Question

One study claimed that 92% 92 % 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 278 278 college students and finds that 247 247 of them identify themselves as procrastinators. does this evidence support the professor’s claim that fewer than 92% 92 % of college students are procrastinators? use a 0.05 0.05 level of significance.

Answer 1

The evidence supports the professor's claim that fewer than 92% of college students are procrastinators. The hypothesis test conducted shows a significant difference between the sample proportion and the hypothesized proportion of 92%. The test statistic falls in the rejection region, leading us to reject the null hypothesis.

Step-by-step explanation:

To determine whether the evidence supports the professor's claim that fewer than 92% of college students are procrastinators, we can conduct a hypothesis test. The null hypothesis (H0) is that the true proportion of college students who identify as procrastinators is 92%, while the alternative hypothesis (Ha) is that the true proportion is less than 92%. We can use a one-sample proportion test with a significance level of 0.05.

First, we calculate the sample proportion, which is the number of college students who identify as procrastinators divided by the total sample size: 247/278 = 0.8881.

Next, we calculate the test statistic, which is z = (sample proportion - hypothesized proportion) / sqrt((hypothesized proportion * (1 - hypothesized proportion)) / sample size). Plugging in the values: z = (0.8881 - 0.92) / sqrt((0.92 * (1 - 0.92)) / 278) ≈ -2.4768.

Finally, we compare the test statistic to the critical value for a one-tailed test at a significance level of 0.05. Since -2.4768 is less than -1.645, we reject the null hypothesis. Therefore, the evidence supports the professor's claim that fewer than 92% of college students are procrastinators.

Answer 2

A hypothesis test is conducted to determine if the evidence supports the professor's claim that fewer than 92% of college students are procrastinators. The test statistic is compared to the critical value to make a decision on the null hypothesis.

To determine if the professor's claim that fewer than 92% of college students are procrastinators is supported by the evidence, we can conduct a hypothesis test. The null hypothesis (H0) is that the proportion of college students who identify as procrastinators is 92%, while the alternative hypothesis (Ha) is that the proportion is less than 92%.

We can use a one-sample proportion test at a significance level of 0.05.

Using the sample data, we calculate the test statistic

z = (0.247 - 0.92) / √(0.92 * (1 - 0.92) / 278) = -16.3.

The critical value for a left-tailed test at a significance level of 0.05 is -1.645.

Since the test statistic is smaller than the critical value, we reject the null hypothesis.

Therefore, the evidence supports the professor's claim that fewer than 92% of college students are procrastinators.