MATH 225N Week 7 Hypothesis Testing Questions and Answers.pdf

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 < ???? = 0.05 we reject Ho and support the Ha

Hypothesis Test for p population

proportion

Level of Significance 0.05 (decimal)

Proportion under H0 0.5000 (decimal)

n 50

Number of Successes 32

Sample Proportion 0.640000

StDev 0.500000

SE 0.070711

Test Statistic (z) 1.979899

One-Sided p-value 0.023852

Two-Sided p-value 0.047704

Right-Tailed (>) 1.644854

Left-Tailed (<) -1.644854

Two-Tailed (≠) ± 1.959964

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 (???? ???????? ???????????? ???????????????????? ???????? ???????????????????? ???????? 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 Proportion 0.540000

StDev 0.500000

SE 0.035355

Test Statistic (z) 1.131371

One-Sided p-value 0.129238

Two-Sided p-value 0.258476

Right-Tailed (>) 1.644854

Left-Tailed (<) -1.644854

Two-Tailed (≠) ± 1.959964

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 (???? ???????? ???????????? ???????????????????? ???????? ???????????????????? ???????? 0.05)

Hypothesis Test for p population

proportion

Level of Significance 0.05 (decimal)

Proportion under H0 0.0580 (decimal)

n 500

Number of Successes 21

Sample Proportion 0.042000

StDev 0.233743

SE 0.010453

Test Statistic (z)

-

1.530613

One-Sided p-value 0.063008

Two-Sided p-value 0.126016

Right-Tailed (>) 1.644854

Left-Tailed (<) -1.644854

Two-Tailed (≠) ± 1.959964

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

Sample Proportion 0.040667

StDev 0.217945

SE 0.005627

Test Statistic (z)

-

1.658577

One-Sided p-value 0.048457

Two-Sided p-value 0.096914

Answer: 0.048