A student has heard that spinning pennies on a table, rather than flipping them in the air, results in tails-side up 60% of the time. She hypothesizes that a spinning penny lands tails-side up even more frequently than 60% of the time, due to the extra metal on the "heads” side of the coin shifting its center of mass. She spins a coin 76 times, and finds the coin coming to rest tails-side up in 55 of those spins. What is the z-statistic and the P-value for the significance test?

Find the z-table here.

The test statistic is z = and the P-value = 0.9930.

The test statistic is z = and the P-value = 0.0070.

PART 2

A medical device company has developed a new needle that they claim reduces the incidence of injection-site reactions for a particular medication. The percentage of patients experiencing injection-site reactions with the current needle is 11%. The new needle is tested on 250 volunteers, and 20 of those volunteers experienced injection-site reactions with the new needle. The z-statistic for the test was calculated to be –1.52. Which of the following statements has the correct P-value and decision for the test at the = 0.10 significance level?

Find the z-table here.

The P-value = 0.08, and the researchers should conclude that the new needle produces fewer injection-site reactions because 0.08 < 0.10.

The P-value = 0.0643, and the researchers should conclude that the new needle produces fewer injection-site reactions because 0.0643 < 0.10.

The P-value = 0.1286, and the researchers should conclude that the new needle does not produce fewer injection-site reactions because 0.1286 > 0.10.

The P-value = 0.0643, and the researchers should conclude that the new needle does not produce fewer injection-site reactions because 0.0643 < 0.10.

The P-value = 0.9357, and the researchers should conclude that the new needle does not produce fewer injection-site reactions because 0.9375 > 0.10.

The test statistic is z = and the P-value = 0.9861.

The test statistic is z = and the P-value = 0.0139.

The test statistic is z = and the P-value = 0.0139.

PART 3

In the US, primary elections are held to determine who will be the presidential candidate for a particular party. Two of the first states to hold primary elections for president are Iowa and New Hampshire. A candidate wishes to know if her popularity in these states is similar. She decides to conducts polls of random samples of voters to determine their voting preference. A recent poll of 825 voters in Iowa shows this candidate has 24% of the vote, while a different poll of 775 voters from New Hampshire shows this candidate has 20% of the vote in New Hampshire.

Do these polls give strong evidence that the candidate’s share of the vote in these two states is different, at the = 0.05 significance level?

Find the z-table here.

The test statistic is z = 1.93 and the P-value = 0.0538. Because 0.0538 > 0.05, there is not sufficient evidence that the candidate’s levels of support are different in the two states.

The test statistic is z = 1.93 and the P-value = 0.0538. Because 0.0538 > 0.05, there is sufficient evidence that the candidate’s levels of support are different in the two states.

The test statistic is z = 1.93 and the P-value = 0.0269. Because 0.0269 < 0.05, there is not sufficient evidence that the candidate’s levels of support are different in the two states.

The test statistic is z = –1.93 and the P-value = 0.0269. Because 0.0269 < 0.05, there is sufficient evidence that the candidate’s levels of support are different in the two states.

The test statistic is z = –1.93 and the P-value = 0.9731. Because 0.9731 > 0.05, there is not sufficient evidence that the candidate’s levels of support are different in the two states.

Z-Table (edgenuity.com)