Many of the ideas behind single sample testing of proportions carry over from the testing of means.
Whenever we are dealing with percentages, we are dealing with proportions. For example, if we take a survey, we might find that 45% of the respondents responded in a favorable manner. Or if we count the number of mutant fruit flies among the wild ones, we might find the percentage to be 4%.
As with means, our goal is to test a claim made about the proportion of the population. There are very few differences between hypotheses testing of means and testing of proportions. One of the differences, thought minor, is how we express the statement of the hypotheses. We use p instead of 𝜇. For example, if the claim is that p equals 60%, we would have,
𝐻
0
: 𝑝 = 0.60
𝐻
1
: 𝑝 ≠ 0.60
The only other major difference is how we compute the test statistic and the critical value or p-value.
Whereas the testing of means uses the Student t distribution, testing of proportions uses the Standard
Normal distribution.
The test statistic is given by, 𝑧 = 𝑝̂ − 𝑝
√ 𝑝𝑞 𝑛 where p is the claimed proportion and 𝑝̂ is the sample proportion. Be careful.
We run into the same problems using NORM.INV that we did before when doing confidence intervals.
The table shows the various adjustments we have to make for a given tail type test. As before, 𝛼 is the significance and is equal to 1 − 𝐶𝐿. 𝑧 ∗ is the critical value and pv is the p-Value.
LT
RT
2T
𝑧 ∗ = 𝑁𝑂𝑅𝑀. 𝑆. 𝐼𝑁𝑉(𝛼) 𝑝𝑣 = 𝑁𝑂𝑅𝑀. 𝑆. 𝐷𝐼𝑆𝑇(𝑧)
𝑧 ∗ = 𝑁𝑂𝑅𝑀. 𝑆. 𝐼𝑁𝑉(1 − 𝛼) 𝑝𝑣 = 1.0 − 𝑁𝑂𝑅𝑀. 𝑆. 𝐷𝐼𝑆𝑇(𝑧)
𝑧 ∗ = 𝑁𝑂𝑅𝑀. 𝑆. 𝐼𝑁𝑉(1 − 𝛼 2 𝑝𝑣 = 2[1.0 − 𝑁𝑂𝑅𝑀. 𝑆. 𝐷𝐼𝑆𝑇(|𝑧|)]
Note: we must use absolute value of z for a two-tail test.
For a complete summary of how to perform One Sample Testing of Proportions, see the chart,
Hypotheses Testing Chart SSP.
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In 1997, a survey of 940 households showed that 168 of them use email. Test the claim that more than 15% of households use email using a 95% confidence level (𝛼 = 0.05) .
Hypotheses Statement:
𝐻
0
: 𝑝 = 0.15
𝐻
1
: 𝑝 > 0.15
Calculate the z-score: 𝑥 𝑝̂ = 𝑛
=
168
940
= 0.1787
𝑧 = 𝑝̂ − 𝑝
√ 𝑝𝑞 𝑛
=
0.1787 − 0.15
= 2.4663
Calculate the critical value:
𝑧 ∗ 𝑧 ∗
= 𝑁𝑂𝑅𝑀. 𝑆. 𝐼𝑁𝑉(1 − 𝛼)
= 𝑁𝑂𝑅𝑀. 𝑆. 𝐼𝑁𝑉(0.95) = 1.6449
Calculate the p-Value: 𝑝𝑣 = 1.0 − 𝑁𝑂𝑅𝑀. 𝑆. 𝐷𝐼𝑆𝑇(𝑧) 𝑝𝑣 = 1.0 − 𝑁𝑂𝑅𝑀. 𝑆. 𝐷𝐼𝑆𝑇(2.4663) = 0.0068
Result:
For critical value: 𝑧 > 𝑧 ∗ , therefore Reject the Null Hypothesis
For p-Value: 𝑝𝑣 < 𝛼, therefore Reject the Null Hypothesis
Conclusion:
There is sufficient evidence to support the claim that more than 15% of households used email in 1997.
Here’s the spreadsheet that performs the above calculations: x
168 n
940 p^ p q z z* pv
0.1787
0.1500
0.8500
2.4663
1.6449
0.0068
If you look under Miscellaneous on the website, Hypotheses Testing Chart SSP is a wonderful summary of this Unit.
Note : Sometimes it can be confusing as to whether rejecting the null hypothesis means supporting the claim or not supporting it. It very much depends on how the claim is worded. The key is carefully analyzing what the alternative hypothesis is saying, and how that relates to the claim. Note the alternative statement in all of the examples below and interpret what is being said. And remember, rejecting the null hypothesis means that the alternative is correct and failing to reject the null hypothesis means that the alternative is not correct. It’s all about the alternative .
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Here are some examples.
Claim: A hit song must be less than 210 seconds. 𝐻
1
: 𝜇 < 210
Reject the null hypothesis: this means that hit songs must be less than 210 seconds and we have
Fail to reject 𝐻
0
: supported the claim. this means that hit songs do not have to be less than 210 seconds and so, we do not support the claim.
Claim: A hit song must be more than 210 seconds. 𝐻
1
: 𝜇 > 210
Reject 𝐻
0
: the claim is supported.
Fail to reject 𝐻
0
: the claim cannot be supported, hit songs can be less than 210 seconds.
Claim: A hit song is 210 seconds long. 𝐻
1
: 𝜇 ≠ 210
Reject 𝐻
0
: 𝐻
1 says that a hit song is not 210 seconds longs. If we reject the null, then we must go with the alternative, and hence cannot support the claim.
Fail to reject 𝐻
0
: the claim is supported.
As you can see, support or not support do not always line up with either just rejecting the null or failing to reject the null. You have to look at the alternative and think about what it means in terms of rejecting or failing to reject the null. Here are two more examples.
Claim: A hit song must be at least 210 seconds long. 𝐻
1
: 𝜇 < 210
Reject 𝐻
0
: Rejecting the null means that according to the statement of the alternative hypothesis, hit song are less than 210 seconds. Hence, we cannot support the
Fail to reject 𝐻
0
: claim.
Failing to reject the null means that the alternative hypothesis is wrong. Hence hit songs are at least (the opposite of less than) 210 seconds long and so, we would support the claim.
Claim: A hit song must be no more than 210 seconds long: 𝐻
1
: 𝜇 > 210
Reject 𝐻
0
: Rejecting the null means that according to the alternative hypothesis, hit song have to be greater than 210 seconds, and hence we cannot support the claim.
Fail to reject 𝐻
0
: This means that the alternative hypothesis is not correct, and so, it is true that hit songs need be no more than (the opposite of greater than) 210 and so, we would support the claim.
I know this is confusing at first, but if you really think about it, you’ll get it.
Now, you try a few.
Question #1
The claim is that passengers and their carryon weigh at most 200 lbs. If we end up rejecting the null hypothesis, do we support or not support the claim? First write out the statement of the hypotheses, draw a line through the null hypothesis, and then interpret exactly what the alternative is saying.
Question #2
The claim is that the mean reading of a batch of thermometers immersed in ice water will read exactly 0 ℃. If we reject the null hypothesis, do we support or not support the claim.
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Left-Tail Test x
140 n
940 p^ p (claim)
0.148936
0.15
q
0.85
In this case, we would fail to reject the null hypothesis.
CL z z* pv
0.95
-0.09134 -1.64485
0.46361
Right-Tail Test x
160 n
940 p^ p (claim)
0.170213
0.15
In this case, we would reject the null hypothesis. q
0.85
CL
0.95
z z* pv
1.73554 1.644854 0.041323
Two-Tail Test x
150 n p^
940 0.159574
p (claim)
0.15
q
0.85
CL z z* pv
0.95 0.822098 1.959964 0.411021
In this case, we would fail to reject the null hypothesis.
This is the end of Unit 25. In class, you will get more practice with these concepts by working exercises in MyMathLab .
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