Question

One study claimed that 86%

of college students identify themselves as procrastinators. A professor believes that the claim regarding college students is too high. The professor conducts a simple random sample of 236
college students and finds that 192
of them identify themselves as procrastinators. Does this evidence support the professor’s claim that fewer than 86%
of college students are procrastinators? Use a 0.05
level of significance.
Step 2 of 3: Compute the value of the test statistic. Round your answer to two decimal places.

Answer 1

[AI-generated answer — Bot GPT 3.5]

To compute the value of the test statistic, we can use the formula for the z-test:

z = (p̂ - p) / √(p(1-p) / n)

where:
p̂ = sample proportion = 192/236 = 0.8136
p = hypothesized population proportion = 0.86
n = sample size = 236

Plugging in the values:

z = (0.8136 - 0.86) / √(0.86(1-0.86) / 236)
z = (-0.0464) / √(0.86 * 0.14 / 236)
z = -0.0464 / √(0.1204 / 236)
z = -0.0464 / √0.0005092
z = -0.0464 / 0.02256
z ≈ -2.0574

Therefore, the value of the test statistic is approximately -2.06.