Calculating a single sample z test by hand

1. Calculating a
Single-Sample Z Test

2. In this presentation you will be shown how to calculate
a single sample z-test by hand.

3. We first determine the z-critical for our question.

4. We will use the following problem:

5. A survey claims that 9 out of 10 doctors recommend aspirin for
their patients with headaches. To test this claim, a random
sample of 100 doctors is obtained. Of these 100 doctors, 82
indicate that they recommend aspirin. Is this claim accurate?
Use alpha = 0.05

6. First, we determine that our decision rule is that we
will reject the null hypothesis. In this case we will
reject the null hypothesis if the p value is less than .05.

7. First, we determine that our decision rule is that we
will reject the null hypothesis. In this case we will
reject the null hypothesis if the p value is less than .05.
Basically, we are saying that we are willing live with the
probability of being wrong 5 times out of 100 (.05) or 1
time out of 20.

8. With a cut off of .05, if we hypothesize that sample has
a higher value than the population then our cut off z-
score would be 1.64 (this can be located in a z-table)

9. With a cut off of .05, if we hypothesize that sample has
a higher value than the population then our cut off z-
score would be 1.64 (this can be located in a z-table)
95%
mean-1σ +1σ-2σ +2σ
Common
+1.64
rare

10. With a cut off of .05, if we hypothesize that sample has
a lower value than the population then our cut off z-
score would be -1.64 (this can be located in a z-table)

11. With a cut off of .05, if we hypothesize that sample has
a lower value than the population then our cut off z-
score would be -1.64 (this can be located in a z-table)
95%
mean-1σ +1σ-2σ +2σ
Common
+1.64
rare

12. With a cut off of .05, if we hypothesize that sample
could have either a lower or higher value than the
population then our cut off z-scores would be -1.96
and +1.96

13. With a cut off of .05, if we hypothesize that sample
could have either a lower or higher value than the
population then our cut off z-scores would be -1.96
and +1.96
rarerare
95%
mean-1σ +1σ-2σ +2σ
Common
-1.96 +1.96

14. So if the z statistic we calculate is less than -1.96
(e.g., -1.99) or greater than +1.96 (e.g., +2.30) then we
will consider this to be a rare event and reject the null
hypothesis and state that there is a statistically
significant difference between .9 (population) and .82
(the sample).

15. So if the z statistic we calculate is less than -1.96
(e.g., -1.99) or greater than +1.96 (e.g., +2.30) then we
will consider this to be a rare event and reject the null
hypothesis and state that there is a statistically
significant difference between .9 (population) and .82
(the sample).
Let’s calculate the z statistic and see where if falls!

16. So if the z statistic we calculate is less than -1.96
(e.g., -1.99) or greater than +1.96 (e.g., +2.30) then we
will consider this to be a rare event and reject the null
hypothesis and state that there is a statistically
significant difference between .9 (population) and .82
(the sample).
Let’s calculate the z statistic and see where if falls!
We do this by using the following equation:

17. So if the z statistic we calculate is less than -1.96
(e.g., -1.99) or greater than +1.96 (e.g., +2.30) then we
will consider this to be a rare event and reject the null
hypothesis and state that there is a statistically
significant difference between .9 (population) and .82
(the sample).
Let’s calculate the z statistic and see where if falls!
We do this by using the following equation:
𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
𝑝 − 𝑝
𝑝(1 − 𝑝)
𝑛

18. So if the z statistic we calculate is less than -1.96
(e.g., -1.99) or greater than +1.96 (e.g., +2.30) then we
will consider this to be a rare event and reject the null
hypothesis and state that there is a statistically
significant difference between .9 (population) and .82
(the sample).
Let’s calculate the z statistic and see where if falls!
We do this by using the following equation:
Zstatistic is what we are trying to find to see if it is
outside or inside the z critical values (-1.96 and +1.96).
𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
𝑝 − 𝑝
𝑝(1 − 𝑝)
𝑛

19. Here’s the problem again:

20. A survey claims that 9 out of 10 doctors recommend aspirin for
their patients with headaches. To test this claim, a random
sample of 100 doctors is obtained. Of these 100 doctors, 82
indicate that they recommend aspirin. Is this claim accurate?
Use alpha = 0.05

21. A survey claims that 9 out of 10 doctors recommend aspirin for
their patients with headaches. To test this claim, a random
sample of 100 doctors is obtained. Of these 100 doctors, 82
indicate that they recommend aspirin. Is this claim accurate?
Use alpha = 0.05

22. 𝒑 is the proportion from the sample that
recommended aspirin to their patients (. 𝟖𝟐)
𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
𝑝 − 𝑝
𝑝(1 − 𝑝)
𝑛

23. 𝒑 is the proportion from the sample that
recommended aspirin to their patients (. 𝟖𝟐)
𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
𝑝 − 𝑝
𝑝(1 − 𝑝)
𝑛
Note – this little
hat ( 𝑝) over the
p means that
this proportion
is an estimate
of a population

24. 𝐩 is the proportion from the population that
recommended aspirin to their patients (.90)
𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
𝑝 − 𝑝
𝑝(1 − 𝑝)
𝑛

25. 𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
𝑝 − 𝑝
𝑝(1 − 𝑝)
𝑛
𝒏 is the size of the sample (100)

26. 𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
𝑝 − 𝑝
𝑝(1 − 𝑝)
𝑛
A survey claims that 9 out of 10 doctors recommend aspirin for
their patients with headaches. To test this claim, a random
sample of 100 doctors is obtained. Of these 100 doctors, 82
indicate that they recommend aspirin. Is this claim accurate?
Use alpha = 0.05
𝒏 is the size of the sample (100)

27. 𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
𝑝 − 𝑝
𝑝(1 − 𝑝)
𝑛
A survey claims that 9 out of 10 doctors recommend aspirin for
their patients with headaches. To test this claim, a random
sample of 100 doctors is obtained. Of these 100 doctors, 82
indicate that they recommend aspirin. Is this claim accurate?
Use alpha = 0.05
𝒏 is the size of the sample (100)

28. 𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
𝑝 − 𝑝
𝑝(1 − 𝑝)
𝑛
Let’s plug in the numbers

29. 𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
.82 − 𝑝
𝑝(1 − 𝑝)
𝑛
Sample Proportion

30. 𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
.82 − 𝑝
𝑝(1 − 𝑝)
𝑛
A survey claims that 9 out of 10 doctors recommend aspirin for
their patients with headaches. To test this claim, a random
sample of 100 doctors is obtained. Of these 100 doctors, 82
indicate that they recommend aspirin. Is this claim accurate?
Use alpha = 0.05
Sample Proportion

31. 𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
.82 − 𝑝
𝑝(1 − 𝑝)
𝑛
A survey claims that 9 out of 10 doctors recommend aspirin for
their patients with headaches. To test this claim, a random
sample of 100 doctors is obtained. Of these 100 doctors, 82
indicate that they recommend aspirin. Is this claim accurate?
Use alpha = 0.05
Sample Proportion

32. 𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
.82 − .90
.90(1 − .90)
𝑛
Population Proportion

33. 𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
.82 − .90
.90(1 − .90)
𝑛
A survey claims that 9 out of 10 doctors recommend aspirin for
their patients with headaches. To test this claim, a random
sample of 100 doctors is obtained. Of these 100 doctors, 82
indicate that they recommend aspirin. Is this claim accurate?
Use alpha = 0.05
Population Proportion

34. 𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
.82 − .90
.90(1 − .90)
𝑛
A survey claims that 9 out of 10 doctors recommend aspirin for
their patients with headaches. To test this claim, a random
sample of 100 doctors is obtained. Of these 100 doctors, 82
indicate that they recommend aspirin. Is this claim accurate?
Use alpha = 0.05
Population Proportion

35. The difference
𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
.82 − .90
.90(1 − .90)
𝑛

36. 𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
−.08
.90(1 − .90)
𝑛
The difference

37. Now for the denominator which is the estimated
standard error. This value will help us know how many
standard error units .82 and .90 are apart from one
another (we already know they are .08 raw units apart)

38. 𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
−.08
.90(1 − .90)
𝑛
Now for the denominator which is the estimated
standard error. This value will help us know how many
standard error units .82 and .90 are apart from one
another (we already know they are .08 raw units apart)

39. 𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
−.08
.90(1 − .90)
𝑛
Note - If the standard error is small then the z statistic
will be larger. The larger the z statistics the more likely
that it will exceed the -1.96 or +1.96 boundaries,
compelling us to reject the null hypothesis. If it is
smaller than we will not.

40. 𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
−.08
.90(1 − .90)
𝑛
Let’s continue our calculations and find out:

41. 𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
−.08
.90(1 − .90)
𝑛
A survey claims that 9 out of 10 doctors recommend aspirin for
their patients with headaches. To test this claim, a random
sample of 100 doctors is obtained. Of these 100 doctors, 82
indicate that they recommend aspirin. Is this claim accurate?
Use alpha = 0.05
Let’s continue our calculations and find out:

42. 𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
−.08
.90(.10)
𝑛
Let’s continue our calculations and find out:

43. 𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
−.08
.09
𝑛
Let’s continue our calculations and find out:

44. 𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
−.08
.09
100
Sample Size:

45. 𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
−.08
.09
100
A survey claims that 9 out of 10 doctors recommend aspirin for
their patients with headaches. To test this claim, a random
sample of 100 doctors is obtained. Of these 100 doctors, 82
indicate that they recommend aspirin. Is this claim accurate?
Use alpha = 0.05
Sample Size:

46. 𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
−.08
.0009
Let‘s continue our calculations:

47. 𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
−.08
.03
Let‘s continue our calculations:

48. 𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 = −2.67
Let‘s continue our calculations:

49. 𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 = −2.67
Let‘s continue our calculations:
Now we have our z statistic.

50. Let’s go back to our distribution:
rarerare
95%
mean-1σ +1σ-2σ +2σ
Common
-1.96 +1.96

51. Let’s go back to our distribution: So, is this result
rare or common?
rarerare
95%
mean-1σ +1σ-2σ +2σ
Common
-1.96 +1.96-2.67

52. Let’s go back to our distribution: So, is this result
rare or common?
rarerare
95%
mean-1σ +1σ-2σ +2σ
Common
-1.96 +1.96
This is the
Z-Statistic we
calculated
-2.67

53. Let’s go back to our distribution: So, is this result
rare or common?
rarerare
95%
mean-1σ +1σ-2σ +2σ
Common
-1.96 +1.96-2.67
This is the
Z – Critical

54. Looks like it is a rare event therefore we will reject the
null hypothesis in favor of the alternative hypothesis:

55. Looks like it is a rare event therefore we will reject the
null hypothesis in favor of the alternative hypothesis:
The proportion of a sample of 100 medical doctors
who recommend aspirin for their patients with
headaches IS statistically significantly different from
the claim that 9 out of 10 doctors recommend aspirin
for their patients with headaches.