MATH 225N WEEK 7 HYPOTHESIS TESTING Q & A
1. Steve listens to his favorite streaming music service when he works out. He wonders
whether the service algorithm does a good job of finding random songs that he will
like more often than not. To test this, he listens to 50 songs chosen by the service at
random and finds that he likes 32 of them.
Use Excel to test whether Steve will like a randomly selected song more than not and then draw a
conclusion in the context of a problem. Use α = 0.05. Type equation here .
Ho: p = ≤ 0.5 (50%) p = 0.5
Ha: p = > 0.5 (strictly ¿>≠ )
P-value = 0.02 which is < α =0.05 we reject Ho and support the Ha
Hypothesis Test for p
population proportion
Level of (decim
Significance 0.05 al)
Proportion (decim
under H0 0.5000 al)
n 50
Number of
Successes 32
Sample 0.6400
Proportion 00
0.5000
StDev 00
0.0707
SE 11
Test Statistic 1.9798
(z) 99
One-Sided p- 0.0238
value 52
Two-Sided p- 0.0477
value 04
, Right-Tailed 1.64485
(>) 4
-
Left-Tailed 1.64485
(<) 4
Two-Tailed 1.95996
(≠) ± 4
Answer: Reject the null hypothesis. There is sufficient evidence to prove that Steve will like a random
selected song more often than not.
2. A magazine regularly tested products and gave the reviews to its customers. In one of
its reviews, it tested 2 types of batteries and claimed that the batteries from company
A outperformed batteries from company B in 108 of the tests. There were 200 tests.
Company B decided to sue the magazine, claiming that the results were not
significantly different from 50% and that the magazine was slandering its good name.
Use Excel to test whether the true proportion of times that Company A’s batteries outperformed
Company B’s batteries is different from 0.5. Identify the p=value rounding it to 3 decimal places.
Ho: p = 0.5 Ha ≠ 0.5 (two tailed test) n = 200 (α is not given so leave it 0.05)
Hypothesis Test for p
population proportion
Level of
Significance 0.05
Proportion
under H0 0.5000
n 200
Number of
Successes 108
Sample 0.54000
Proportion 0
0.50000
StDev 0
0.03535
SE 5
Test Statistic 1.13137
(z) 1
One-Sided p- 0.12923
value 8
Two-Sided p- 0.25847
, value 6
Right-Tailed 1.64485
(>) 4
-
Left-Tailed 1.64485
(<) 4
Two-Tailed 1.95996
(≠) ± 4
Answer: 0.258 (because it is a two tailed test). We are not rejecting the null hypothesis and we do not
have evidence to support the alternative hypothesis.
3. A candidate in an election lost by 5.8% of the vote. The candidate sued the state and
said that more than 5.8% of the ballots were defective and not counted by the voting
machine, so a full recount would need to be done. His opponent wanted to ask for the
case to be dismissed, so she had a government official from the state randomly select
500 ballots and count how many were defective. The official found 21 defective ballots.
Use Excel to test if the candidates claim is true and that < 5.8% of the ballots were defective. Identify
the p=value rounding to 3 decimal places.
Ho: p = ≥ 0.058 Ha ¿ 0.058 (one tailed test) n = 500 (α is not given so leave it 0.05)
Hypothesis Test for p population
proportion
Level of (decima
Significance 0.05 l)
Proportion (decima
under H0 0.0580 l)
n 500
Number of
Successes 21
Sample 0.0420
Proportion 00
0.2337
StDev 43
0.0104
SE 53
-
Test Statistic 1.5306
(z) 13
One-Sided p- 0.0630
value 08
Two-Sided p- 0.1260
, value 16
Right-Tailed 1.64485
(>) 4
-
Left-Tailed 1.64485
(<) 4
Two-Tailed 1.95996
(≠) ± 4
Answer: 0.063
4. A researcher claims that the incidence of a certain type of cancer is < 5%. To test this
claim, a random sample of 4000 people are checked and 170 are found to have the
cancer.
The following is the set up for the hypothesis:
Ho = 0.05
Ha = < 0.05
In the example the p-value was determined to be 0.015.
Come to a conclusion and interpret the results of this hypothesis test for a proportion (use a significance
level of 5%)
Answer: The decision is to reject the null hypothesis. The conclusion is that there is enough evidence
to support the claim.
5. A researcher is investigating a government claim that the unemployment rate is < 5%.
TO test this claim, a random sample of 1500 people is taken and it is determined that
61 people were unemployed.
Ho: p = 0.05 Ha: p < 0.05
Find the p-value for this hypothesis test for a proportion & round to 3 decimal places.
Hypothesis Test for p
population proportion
Level of
Significance 0.05
Proportion
under H0 0.0500
n 1500
Number of
Successes 61
1. Steve listens to his favorite streaming music service when he works out. He wonders
whether the service algorithm does a good job of finding random songs that he will
like more often than not. To test this, he listens to 50 songs chosen by the service at
random and finds that he likes 32 of them.
Use Excel to test whether Steve will like a randomly selected song more than not and then draw a
conclusion in the context of a problem. Use α = 0.05. Type equation here .
Ho: p = ≤ 0.5 (50%) p = 0.5
Ha: p = > 0.5 (strictly ¿>≠ )
P-value = 0.02 which is < α =0.05 we reject Ho and support the Ha
Hypothesis Test for p
population proportion
Level of (decim
Significance 0.05 al)
Proportion (decim
under H0 0.5000 al)
n 50
Number of
Successes 32
Sample 0.6400
Proportion 00
0.5000
StDev 00
0.0707
SE 11
Test Statistic 1.9798
(z) 99
One-Sided p- 0.0238
value 52
Two-Sided p- 0.0477
value 04
, Right-Tailed 1.64485
(>) 4
-
Left-Tailed 1.64485
(<) 4
Two-Tailed 1.95996
(≠) ± 4
Answer: Reject the null hypothesis. There is sufficient evidence to prove that Steve will like a random
selected song more often than not.
2. A magazine regularly tested products and gave the reviews to its customers. In one of
its reviews, it tested 2 types of batteries and claimed that the batteries from company
A outperformed batteries from company B in 108 of the tests. There were 200 tests.
Company B decided to sue the magazine, claiming that the results were not
significantly different from 50% and that the magazine was slandering its good name.
Use Excel to test whether the true proportion of times that Company A’s batteries outperformed
Company B’s batteries is different from 0.5. Identify the p=value rounding it to 3 decimal places.
Ho: p = 0.5 Ha ≠ 0.5 (two tailed test) n = 200 (α is not given so leave it 0.05)
Hypothesis Test for p
population proportion
Level of
Significance 0.05
Proportion
under H0 0.5000
n 200
Number of
Successes 108
Sample 0.54000
Proportion 0
0.50000
StDev 0
0.03535
SE 5
Test Statistic 1.13137
(z) 1
One-Sided p- 0.12923
value 8
Two-Sided p- 0.25847
, value 6
Right-Tailed 1.64485
(>) 4
-
Left-Tailed 1.64485
(<) 4
Two-Tailed 1.95996
(≠) ± 4
Answer: 0.258 (because it is a two tailed test). We are not rejecting the null hypothesis and we do not
have evidence to support the alternative hypothesis.
3. A candidate in an election lost by 5.8% of the vote. The candidate sued the state and
said that more than 5.8% of the ballots were defective and not counted by the voting
machine, so a full recount would need to be done. His opponent wanted to ask for the
case to be dismissed, so she had a government official from the state randomly select
500 ballots and count how many were defective. The official found 21 defective ballots.
Use Excel to test if the candidates claim is true and that < 5.8% of the ballots were defective. Identify
the p=value rounding to 3 decimal places.
Ho: p = ≥ 0.058 Ha ¿ 0.058 (one tailed test) n = 500 (α is not given so leave it 0.05)
Hypothesis Test for p population
proportion
Level of (decima
Significance 0.05 l)
Proportion (decima
under H0 0.0580 l)
n 500
Number of
Successes 21
Sample 0.0420
Proportion 00
0.2337
StDev 43
0.0104
SE 53
-
Test Statistic 1.5306
(z) 13
One-Sided p- 0.0630
value 08
Two-Sided p- 0.1260
, value 16
Right-Tailed 1.64485
(>) 4
-
Left-Tailed 1.64485
(<) 4
Two-Tailed 1.95996
(≠) ± 4
Answer: 0.063
4. A researcher claims that the incidence of a certain type of cancer is < 5%. To test this
claim, a random sample of 4000 people are checked and 170 are found to have the
cancer.
The following is the set up for the hypothesis:
Ho = 0.05
Ha = < 0.05
In the example the p-value was determined to be 0.015.
Come to a conclusion and interpret the results of this hypothesis test for a proportion (use a significance
level of 5%)
Answer: The decision is to reject the null hypothesis. The conclusion is that there is enough evidence
to support the claim.
5. A researcher is investigating a government claim that the unemployment rate is < 5%.
TO test this claim, a random sample of 1500 people is taken and it is determined that
61 people were unemployed.
Ho: p = 0.05 Ha: p < 0.05
Find the p-value for this hypothesis test for a proportion & round to 3 decimal places.
Hypothesis Test for p
population proportion
Level of
Significance 0.05
Proportion
under H0 0.0500
n 1500
Number of
Successes 61
