calculating a single sample z test by hand

Calculating a Single-Sample Z Test

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

We first determine the z-critical for our question.

We will use the following problem:

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

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.

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.

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)

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

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)

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

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

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

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).

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!

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:

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 (1−𝑝)𝑛

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 (1−𝑝)𝑛

Here’s the problem again:

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

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

𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄=�̂�−𝑝

√𝑝 (1−𝑝)𝑛

𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄=�̂�−𝑝

√𝑝 (1−𝑝)𝑛

Note – this little hat () over the p means that this proportion is an

estimate of a population

(.90)

𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄=�̂�−𝑝

√𝑝 (1−𝑝)𝑛

𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄=�̂�−𝑝

√𝑝 (1−𝑝)𝑛

(100)

𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄=�̂�−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

(100)

𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄=�̂�−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

(100)

𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄=�̂�−𝑝

√𝑝 (1−𝑝)𝑛

Let’s plug in the numbers

𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄=.82−𝑝

√𝑝 (1−𝑝)𝑛

Sample Proportion

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

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

𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄=.82− .90

√ .90(1− .90)𝑛

Population Proportion

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

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

The difference

𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄=.82− .90

√ .90(1− .90)𝑛

𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄=− .08

√ .90(1− .90)𝑛

The difference

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)

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)

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.

𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄=− .08

√ .90(1− .90)𝑛

Let’s continue our calculations and find out:

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:

𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄=− .08

√ .90( .10)𝑛

Let’s continue our calculations and find out:

𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄=− .08

√ .09𝑛

Let’s continue our calculations and find out:

𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄=− .08

√ .09100

Sample Size:

𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄=− .08

√ .09100

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:

𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄=− .08

√.0009

Let‘s continue our calculations:

𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄=− .08.03

Let‘s continue our calculations:

𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄=−2.67

Let‘s continue our calculations:

𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄=−2.67

Let‘s continue our calculations:

Now we have our z statistic.

Let’s go back to our distribution:

rarerare

95%

mean-1σ +1σ-2σ +2σ

Common

-1.96 +1.96

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

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

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

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

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.