Proportion Z Test

On average, 86% of all enrolled college students are undergraduates. A random sample of 500 college students revealed that 420 were undergraduates. At α = 0.10 level, is there enough evidence to conclude that the proportion is lower than the national average? Find the z-score.

• z = 1.29

• z = 0.09

• z = -1.29

• z = -0.09

On average, 86% of all enrolled college students are undergraduates. A random sample of 500 college students revealed that 420 were undergraduates. At α = 0.10 level, is there enough evidence to conclude that the proportion is lower than the national average? Which is closest to the p-value?

• 0.0099

• 0.00099

• 0.99

• 0.099

On average, 86% of all enrolled college students are undergraduates. A random sample of 500 college students revealed that 420 were undergraduates. At α = 0.10 level, is there enough evidence to conclude that the proportion is lower than the national average? What is the correct conclusion?

• Fail to reject the null hypothesis

• Reject the null hypothesis

On average, 86% of all enrolled college students are undergraduates. A random sample of 500 college students revealed that 420 were undergraduates. At α = 0.10 level, is there enough evidence to conclude that the proportion is lower than the national average?

• There is not enough evidence to conclude that the proportion is lower than the national average

• There is enough evidence to conclude that the proportion is lower than the national average

This is the same write-up as found in the previous question:

A recent newspaper article claimed that the President’s approval rating is 46%; however, a sample of 80 U.S. citizens shows that 43 of them approve of the President’s performance. At a 0.10 level of significance, you are interested in determining if there is evidence that the President’s approval rating is actually higher.

What is the test statistic for this hypothesis test?

• t = -1.39

• z = 1.39

• t = -.538

• z = -.538

This is the same write-up as found in the previous question:

A recent newspaper article claimed that the President’s approval rating is 46%; however, a sample of 80 U.S. citizens shows that 43 of them approve of the President’s performance. At a 0.10 level of significance, you are interested in determining if there is evidence that the President’s approval rating is actually higher.

What is the p-value for this hypothesis test?

• .082

• .164

• .702

• .298

A large-scale study found that 42% of Americans play video games. A random sample of 1100 teenagers were asked whether they play games online; 575 said they did. Is there sufficient evidence to suggest that the proportion of teens who play video games is different from the national figure? Use a 5% level of significance.

• Snice p-value = 5.08 x 10 ^-12 < 0.05, I reject H0. There is enough evidence to suggest a difference.

• Snice p-value = 5.08 x 10 ^-12 < 0.05, I fail to reject H0. There is enough evidence to suggest a difference.

• Snice p-value = 5.08 > 0.05, I fail to reject H0. There is not enough evidence to suggest a difference.

• Snice p-value = 5.08 > 0.05, I reject H0. There is enough evidence to suggest a difference.

Sally reads a newspaper report claiming that 12% of all adults in the U.S. are left-handed. She decides to test this claim. Sally selects a random sample of 55 classmates and finds that 9 were left-handed. Does this provide sufficient evidence that the proportion of lefties in Sally’s school is different than the national figure? Use α = 0.01.

• 0.160

• 0.319

• 0.508

• 0.492

Sally reads a newspaper report claiming that 12% of all adults in the U.S. are left-handed. She decides to test this claim. Sally selects a random sample of 55 classmates and finds that 9 were left-handed. Does this provide sufficient evidence that the proportion of lefties in Sally’s school is different than the national figure? Use α = 0.01.

• Since p-value = 0.319 < 0.01, I reject H0. There is enough evidence to suggest a difference.

• Since p-value = 0.319 > 0.01, I reject H0. There is enough evidence to suggest a difference.

• Since p-value = 0.319 > 0.01, I fail to reject H0. There is not enough evidence to suggest a difference.

• Since p-value = 0.319 < 0.01, I fail to reject H0. There is enough evidence to suggest a difference.

We want to estimate p using a confidence interval that has a 3% margin of error, and we want to have 99% confidence that our interval captures p. We think that p is between .4 and .6. What sample size should we use?

• (1.96/.03)²(.5) ²

• (2.58/.03)²(.5) ²

• (1.96/.5)²(.03)(.97)

• (2.58/.5)² (.03)(.97)