Question

ch 17 a statistician wishing to test a hypothesis that students score at most 75% on the final exam in an introductory statistics course decides to randomly select 20 students in the class and have them take the exam early. the average score of the 20 students on the exam was 72% and the standard deviation in the population is known to be σ = 15%. the statistician calculates the test statistic to be -0.8944. if the statistician chose to do a two - sided alternative, the p - value would be calculated by finding the area to the left of -.8944 and doubling it. finding the area to the right of -.8944 and doubling it. finding the area to the right of the absolute value of -.8944 and dividing it by two. finding the area to the left of -.8944. question 19 1 pts the following are the conditions to be met to test for the population mean except: σ known normal distribution (or n≥30) srs μ known

Explanation:

Question 1

Step1: Recall p - value for two - sided test

For a two - sided test with a test statistic \(z\) (in a normal distribution context), the p - value is the probability of getting a value as extreme or more extreme than the observed test statistic in either tail. When the test statistic \(z=- 0.8944\), we first find the area to the left of \(z =-0.8944\) (this represents the probability of getting a value less than the observed in the left - hand tail). Then, since it's a two - sided test, we double this area to account for the extreme values in the right - hand tail as well.

When testing for the population mean \(\mu\):

• If the population standard deviation \(\sigma\) is known, we can use the z - test under certain conditions.
• We need the data to come from a simple random sample (SRS). Also, the population should be normally distributed or the sample size \(n\geq30\) (by the Central Limit Theorem). The population mean \(\mu\) is the parameter we are trying to test or estimate, and we do not need to know its value in advance to conduct the test.

Answer:

finding the area to the left of -.8944 and doubling it.

Question 2