single sample z test - explain (final)

Single-Sample Z TestTheoretical Explanation

A one-sample Z-test for proportions is a test that helps us compare a population proportion with a sample proportion.

A one-sample Z-test for proportions is a test that helps us compare a population proportion with a sample proportion.

Sample Population

30% 32%

A one-sample Z-test for proportions is a test that helps us compare a population proportion with a sample proportion.

Here is our question: Are the population and the sample proportions (which supposedly have the same general characteristics as the population) statistically significantly the same or different?

Sample Population30% 32%

Consider the following example:

Consider the following example: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

Consider the following example:

Which is the sample?

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

Consider the following example:

Which is the sample? What is the population?

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

Consider the following example:

Which is the sample? What is the population?The sample proportion is .82 – 82 out of doctors recommend aspirin.

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

Consider the following example:

Which is the sample? What is the population?The sample proportion is .82 – 82 out of doctors recommend aspirin.The population proportion .90 (the claim that 9 out of 10 doctors recommend aspirin).

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

We begin by stating the null hypothesis:

We begin by stating the null hypothesis:

The alternative hypothesis would be:

The proportion of a sample of 100 medical doctors who recommend aspirin for their patients with headaches IS NOT statistically significantly different from the claim that 9 out of 10 doctors recommend aspirin for their patients with headaches.

We begin by stating the null hypothesis:

The alternative hypothesis would be:

The proportion of a sample of 100 medical doctors who recommend aspirin for their patients with headaches IS NOT statistically significantly different from the claim that 9 out of 10 doctors recommend aspirin for their patients with headaches.

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.

State the decision rule: We will calculate what is called the z statistic which will make it possible to determine the likelihood that the sample proportion (.82) is a rare or common occurrence with reference to the population proportion (.90).

State the decision rule: We will calculate what is called the z statistic which will make it possible to determine the likelihood that the sample proportion (.82) is a rare or common occurrence with reference to the population proportion (.90).

If the z-statistic falls outside of the 95% common occurrences and into the 5% rare occurrences then we will conclude that it is a rare event and that the sample is different from the population and therefore reject the null hypothesis.

Before we calculate this z-statistic, we must locate the z critical values.

Before we calculate this z-statistic, we must locate the z critical values.

What are the z critical values? These are the values that demarcate what is the rare and the common occurrence.

Let’s look at the normal distribution:

It has some important properties that make it possible for us to locate the z statistic and compare them to the z criticals.

Here is the mean and the median of a normal distribution.

50% of the values are above and below the orange line.

50% - 50% +

68% of the values fall between +1 and -1 standard deviations from the mean.

34% - 34% +

68% of the values fall between +1 and -1 standard deviations from the mean.

34% - 34% +

68%

mean-1σ +1σ

68% of the values fall between +1 and -1 standard deviations from the mean.

34% - 34% +

95%

mean-1σ +1σ-2σ +2σ

Since our decision rule is .05 alpha, this means that if the z value falls outside of the 95% common occurrences we will consider it a rare occurrence.

34% - 34% +

95%

mean-1σ +1σ-2σ +2σ

Since are decision rule is .05 alpha we will see if the z statistic is rare using this visual

95%

mean-1σ +1σ-2σ +2σ

2.5% 2.5%

rarerare

Or common

95%

mean-1σ +1σ-2σ +2σ

Common

Before we can calculate the z – statistic to see if it is rare or common we first must determine the z critical values that are associated with -2σ and +2σ.

95%

mean-1σ +1σ-2σ +2σ

Common

We look these up in the Z table and find that they are -1.96 and +1.96

95%

mean-1σ +1σ-2σ +2σ

Common

-1.96 +1.96Z values

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

So if the z statistic we calculate is less than -1.96 or greater than +1.96 then we will consider this to be rare and reject the null hypothesis and state that there is 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 or greater than +1.96 then we will consider this to be rare and reject the null hypothesis and state that there is 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 or greater than +1.96 then we will consider this to be rare and reject the null hypothesis and state that there is 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 or greater than +1.96 then we will consider this to be rare and reject the null hypothesis and state that there is 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

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

(.90)

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

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

Let’s plug in the numbers

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

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=− .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

The difference

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

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.

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

If the standard error is small then the z statistic will be larger. If it exceeds the -1.96 or +1.96 boundaries, then we will reject the null hypothesis. If it is smaller than we will not.

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

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

√ .09𝑛

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

√ .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

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:

𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄=− .08.03

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:

𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄=−2.67

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:

𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄=−2.67

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:

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 +1.96-2.67

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.