Question

1)

H0:p=0.54H0:p=0.54
H1:p<0.54H1:p<0.54

Your sample consists of 88 subjects, with 47 successes. Calculate the test statistic, rounded to 2 decimal places

z=

2)  A well-known brokerage firm executive claimed that at least 43 % of investors are currently confident of meeting their investment goals. An XYZ Investor Optimism Survey, conducted over a two week period, found that out of 106 randomly selected people, 41 of them said they are confident of meeting their goals.

Suppose you are have the following null and alternative hypotheses for a test you are running:

H0:p=0.43H0:p=0.43
Ha:p<0.43Ha:p<0.43

Calculate the test statistic, rounded to 3 decimal places

3)In a recent Super Bowl, a TV network predicted that 58 % of the audience would express an interest in seeing one of its forthcoming television shows. The network ran commercials for these shows during the Super Bowl. The day after the Super Bowl, and Advertising Group sampled 58 people who saw the commercials and found that 34 of them said they would watch one of the television shows.

Suppose you are have the following null and alternative hypotheses for a test you are running:

H0:p=0.58H0:p=0.58
Ha:p>0.58Ha:p>0.58

Calculate the test statistic, rounded to 3 decimal places

Answer 1

1)

H0: p = 0.54

Ha: p < 0.54 (left tailed)

Sample proportion = 47 / 88 = 0.5341

Test statistics

z = - p / sqrt( p ( 1 -p ) / n)

= 0.5341 - 0.54 / sqrt( 0.54 * 0.46 / 88)

= -0.11

Test statistics z = -0.11

2)

H0: p = 0.43

Ha: p < 0.43 (left tailed)

Sample proportion = 41 / 106 = 0.3868

Test statistics

z = - p / sqrt( p ( 1 - p) / n)

= 0.3868 - 0.43 / sqrt( 0.43 * 0.57 / 106)

= -0.898

Test statistics = -0.898

3)

H0: p = 0.58

Ha: p > 0.58 (Right tailed)

Sample proportion = 34/58 = 0.5862

Test statistics

z = - p / sqrt( p( 1 - p) / n)

= 0.5862 - 0.58 / sqrt( 0.58 * 0.42 / 58)

= 0.096

Test statistics z = 0.096