Question

It is known that nationally, students attending four-year colleges get an average of 6.75 hours of sleep per night, with a standard deviation of 1.7 hours. You are interested in seeing if students who attend commuter colleges (colleges where students must commute to school because there are no dorms on campus), such as John Jay College of Criminal Justice, differ significantly from this national average. Complete the following steps in order: What are the population mean and population standard deviation? Population mean (μ) = Population standard deviation (σ)= What do we expect the mean of our sample to be? Write your Null Hypothesis and Alternative Hypothesis Statements for a two-tailed test at an α = 0.05. H0: H1: What are the Z-critical values for a two-tailed test at an α = 0.05? Z_c= Using the Day 01 Survey Data Set, calculate the valid sample size (n) for the variable Sleep. Valid sample size (n) = What is the standard deviation for sample means (σ_x ̅ )? Round your answer to the second decimal place. Standard deviation for sample means (σ_x ̅ ) = Calculate the 95% Confidence Interval for the sample based on the known population mean (if the sample mean is not what we expect, 95% of the time it will fall in what interval?). Round your answers to the second decimal place. 95% CI = Using the Day 01 Survey Data Set, calculate the sample mean (x ̅) for the variable Sleep. Sample mean (x ̅) = Calculate the Z-test for your sample mean. Round your answer to the second decimal place. Z = Calculate the p-value (area below) for your Z-score (remember that you are conducting a two-tailed test). p-value = Does the actual sample mean fall within the 95% Confidence Interval, yes or no? State whether the Null Hypothesis should be Accepted or Rejected. Write a statement about your findings, include the proper statistical notation. Based on your answer to question 12, what type of error might you have made, Type I or Type II? 6.35 sample mean

Answer 1

Here we have

Hypotheses are:

$H_{0}:\mu=6.75,H_{a}:\mu\neq 6.75$

Test is two tailed.

Critical values of z for 0.05 signficance level are -1.96 and 1.96.

Here we have

$\bar{x}=6.35,n=8$

The standard deviation of sampling distribution is:

$\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{1.7}{\sqrt{8}}=0.60$

The required confidence interval is:

$\bar{x}\pm z_{c}\sigma_{\bar{x}}=6.35\pm 1.96\cdot 0.60=6.35\pm 1.176=(5.174,7.526)$

The confidence interval is (5.17, 7.53)

The test statistics is:

$z=\frac{6.35-6.75}{0.60}=-0.67$

The p-avlue is:

p-value = 2 * P(z < -0.67)= 0.5028

Since p-value is greater than 0.05 sowe fail to reject the null hypothesis.

Hypothesis should be Accepted.

Since we fail to reject the null hypothesis so type II error is possible.