A student group claims that first year students at a university study 2.5 hours (150 minutes) per night during the school week. A skeptic suspects that they study less than that on average. He takes a random sample of 30 first year students and finds that X=137 minutes and sx=45 minutes. A graph of the data shows no outliers but some skewness. Carry out an appropriate significance test at the 5% significance level. What conclusions do you draw?
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
However, depending on the degree and direction of the skew, the probability may not apply.
To carry out a significance test, we will need to follow the steps of hypothesis testing:
Step 1: State the hypotheses.
The skeptic's suspicion is that the first-year students study less than claimed. Let's denote the mean study time of first-year students as μ.
Null hypothesis (H0): μ = 150 (the claimed mean study time)
Alternate hypothesis (H1): μ < 150 (the skeptic's suspicion)
Step 2: Choose an appropriate test statistic.
Since we're comparing the mean of a sample to a known value, we can use a t-test for this problem. The test statistic is given by:
t = (X̄ - μ) / (s / √n)
where X̄ is the sample mean (137), μ is the population mean (150), s is the sample standard deviation (45), and n is the sample size (30).
Step 3: Determine the significance level.
The significance level (α) is given in the question as 5% or 0.05.
Step 4: Calculate the p-value.
The p-value is the probability of obtaining a result as extreme as (or more extreme than) the observed sample mean, assuming the null hypothesis is true. Since the alternative hypothesis is one-sided (μ<150), we need to find the area in the left tail of the t-distribution.
To calculate the p-value, we can use a t-table or statistical software. Given the sample size of 30, the degrees of freedom are (n-1) = 30-1 = 29. Looking up a t-table with 29 degrees of freedom and a one-sided test, we find the critical value for a 0.05 significance level to be -1.699.
Using the formula, t = (X̄ - μ) / (s / √n), we can calculate the test statistic:
t = (137 - 150) / (45 / √30) ≈ -1.448
Step 5: Make a decision.
Since the calculated test statistic (-1.448) is greater than the critical value (-1.699), we fail to reject the null hypothesis.
Step 6: State the conclusion.
Based on the significance test, at the 5% significance level, there is not enough evidence to conclude that first-year students study less than the claimed average of 150 minutes per night during the school week.
In summary, the skeptic's suspicion is not supported by the data, and we conclude that there is no significant evidence to suggest that first-year students study less than 150 minutes per night on average.