Section 8.3

#9. Given info: n = 325, p-hat = 295/325 = 0.908

We are to "test the claim that ... the probability of a baby being a girl is greater than 0.5," so one of our hypotheses must be "p>0.50." This can't be H_{0}, since H_{0} must have an equal sign in it, so this must be H_{1}. That tells us, then, that H_{0} must be "p = 0.50."

Our test (using the classical approach):

H_{0}: p = 0.50 , H_{1}: p > 0.50 (one tail test)

Choose alpha = 0.01 (that was actually given)

DR: If z_{p-hat} > 2.33 then reject H_{0} (2.33 was the z-score that chopped off the highest 1% of area)

We find z_{p-hat} = (p-hat - p)/sqrt(pq/n) = (0.908 - 0.50)/sqrt(.50*.50/325) = 0.408/0.0277 = 14.73.

Since 14.73 is WELL beyond 2.33, we reject H_{0}.

The proportion is higher than 50%. The method appears to work.

#11. Given info: n = 11,000, p-hat = 5720/11,000 = 0.52

We are to "test the claim that more than 50% of car crashes ...," so one of our hypotheses must be "p>0.50." This can't be H_{0}, since H_{0} must have an equal sign in it, so this must be H_{1}. That tells us, then, that H_{0} must be "p = 0.50."

Our test (using the classical approach):

H_{0}: p = 0.50 , H_{1}: p > 0.50 (one tail test)

Choose alpha = 0.01 (that was actually given)

DR: If z_{p-hat} > 2.33 then reject H_{0} (2.33 was the z-score that chopped off the highest 1% of area)

We find z_{p-hat} = (p-hat - p)/sqrt(pq/n) = (0.52 - 0.50)/sqrt(.50*.50/11000) = 0.02/0.0048 = 4.17.

Since 4.17 is beyond 2.33, we reject H_{0}.

The proportion is higher than 50%.

#12. Given info: n = 734, p-hat = 360/734 = 0.490

We are to "test the claim that among Internet users, less than 50% use it ...," so one of our hypotheses must be "p<0.50." This can't be H_{0}, since H_{0} must have an equal sign in it, so this must be H_{1}. That tells us, then, that H_{0} must be "p = 0.50."

Our test (using the classical approach):

H_{0}: p = 0.50 , H_{1}: p < 0.50 (one tail test)

Choose alpha = 0.01 (that was actually given)

DR: If z_{p-hat} < -2.33 then reject H_{0} (-2.33 was the z-score that chopped off the lowest 1% of area)

We find z_{p-hat} = (p-hat - p)/sqrt(pq/n) = (0.490 - 0.50)/sqrt(.50*.50/734) = -0.01/0.01846 = -0.5417.

Since -0.5417 is NOT less than -2.33, we fail to reject H_{0}.

So 50% of Internet users use the Internet to make travel plans.

#13. Given info: n = 880, p-hat = 149/880 = 0.1693

We are to "test the claim that more than 15% of U.S. households use e-mail," so one of our hypotheses must be "p>0.15." This can't be H_{0}, since H_{0} must have an equal sign in it, so this must be H_{1}. That tells us, then, that H_{0} must be "p = 0.15."

Our test (using the classical approach):

H_{0}: p = 0.15 , H_{1}: p > 0.15 (one tail test)

Choose alpha = 0.05 (that was actually given)

DR: If z_{p-hat} > 1.645 then reject H_{0} (1.645 was the z-score that chopped off the highest 5% of area)

We find z_{p-hat} = (p-hat - p)/sqrt(pq/n) = (0.1693 - 0.15)/sqrt(.15*.85/880) = 0.0193/0.0120 = 1.61.

Since 1.61 is NOT greater than 1.645, we fail to reject H_{0}.

The household usage of email is not more than 15%.

Note: This conclusion is probably NOT valid today, since the usage of email has exploded since 1997.

#14. Given info: n = 1520, p-hat = 58/1520 = 0.038

We are given that the rate was 0.058, and we are to "test the claim that the failure rate is now lower," so one of our hypotheses must be "p<0.058." This can't be H_{0}, since H_{0} must have an equal sign in it, so this must be H_{1}. That tells us, then, that H_{0} must be "p = 0.058."

Our test (using the classical approach):

H_{0}: p = 0.058 , H_{1}: p < 0.058 (one tail test)

Choose alpha = 0.01 (that was actually given)

DR: If z_{p-hat} < -2.33 then reject H_{0} (-2.33 was the z-score that chopped off the lowest 1% of area)

We find z_{p-hat} = (p-hat - p)/sqrt(pq/n) = (0.038 - 0.058)/sqrt(.058*.942/1520) = -0.020/0.006 = -3.33.

Since -3.33 is < -2.33, we reject H_{0}.

There is evidence that fewer job applicants now use drugs.

#17. Given info: n = 1234, p-hat = 20/1234 = 0.0162

We are given that the overcharge rate was 0.01, and we are to decide if scanners help consumers avoid overcharges, which could be translated to "is p<0.01 with scanners?" This must be our H_{1}, so H_{0} must be "p = 0.01."

Our test (using the classical approach):

H_{0}: p = 0.01 , H_{1}: p < 0.01 (one tail test)

Choose alpha = 0.05 (that was actually given)

DR: If z_{p-hat} < -1.645 then reject H_{0} (-1.645 was the z-score that chopped off the lowest 5% of area)

We find z_{p-hat} = (p-hat - p)/sqrt(pq/n) = (0.0162 - 0.01)/sqrt(.01*.99/1234) = 0.0062/0.0028 = 2.21.

Since 2.21 is NOT < -1.645, we fail to reject H_{0}.

There is no evidence that scanners are helping.

#18. Given info: n = 6062+5938 = 12,000, p-hat = proportion in the week prior to Thanksgiving = 6062/12,000 = 0.5052

We are told to test that "the proportion of deaths in the week before Thanksgiving is less than 0.5," so one of our hypotheses must be "p<0.50." This can't be H_{0}, since H_{0} must have an equal sign in it, so this must be H_{1}. That tells us, then, that H_{0} must be "p = 0.50."

Our test (using the classical approach):

H_{0}: p = 0.50 , H_{1}: p < 0.50 (one tail test)

Choose alpha = 0.05 (that was actually given)

DR: If z_{p-hat} < -1.645 then reject H_{0} (-1.645 was the z-score that chopped off the lowest 5% of area)

We find z_{p-hat} = (p-hat - p)/sqrt(pq/n) = (0.5052 - 0.500)/sqrt(.50*.50/12000) = 0.0052/0.00456 = 1.14.

Since 1.14 is NOT < -1.645, we fail to reject H_{0}.

There is no evidence that people postpone their deaths.

#19. Given info: n = 976, p-hat = 312/976 = 0.320

We are told to test that "less than 1/3 of adults ...," so one of our hypotheses must be "p<1/3." This can't be H_{0}, since H_{0} must have an equal sign in it, so this must be H_{1}. That tells us, then, that H_{0} must be "p = 1/3."

Our test (using the classical approach):

H_{0}: p =1/3 , H_{1}: p < 1/3 (one tail test)

Choose alpha = 0.05 (that was actually given)

DR: If z_{p-hat} < -1.645 then reject H_{0} (-1.645 was the z-score that chopped off the lowest 5% of area)

We find z_{p-hat} = (p-hat - p)/sqrt(pq/n) = (0.320 - 1/3)/sqrt(1/3*2/3/976) = -0.0133/0.0151 = -0.88.

Since -0.88 is NOT < -1.645, we fail to reject H_{0}.

We stick with 1/3 as the true percentage.

#20. Given info: n = 1018, p-hat = 0.22

We are told to test that "less than 25% of adults ...," so one of our hypotheses must be "p<0.25." This can't be H_{0}, since H_{0} must have an equal sign in it, so this must be H_{1}. That tells us, then, that H_{0} must be "p = 0.25."

Our test (using the classical approach):

H_{0}: p = 0.25 , H_{1}: p < 0.25 (one tail test)

Choose alpha = 0.05 (that was actually given)

DR: If z_{p-hat} < -1.645 then reject H_{0} (-1.645 was the z-score that chopped off the lowest 5% of area)

We find z_{p-hat} = (p-hat - p)/sqrt(pq/n) = (0.22 - 0.25)/sqrt(.25*.75/1018) = -0.03/0.0136 = -2.21.

Since -2.21 is < -1.645, we reject H_{0}.

There is evidence that the real percentage is less than 25%.

#21. Given info: n = 1125, p-hat = 0.47

We are told to test that the percentage is never or rarely 50%, so one of our hypotheses must be "p≠0.50." This can't be H_{0}, since H_{0} must have an equal sign in it, so this must be H_{1}. That tells us, then, that H_{0} must be "p = 0.50."

Our test (using the classical approach):

H_{0}: p = 0.50 , H_{1}: p ≠ 0.50 (two tail test)

Choose alpha = 0.05 (that was actually given)

DR: If z_{p-hat} > 1.96 or z_{p-hat} < -1.96 then reject H_{0} (1.96 was the z-score that chopped off the highest and lowest 5% of area)

We find z_{p-hat} = (p-hat - p)/sqrt(pq/n) = (0.47 - 0.50)/sqrt(.50*.50/1125) = -0.03/0.0149 = -2.01.

Since -2.01 is < -1.96, we reject H_{0}.

There is evidence that the real percentage is not 50%.