Calculating a single sample z test by hand
-
Upload
byu-center-for-teaching-learning -
Category
Education
-
view
218 -
download
0
description
Transcript of 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-1Ο +1Ο-2Ο +2Ο
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-1Ο +1Ο-2Ο +2Ο
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-1Ο +1Ο-2Ο +2Ο
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).
Letβs 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).
Letβs 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).
Letβs calculate the z statistic and see where if falls!
We do this by using the following equation:π πππππππππ=
οΏ½ΜοΏ½βπ
βπ (1βπ)π
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).
Letβs 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).
π πππππππππ=οΏ½ΜοΏ½βπ
βπ (1βπ)π
Hereβs 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
π πππππππππ=οΏ½ΜοΏ½βπ
βπ (1βπ)π
π πππππππππ=οΏ½ΜοΏ½βπ
βπ (1βπ)π
Note β this little hat () over the p means that this proportion is an
estimate of a population
(.90)
π πππππππππ=οΏ½ΜοΏ½βπ
βπ (1βπ)π
π πππππππππ=οΏ½ΜοΏ½βπ
βπ (1βπ)π
(100)
π πππππππππ=οΏ½ΜοΏ½βπ
βπ (1βπ)π
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)
π πππππππππ=οΏ½ΜοΏ½βπ
βπ (1βπ)π
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)
π πππππππππ=οΏ½ΜοΏ½βπ
βπ (1βπ)π
Letβs plug in the numbers
π πππππππππ=.82βπ
βπ (1βπ)π
Sample Proportion
π πππππππππ=.82βπ
βπ (1βπ)π
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
π πππππππππ=.82βπ
βπ (1βπ)π
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
π πππππππππ=.82β .90
β .90(1β .90)π
Population Proportion
π πππππππππ=.82β .90
β .90(1β .90)π
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
π πππππππππ=.82β .90
β .90(1β .90)π
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
π πππππππππ=.82β .90
β .90(1β .90)π
π πππππππππ=β .08
β .90(1β .90)π
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)
π πππππππππ=β .08
β .90(1β .90)π
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)
π πππππππππ=β .08
β .90(1β .90)π
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.
π πππππππππ=β .08
β .90(1β .90)π
Letβs continue our calculations and find out:
π πππππππππ=β .08
β .90(1β .90)π
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
Letβs continue our calculations and find out:
π πππππππππ=β .08
β .90( .10)π
Letβs continue our calculations and find out:
π πππππππππ=β .08
β .09π
Letβs continue our calculations and find out:
π πππππππππ=β .08
β .09100
Sample Size:
π πππππππππ=β .08
β .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:
π πππππππππ=β .08
β.0009
Letβs continue our calculations:
π πππππππππ=β .08.03
Letβs continue our calculations:
π πππππππππ=β2.67
Letβs continue our calculations:
π πππππππππ=β2.67
Letβs continue our calculations:
Now we have our z statistic.
Letβs go back to our distribution:
rarerare
95%
mean-1Ο +1Ο-2Ο +2Ο
Common
-1.96 +1.96
Letβs go back to our distribution: So, is this result rare or common?
rarerare
95%
mean-1Ο +1Ο-2Ο +2Ο
Common
-1.96 +1.96-2.67
Letβs go back to our distribution: So, is this result rare or common?
rarerare
95%
mean-1Ο +1Ο-2Ο +2Ο
Common
-1.96 +1.96
This is the Z-Statistic we
calculated
-2.67
Letβs go back to our distribution: So, is this result rare or common?
rarerare
95%
mean-1Ο +1Ο-2Ο +2Ο
Common
-1.96 +1.96-2.67
This is the Z β 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.