Test for

Test forMEAN

WHY test the mean?

Test forMEANTwo Statistical Test Concerning Means

• z test • t test

Test forMEANGeneral Formula for Computation of Test

Statistic• observed value (sample mean)

• expected value (population mean)

Test forMEANStandard Error for the

Mean (Pop. SD)

Test forMEANStandard Error for the Mean (Sample. SD)

Test forMEANz Test for Mean ( is

known or • This is a statistical test for population

mean. The z test can be used when the population is normal and is known

or when the sample size is greater than or equal to 30.

Test forMEANz Test for Mean ( is

known or 𝑇𝑒𝑠𝑡 𝑆𝑡𝑎𝑡𝑖𝑐=

𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑𝑉𝑎𝑙𝑢𝑒−𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑𝑉𝑎𝑙𝑢𝑒𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝐸𝑟𝑟𝑜𝑟

𝑧=𝑠𝑎𝑚𝑝𝑙𝑒𝑚𝑒𝑎𝑛−𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛𝑚𝑒𝑎𝑛

𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝐸𝑟𝑟𝑜𝑟

𝑧=𝑥−𝜇0𝜎𝑥

√𝑛

Test forMEANCommon Levels of Significance

and Critical ValuesONE-TAILED TEST TWO-TAILED TEST

LEFT (<) RIGHT (>)

0.100.050.01

Test forMEAN

Test forMEANEXAMPLE 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 .

Test forMEANEXAMPLE 1

1.

2.

3.

Test forMEAN EXAMPLE 1

Sample Mean: 2.5083

𝑧=𝑥−𝜇0𝜎𝑥

√𝑛

𝑧=2.5083−2.5

0.2√12

𝑧=0.1438

Test forMEAN 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.

Test forMEANEXAMPLE 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.

Test forMEANEXAMPLE 2

1.

2. 1

3.

Test forMEAN EXAMPLE 2

Sample Mean: 190

𝑧=𝑥−𝜇0𝜎𝑥

√𝑛

𝑧=190−20030

√100𝑧=−3.33

Test forMEAN 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.

Test forMEANt Test for Mean ( is

unknown or • This is a statistical test for population

mean. The t test can be used when the population is normal and is

unknown or when the sample size is less than 30.

• uses the degree of freedom

Test forMEANt Test for Mean ( is

unknown or t

Test forMEAN EXAMPLE 1

Find the critical value(s) when with degrees of freedom df=16 for

a. a right-tailed t testb. a left-tailed t testc. a two-tailed test

a. b. c.

Test forMEAN 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?

Test forMEANEXAMPLE 1

1.

2.

3. , df=19

Test forMEAN EXAMPLE 1

Sample Mean: 49.7

𝑡=𝑥−𝜇0𝑠

√𝑛

𝑡=49.7−500.35√20

𝑡=−3.8333

Test forMEAN 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.

Test forMEAN 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 seconds at the

0.05 level of significance?

Test forMEANEXAMPLE 2

1.

2.

3. , df=17

Test forMEAN EXAMPLE 2

Sample Mean: 98.2

𝑡=98.2−980.4√18

𝑡=2.1213

Test forMEAN 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.