The document describes how to calculate a two-sample z-test by hand to determine if there is a statistically significant difference between the reported anxiety symptoms of patients taking a new anti-anxiety medication versus a placebo. It provides the formula for the z-statistic and walks through calculating it step-by-step for a sample problem where 64 out of 200 patients taking the medication reported anxiety symptoms compared to 92 out of 200 patients taking the placebo. The calculated z-statistic is then compared to critical values to determine whether to reject or fail to reject the null hypothesis that there is no difference between the groups.

1. Calculating a Two-Sample Z
Test by Hand

2. Here is the problem you just saw in a previous
presentation. Follow along doing the calculations on
your own problem as you view this one.

3. Here is the problem you just saw in a previous
presentation. Follow along doing the calculations on
your own problem.
Researchers want to test the effectiveness of a new anti-
anxiety medication. In clinical testing, 64 out of 200 people
taking the medication report symptoms of anxiety. Of the
people receiving a placebo, 92 out of 200 report symptoms of
anxiety. Is the medication working any differently than the
placebo? Test this claim using alpha = 0.05.

4. What are the researchers asking here?

5. What are the researchers asking here?
They are trying to determine if there is a difference in
reported anxiety symptoms between those taking the
new anti-anxiety drug and those taking the placebo.

6. What are the researchers asking here?
They are trying to determine if there is a difference in
reported anxiety symptoms between those taking the
new anti-anxiety drug and those taking the placebo.
What makes this question one that can be answered
using a two-sample Z test is the fact that we are
examining the difference between proportions (64 out
of 200 or .32 compared to 92 out 200 or .46).

7. What are the researchers asking here?
They are trying to determine if there is a difference in
reported anxiety symptoms between those taking the
new anti-anxiety drug and those taking the placebo.
What makes this question one that can be answered
using a two-sample Z test is the fact that we are
examining the difference between proportions (64 out
of 200 or .32 compared to 92 out 200 or .46).
Are these differences similar enough to make any
differences we would find if we were to repeat the
experiment to be due to chance or not?

8. With that in mind, we are able to state the null-
hypothesis and then determine if we will reject or fail
to reject it:

9. With that in mind, we are able to state the null-
hypothesis and then determine if we will reject or fail
to reject it:
There is NO significant proportional difference in
reported anxiety symptoms between a sample of
participants who took a new anti-anxiety drug and a
sample who took a placebo.

10. Since the test uses a .05 alpha value, that is what we
will use to determine if the probability of a meaningful
difference is rare or common.

11. Since the test uses a .05 alpha value, that is what we
will use to determine if the probability of a meaningful
difference is rare or common.
The alpha value makes it possible to determine what is
called the z critical. If the z statistic that we are about
to calculate from the data in the question is outside of
the z critical [for a one-tailed test (e.g., +1.64) or a two-
tailed test (e.g., -1.96 or +1.96]

12. One tailed test visual depiction:
rarecommon
+1.64
If the z value we
are about to
calculate lands
above this point,
we will reject the
null hypothesis

13. One tailed test visual depiction:
rarecommon
+1.64
If the z value lands
below this point
we will fail to
reject the null
hypothesis

14. A one tailed test could also go the other direction if we
are testing the probability of one sample having a
smaller proportion than another.
rare common
-1.64

15. A two-tailed test implies that we are not sure as to
which direction it will go. We don’t know if the
placebo or the new anxiety medicine will have better
results.

16. Here is a visual depiction of the two-tailed test.
rare
-1.96
Common rare
+1.96

17. All we have left to do now is calculate the z statistic.

18. All we have left to do now is calculate the z statistic.
Here is the formula for the z statistic for a Two-Sample
Z-Test:
𝒁 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
( 𝑝1 − 𝑝2)
𝑝 1 − 𝑝
1
𝑛1
+
1
𝑛2

19. In the numerator we are subtracting one sample
proportion from another sample proportion:
𝒁 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
( 𝑝1 − 𝑝2)
𝑝 1 − 𝑝
1
𝑛1
+
1
𝑛2

20. In the numerator we are subtracting one sample
proportion from another sample proportion:
𝒁 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
( 𝑝1 − 𝑝2)
𝑝 1 − 𝑝
1
𝑛1
+
1
𝑛2
Researchers want to test the effectiveness of a new anti-
anxiety medication. In clinical testing, 64 out of 200 people
taking the medication report symptoms of anxiety. Of the
people receiving a placebo, 92 out of 200 report symptoms of
anxiety. Is the medication working any differently than the
placebo? Test this claim using alpha = 0.05.

21. In the numerator we are subtracting one sample
proportion from another sample proportion:
𝒁 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
( 𝑝1 − 𝑝2)
𝑝 1 − 𝑝
1
𝑛1
+
1
𝑛2
Researchers want to test the effectiveness of a new anti-
anxiety medication. In clinical testing, 64 out of 200 people
taking the medication report symptoms of anxiety. Of the
people receiving a placebo, 92 out of 200 report symptoms of
anxiety. Is the medication working any differently than the
placebo? Test this claim using alpha = 0.05.

22. In the numerator we are subtracting one sample
proportion from another sample proportion:
𝒁 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
( 𝑝1 − 𝑝2)
𝑝 1 − 𝑝
1
𝑛1
+
1
𝑛2
Researchers want to test the effectiveness of a new anti-
anxiety medication. In clinical testing, 64 out of 200 people
taking the medication report symptoms of anxiety. Of the
people receiving a placebo, 92 out of 200 report symptoms of
anxiety. Is the medication working any differently than the
placebo? Test this claim using alpha = 0.05.
92/200=.46
64/200=.32

23. In the numerator we are subtracting one sample
proportion from another sample proportion:
𝒁 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
(.32 − .46)
𝑝 1 − 𝑝
1
𝑛1
+
1
𝑛2
Researchers want to test the effectiveness of a new anti-
anxiety medication. In clinical testing, 64 out of 200 people
taking the medication report symptoms of anxiety. Of the
people receiving a placebo, 92 out of 200 report symptoms of
anxiety. Is the medication working any differently than the
placebo? Test this claim using alpha = 0.05.
92/200=.46
64/200=.32

24. In the numerator we are subtracting one sample
proportion from another sample proportion:
𝒁 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
(−. 𝟏𝟐)
𝑝 1 − 𝑝
1
𝑛1
+
1
𝑛2
Researchers want to test the effectiveness of a new anti-
anxiety medication. In clinical testing, 64 out of 200 people
taking the medication report symptoms of anxiety. Of the
people receiving a placebo, 92 out of 200 report symptoms of
anxiety. Is the medication working any differently than the
placebo? Test this claim using alpha = 0.05.

25. The denominator is the estimated standard error:
𝒁 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
(−.12)
𝑝 1 − 𝑝
1
𝑛1
+
1
𝑛2
Researchers want to test the effectiveness of a new anti-
anxiety medication. In clinical testing, 64 out of 200 people
taking the medication report symptoms of anxiety. Of the
people receiving a placebo, 92 out of 200 report symptoms of
anxiety. Is the medication working any differently than the
placebo? Test this claim using alpha = 0.05.
This proportion is
called the pooled
standard deviation.
It is the same value
we use with
independent
sample t-tests.

26. The denominator is the estimated standard error:
𝒁 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
(−.12)
𝑝 1 − 𝑝
1
𝑛1
+
1
𝑛2
It is interesting to note that when you multiply a
proportion (.80) by its complement (1-.8 = .20)
you get the variance (.80*.20 = .16).
If you square that amount it comes to .04 and that
is the standard deviation.

27. Why is this important? Because the standard deviation
divided by the square root of the sample size is the
standard error.
𝒁 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
(−.12)
𝑝 1 − 𝑝
1
𝑛1
+
1
𝑛2
It is interesting to note that when you multiply a
proportion (.80) by its complement (1-.8 = .20)
you get the variance (.80*.20 = .16).
If you square that amount it comes to .04 and that
is the standard deviation.

28. And the larger the standard error the less likely the two
groups will be statistically significantly different.
𝒁 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
(−.12)
𝑝 1 − 𝑝
1
𝑛1
+
1
𝑛2
It is interesting to note that when you multiply a
proportion (.80) by its complement (1-.8 = .20)
you get the variance (.80*.20 = .16).
If you square that amount it comes to .04 and that
is the standard deviation.

29. Conversely, the smaller the standard error the more
likely the two groups will be statistically significantly
different.
𝒁 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
(−.12)
𝑝 1 − 𝑝
1
𝑛1
+
1
𝑛2
It is interesting to note that when you multiply a
proportion (.80) by its complement (1-.8 = .20)
you get the variance (.80*.20 = .16).
If you square that amount it comes to .04 and that
is the standard deviation.

30. So, let’s compute the pooled standard deviation:
𝒑 1 − 𝒑

31. So, let’s compute the pooled standard deviation:
𝒑 1 − 𝒑 𝒑 =
𝑥1 + 𝑥2
𝑛1 + 𝑛2

32. So, let’s compute the pooled standard deviation:
𝒑 1 − 𝒑 𝒑 =
𝑥1 + 𝑥2
𝑛1 + 𝑛2
Researchers want to test the effectiveness of a new anti-
anxiety medication. In clinical testing, 64 out of 200 people
taking the medication report symptoms of anxiety. Of the
people receiving a placebo, 92 out of 200 report symptoms of
anxiety. Is the medication working any differently than the
placebo? Test this claim using alpha = 0.05.
Here’s the problem again:

33. So, let’s compute the pooled standard deviation:
𝒑 1 − 𝒑 𝒑 =
𝑥1 + 𝑥2
𝑛1 + 𝑛2
Researchers want to test the effectiveness of a new anti-
anxiety medication. In clinical testing, 64 out of 200 people
taking the medication report symptoms of anxiety. Of the
people receiving a placebo, 92 out of 200 report symptoms of
anxiety. Is the medication working any differently than the
placebo? Test this claim using alpha = 0.05.

34. So, let’s compute the pooled standard deviation:
𝒑 1 − 𝒑 𝒑 =
64 + 𝑥2
𝑛1 + 𝑛2
Researchers want to test the effectiveness of a new anti-
anxiety medication. In clinical testing, 64 out of 200 people
taking the medication report symptoms of anxiety. Of the
people receiving a placebo, 92 out of 200 report symptoms of
anxiety. Is the medication working any differently than the
placebo? Test this claim using alpha = 0.05.

35. So, let’s compute the pooled standard deviation:
𝒑 1 − 𝒑 𝒑 =
64 + 𝑥2
𝑛1 + 𝑛2
Researchers want to test the effectiveness of a new anti-
anxiety medication. In clinical testing, 64 out of 200 people
taking the medication report symptoms of anxiety. Of the
people receiving a placebo, 92 out of 200 report symptoms of
anxiety. Is the medication working any differently than the
placebo? Test this claim using alpha = 0.05.

36. So, let’s compute the pooled standard deviation:
𝒑 1 − 𝒑 𝒑 =
64 + 𝑥2
200 + 𝑛2
Researchers want to test the effectiveness of a new anti-
anxiety medication. In clinical testing, 64 out of 200 people
taking the medication report symptoms of anxiety. Of the
people receiving a placebo, 92 out of 200 report symptoms of
anxiety. Is the medication working any differently than the
placebo? Test this claim using alpha = 0.05.

37. So, let’s compute the pooled standard deviation:
𝒑 1 − 𝒑 𝒑 =
64 + 𝑥2
200 + 𝑛2
Researchers want to test the effectiveness of a new anti-
anxiety medication. In clinical testing, 64 out of 200 people
taking the medication report symptoms of anxiety. Of the
people receiving a placebo, 92 out of 200 report symptoms of
anxiety. Is the medication working any differently than the
placebo? Test this claim using alpha = 0.05.

38. So, let’s compute the pooled standard deviation:
𝒑 1 − 𝒑 𝒑 =
64 + 92
200 + 𝑛2
Researchers want to test the effectiveness of a new anti-
anxiety medication. In clinical testing, 64 out of 200 people
taking the medication report symptoms of anxiety. Of the
people receiving a placebo, 92 out of 200 report symptoms of
anxiety. Is the medication working any differently than the
placebo? Test this claim using alpha = 0.05.

39. So, let’s compute the pooled standard deviation:
𝒑 1 − 𝒑 𝒑 =
64 + 92
200 + 𝑛2
Researchers want to test the effectiveness of a new anti-
anxiety medication. In clinical testing, 64 out of 200 people
taking the medication report symptoms of anxiety. Of the
people receiving a placebo, 92 out of 200 report symptoms of
anxiety. Is the medication working any differently than the
placebo? Test this claim using alpha = 0.05.

40. So, let’s compute the pooled standard deviation:
𝒑 1 − 𝒑 𝒑 =
64 + 92
200 + 200
Researchers want to test the effectiveness of a new anti-
anxiety medication. In clinical testing, 64 out of 200 people
taking the medication report symptoms of anxiety. Of the
people receiving a placebo, 92 out of 200 report symptoms of
anxiety. Is the medication working any differently than the
placebo? Test this claim using alpha = 0.05.

41. Add the fractions:
𝒑 1 − 𝒑 𝒑 =
64 + 92
200 + 200

42. Add the fractions:
𝒑 1 − 𝒑 𝒑 =
156
200 + 200

43. Add the fractions:
𝒑 1 − 𝒑 𝒑 =
156
200 + 200

44. Add the fractions:
𝒑 1 − 𝒑 𝒑 =
156
200

45. Add the fractions:
𝒑 1 − 𝒑 𝒑 = .78

46. Now we plug this pooled proportion:
𝒑 1 − 𝒑 𝒑 = .78

47. Now we plug this pooled proportion:
into the standard error formula in the denominator
𝒑 1 − 𝒑 𝒑 = .78

48. Now we plug this pooled proportion:
into the standard error formula in the denominator
𝒑 1 − 𝒑 𝒑 = .78

49. Now we plug this pooled proportion:
into the standard error formula in the denominator
.78 1−. 𝟕𝟖 𝒑 = .78

50. Now we plug this pooled proportion:
into the standard error formula in the denominator
.78 .22

51. Now we plug this pooled proportion:
into the standard error formula in the denominator
. 𝟏𝟕𝟐

52. Now we plug this pooled proportion:
Researchers want to test the effectiveness of a new anti-
anxiety medication. In clinical testing, 64 out of 200 people
taking the medication report symptoms of anxiety. Of the
people receiving a placebo, 92 out of 200 report symptoms of
anxiety. Is the medication working any differently than the
placebo? Test this claim using alpha = 0.05.
𝑍𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 =
(−. 𝟏𝟒)
. 𝟏𝟕𝟐
1
𝒏 𝟏
+
1
𝑛2

53. Now we need to calculate n or the sample size:
Researchers want to test the effectiveness of a new anti-
anxiety medication. In clinical testing, 64 out of 200 people
taking the medication report symptoms of anxiety. Of the
people receiving a placebo, 92 out of 200 report symptoms of
anxiety. Is the medication working any differently than the
placebo? Test this claim using alpha = 0.05.
𝑍𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 =
(−. 𝟏𝟒)
.172
1
200
+
1
𝑛

54. Now we need to calculate n or the sample size:
Researchers want to test the effectiveness of a new anti-
anxiety medication. In clinical testing, 64 out of 200 people
taking the medication report symptoms of anxiety. Of the
people receiving a placebo, 92 out of 200 report symptoms of
anxiety. Is the medication working any differently than the
placebo? Test this claim using alpha = 0.05.
𝑍𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 =
(−. 𝟏𝟒)
.172
1
200
+
1
200

55. Researchers want to test the effectiveness of a new anti-
anxiety medication. In clinical testing, 64 out of 200 people
taking the medication report symptoms of anxiety. Of the
people receiving a placebo, 92 out of 200 report symptoms of
anxiety. Is the medication working any differently than the
placebo? Test this claim using alpha = 0.05.
𝑍𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 =
(−. 𝟏𝟒)
.172
1
200
+
1
200
Now we need to calculate the Zstatistic

56. 𝑍𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 =
(−. 𝟏𝟒)
.172
2
200
Now we need to calculate the Zstatistic

57. 𝑍𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 =
(−. 𝟏𝟒)
.172
1
100
Now we need to calculate the Zstatistic

58. 𝑍𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 =
(−. 𝟏𝟒)
.172 .01
Now we need to calculate the Zstatistic

59. 𝑍𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 =
(−. 𝟏𝟒)
.172 (.1)
Now we need to calculate the Zstatistic

60. 𝑍𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 =
(−. 𝟏𝟒)
(.414) (.1)
Now we need to calculate the Zstatistic

61. 𝑍𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 =
(−. 𝟏𝟒)
.0414
Now we need to calculate the Zstatistic

62. Now we need to calculate the Zstatistic
𝑍𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 = −3.379

63. Let’s see where this lies in the distribution:
rare
-1.96
Common rare
+1.96

64. Let’s see where this lies in the distribution:
rare
-1.96
Common rare
+1.96-3.38

65. Because -3.38 is outside the -1.96 range we consider it
to be a rare occurrence and therefore we will reject the
null hypothesis and accept the alternative hypothesis:

66. Because -3.38 is outside the -1.96 range we consider it
to be a rare occurrence and therefore we will reject the
null hypothesis and accept the alternative hypothesis:
There was a significant difference in effectiveness between the
medication group and the placebo group, z = -3.379, p < 0.05.