' gender, while one column contains the students' answer to the horoscope question. The males were coded as a 1, while the females were coded as a 2. Yes responses were coded as a 1, while No responses were coded as a 0.

Because gender forms only two independent groups -- males and females -- we can use either the chi-square test or the 2-proportions Z-test to see if there is a relationship between gender and horoscope reading. We can specify what we are testing either of the following ways.

Way #1

H₀: There is no relationship between gender and horoscope reading.
H_A: There is a relationship between gender and horoscope reading.

Way #2

H₀: The proportion of males who read their horoscope regularly is the same as the proportion of females who do. That is, p_M = p_F.
H_A: The proportion of males who read their horoscope regularly is not the same as the proportion of females who do. That is, p_M p_F.

The Minitab output for the chi-square test looks like:

Tabulated Statistics



 Rows: gender     Columns: horoscop

 

           0        1      All


 1        32        5       37
       86.49    13.51   100.00
          32        5       37
       26.96    10.04    37.00


 2        19       14       33
       57.58    42.42   100.00
          19       14       33
       24.04     8.96    33.00


 All      51       19       70
       72.86    27.14   100.00
          51       19       70
       51.00    19.00    70.00

Chi-Square = 7.372, DF = 1, P-Value = 0.007

 
 Cell Contents --

                  Count
                  % of Row
                  Count
                  Exp Freq

Note that Minitab tells us what each of the numbers in each of the cells means. The first number in each cell is the number of the people in the sample having the two characteristics defined by the cell. The second number in each cell is the row percentage. In this case, in the (1,1) cell, it is the percentage of males who read their horoscope regularly. The last number in each cell is the number of people in the sample we'd expect to fall in the cell if the null hypothesis were true, i.e. if there were no relationship between gender and horoscope reading.

The output tells us:

5 out of 37, or 13.51% of the males read their horoscope regularly
14 out of 33, or 42.42% of the females read their horoscope regularly
The chi-square statistic, which quantifies how much the observed and expected counts differ, is 7.372. The P-value of 0.007 tells us it is unlikely that we'd get such a large chi-square statistic if there were no relationship between gender and horoscope reading. We can reject the null hypothesis and conclude that the proportions of males and females who read their horoscope differ.

The corresponding Minitab output for the 2-proportions Z-test and interval is:

Test and Confidence Interval for Two Proportions


Success = 1


gender            X      N  Sample p

1                 5     37  0.135135
2                14     33  0.424242


Estimate for p(1) - p(2):  -0.289107
95% CI for p(1) - p(2):  (-0.490522, -0.0876921)
Test for p(1) - p(2) = 0 (vs not = 0):  Z = -2.81  P-Value = 0.005

The output tells us:

There are N = 37 males in the sample and N = 33 females in the sample.
Minitab is counting the number of people who read their horoscope regulary ("Success = 1"). So, 5 of the males in the sample read their horoscope regularly and 14 of the females read their horoscope regularly.
13.5% (from 5/37) of the males in the sample read their horoscope regularly and 42.4% (from 14/33) of the females read their horoscope regularly. Note that the sample percentages are the same as calculated in the above chi-square test output.
The difference in the sample proportions (males (1) - females (2)) is -0.289 (from 0.135 - 0.424).
We asked Minitab to test the null hypothesis H₀: p₁ = p₂ against the alternative hypothesis H_A: p₁p₂.
The Z-statistic ("Z") for the hypothesis test is -2.81, which translates into a P-value of 0.005. (That is, 2 × the area under the normal curve to the left of -2.81 is 0.005.) Note that the P-value is very close to the P-value we got from the chi-square test (P = 0.007).
We can be 95% confident that the difference in the population proportions is between -0.491 and -0.088. That is, we can be confident that between 8.8% and 49.1% more females read their horoscope regularly than males.

Two numerical variables

Use a scatter plot to identify outliers and to view the potential relationship between the two variables.
Calculate a least squares regression line to estimate the linear relationship between the two variables.
Use the least squares regression line to predict future values of the response variable.