Statistics and Probability Grade XI

2. Test for
MEAN
WHY test
the mean?

3. Test for
MEAN
Two Statistical Test
Concerning Means
• z test 𝑖𝑓 𝜎 𝑖𝑠 𝑘𝑛𝑜𝑤𝑛
• t test 𝑖𝑓 𝜎 𝑖𝑠 𝑛𝑜𝑡 𝑘𝑛𝑜𝑤𝑛

4. Test for
MEAN
General Formula for
Computation of Test
Statistic
𝑇𝑒𝑠𝑡 𝑆𝑡𝑎𝑡𝑖𝑐 =
𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑉𝑎𝑙𝑢𝑒 − 𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑉𝑎𝑙𝑢𝑒
𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝐸𝑟𝑟𝑜𝑟
• observed value (sample mean)
• expected value (population mean)

5. Test for
MEAN
Standard Error for the
Mean (Pop. SD)
𝜎𝑥 =
𝜎𝑥
𝑛

6. Test for
MEAN
Standard Error for the
Mean (Sample. SD)
𝜎𝑥 =
𝑠
𝑛

7. Test for
MEAN
z Test for Mean (𝜎 is
known or 𝑛 ≥ 30)
• This is a statistical test for population
mean. The z test can be used when
the population is normal and 𝜎 is
known or when the sample size is
greater than or equal to 30.

8. Test for
MEAN
z Test for Mean (𝜎 is
known or 𝑛 ≥ 30)
𝑇𝑒𝑠𝑡 𝑆𝑡𝑎𝑡𝑖𝑐 =
𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑉𝑎𝑙𝑢𝑒 − 𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑉𝑎𝑙𝑢𝑒
𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝐸𝑟𝑟𝑜𝑟
𝑧 =
𝑠𝑎𝑚𝑝𝑙𝑒 𝑚𝑒𝑎𝑛 − 𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑚𝑒𝑎𝑛
𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝐸𝑟𝑟𝑜𝑟
𝑧 =
𝑥 − 𝜇0
𝜎𝑥
𝑛

9. Test for
MEAN
Common Levels of Significance
and Critical Values
𝛼 ONE-TAILED TEST TWO-TAILED TEST
LEFT (<) RIGHT (>)
0.10 𝑧 𝑎 = −1.28 𝑧 𝑎 = 1.28 𝑧 𝑎/2 = ±1.645
0.05 𝑧 𝑎 = −1.645 𝑧 𝑎 = 1.645 𝑧 𝑎/2 = ±1.960
0.01 𝑧 𝑎 = −2.33 𝑧 𝑎 = 2.33 𝑧 𝑎/2 = ±2.575

10. Test for
MEAN

11. Test for
MEAN
EXAMPLE 1
1. A random sample of 12 babies born in a
charity ward of IDH was taken with their
weights (in kg) recorded as follows: 2.3,
2.4, 2.4, 2.5, 2.6, 2.5,
2.8, 2.4, 2.7, 2.3, 2.2, 3.0
Assuming that this sample came from a
normal population, investigate the claim that
the mean weight is greater than 2.5 kg. The
population SD is 0.2 kg. Use 𝛼 = 0.05.

12. Test for
MEAN
EXAMPLE 1
1. 𝐻0: 𝜇 = 2.5
𝐻1: 𝜇 > 2.5
2. 𝛼 = 0.05
3. 𝑧 𝑎 = 1.645

13. Test for
MEAN
EXAMPLE 1
Sample Mean: 2.5083
𝑧 =
𝑥 − 𝜇0
𝜎𝑥
𝑛
𝑧 =
2.5083 − 2.5
0.2
12
𝑧 = 0.1438

14. Test for
MEAN
EXAMPLE 1
Since z = 0.1438 is less than the critical value
z = 1.645, the test statistic is NOT in the
critical (rejection) region. Thus, the null
hypothesis is not rejected.
CONCLUSION: At 5% level of significance,
there is enough evidence to reject the claim
that the mean weight of the babies in the ward
is greater than 2.5 kg.

15. Test for
MEAN
EXAMPLE 2
2. The X Last Company has developed a new
battery. The engineering department of the
company claims that each battery lasts for 200
minutes. In order to test this claim, the
company selects a random sample of 100 new
batteries so that this sample has a mean of
190 minutes with a standard deviation of 30
minutes. Test the claim using 0.01 level of
significance.

16. Test for
MEAN
EXAMPLE 2
1. 𝐻0: 𝜇 = 200
𝐻1: 𝜇 ≠ 200
2. 𝛼 = 0.01
3. 𝑧 𝑎 = ±2.575

17. Test for
MEAN
EXAMPLE 2
Sample Mean: 190
𝑧 =
𝑥 − 𝜇0
𝜎𝑥
𝑛
𝑧 =
190 − 200
30
100
𝑧 = −3.33

18. Test for
MEAN
EXAMPLE 2
Since z = -3.33 is less than the critical value
z = -2.575 the test statistic is in the critical
(rejection) region. Thus, the null hypothesis is
rejected.
CONCLUSION: At 1% level of significance,
there is enough evidence to reject the claim
that the mean number of minutes that the new
batteries last is 200 minutes.

19. Test for
MEAN
t Test for Mean (𝜎 is
unknown or 𝑛 < 30)
• This is a statistical test for population
mean. The t test can be used when
the population is normal and 𝜎 is
unknown or when the sample size is
less than 30.
• uses the degree of freedom

20. Test for
MEAN
t Test for Mean (𝜎 is
unknown or 𝑛 < 30)
t=
𝑥−𝜇0
𝑠
𝑛

21. Test for
MEAN
EXAMPLE 1
Find the critical value(s) 𝑡∝ when ∝= 0.05 with
degrees of freedom df=16 for
a. a right-tailed t test
b. a left-tailed t test
c. a two-tailed test
a. 𝑡∝ = 1.746
b. 𝑡∝ = −1.746
c. 𝑡∝ = ±2.120

22. Test for
MEAN
EXAMPLE 1
A government agency is investigating a
complaint from some concerned citizens who
said that there is short-weight selling of rice in
a certain town. An agent manufacturer took a
random sample of 20 sacks of “50-kilo” sacks
of rice from a large shipment and finds that the
mean weight is 49.7 kilos with a standard
deviation of 0.35 kilo. Is this an evidence of
short-weighing at the 0.01 level of
significance?

23. Test for
MEAN
EXAMPLE 1
1. 𝐻0: 𝜇 = 50
𝐻1: 𝜇 < 50
2. 𝛼 = 0.01
3. 𝛼 = 0.01, df=19
𝑡∝ = −2.539

24. Test for
MEAN
EXAMPLE 1
Sample Mean: 49.7
𝑡 =
𝑥 − 𝜇0
𝑠
𝑛
𝑡 =
49.7 − 50
0.35
20
𝑡 = −3.8333

25. Test for
MEAN
EXAMPLE 1
Since t = -3.8333 is less than the critical value
t = -2.539 the test statistic is in the critical
(rejection) region. Thus, the null hypothesis is
rejected.
CONCLUSION: At 1% level of significance,
there is enough evidence to reject the claim
that the mean weight of each sack is equal 50
kilograms.

26. Test for
MEAN
EXAMPLE 2
A sports trainer wants to know whether the
true average time of his athletes who do 100
meter sprint is 98 seconds. He recorded 18
trials of his team and found that the average
time is 98.2 seconds with a standard deviation
of 0.4 second. Is there a sufficient evidence to
reject the null hypothesis that 𝜇 = 98 seconds
at the 0.05 level of significance?

27. Test for
MEAN
EXAMPLE 2
1. 𝐻0: 𝜇 = 98
𝐻1: 𝜇 ≠ 98
2. 𝛼 = 0.05
3. 𝛼/2 = 0.025, df=17
𝑡∝ = ±2.110

28. Test for
MEAN
EXAMPLE 2
Sample Mean: 98.2
𝑡 =
98.2 − 98
0.4
18
𝑡 = 2.1213

29. Test for
MEAN
EXAMPLE 2
Since t = 2.1213 is within the rejection region,
the null hypothesis is rejected.
CONCLUSION: At 5% level of significance,
there is enough evidence to reject the claim
that the mean time is equal to 98 seconds.