Question

On the SAT exam a total of 25 minutes is allotted for students to answer 20 math questions without the use of a calculator. A guidance counselor would like to know if the students in his school are prepared to complete this portion of the exam in the timeframe allotted. To investigate, the counselor selects a random sample of 35 students and administers this portion of the test. The students are instructed to turn in their test as soon as they have completed the questions. The mean amount of time taken by the students is 23.5 minutes with a standard deviation of 4.8 minutes. The counselor would like to know if the data provide convincing evidence that the true mean amount of time needed for all students of this school to complete this portion of the test is less than 25 minutes and therefore tests the hypotheses H0: μ = 25 versus Ha: μ < 25, where μ = the true mean amount of time needed by students at this school to complete this portion of the exam. The conditions for inference are met. What are the appropriate test statistic and P-value? Find the t-table here. t = –1.85; the P-value is 0.9678. t = –1.85; the P-value is between 0.025 and 0.05. t = 1.85; the P-value is 0.9678. t = 1.85; the P-value is between 0.025 and 0.05.

Answer 1

Using the t-distribution, the correct option regarding the test statistic and the p-value is given as follows:

t = –1.85; the P-value is between 0.025 and 0.05.

What are the hypothesis tested?

The null hypothesis is:

$H_0: \mu = 25$

The alternative hypothesis is:

$H_1: \mu < 25$

What are the test statistic and the p-value?

The test statistic is given by:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

The parameters are:

• $\overline{x}$ is the sample mean.

• $\mu$ is the value tested at the null hypothesis.

• s is the standard deviation of the sample.

• n is the sample size.

In this problem, the values of the parameters are given as follows:

$\overline{x} = 23.5, \mu = 25, s = 4.8, n = 35$

Hence the value of the test statistic is:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{23.5 - 25}{\frac{4.8}{\sqrt{35}}}$

t = -1.85.

Using a t-distribution calculator, with a left-tailed test, as we are testing if the mean is less than a value and 35 - 1 = 34 df, the p-value is of 0.0365.

Hence the correct statement is:

t = –1.85; the P-value is between 0.025 and 0.05.

More can be learned about the t-distribution at https://brainly.com/question/16194574

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