A business journal investigation of the performance and timing of corporate acquisitions discovered that in a random sample of 2,684 firms, 715 announced one or more acquisitions during the year 2000. Does the sample provide sufficient evidence to indicate that the true percentage of all firms that announced one or more acquisitions during the year 2000 is less than 29%? Use alpha equals0.05 to make your decision.

Calculate the value of the z-statistic for this test.

z = 1.960

Step-by-step explanation:

The sample proportion is:

p = 715 / 2684 = 0.2664

The standard error is:

σ = √(pq/n)

σ = √(0.266 × 0.734 / 2684)

σ = 0.0085

For α = 0.05, the confidence level is 95%.  The z-statistic at 95% confidence is 1.960.

The margin of error is 1.960 × 0.0085 = 0.0167.

The confidence interval is 0.2664 ± 0.0167 = (0.2497, 0.2831).

The upper limit is 28.3%, so the journal can conclude with 95% confidence that the true percentage is less than 29%.

i really dont know

step-by-step explanation:

let's solve your equation step-by-step.

24−3x=−27

step 1: simplify both sides of the equation.

24−3x=−27

24+−3x=−27

−3x+24=−27

step 2: subtract 24 from both sides.

−3x+24−24=−27−24

−3x=−51

step 3: divide both sides by -3.

−3x

−3

=

−51

−3

x=17

answer:

x=17

step-by-step explanation:

mark brainliest and have a great day!