Q.76 - The Practice of Statistics for AP

Question

A student group claims that first-year students at a university study 2.5 hours 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 $\stackrel{-}{x}$ =137 minutes and $s_x$=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 conclusion do you draw?

Step-by-Step Solution

Answer

The students in the university on an average study $2.5$ hours on an average night.

Step 1: Introduction

The significance level of an occasion (like a statistical test) is the probability that the occasion might have happened by some coincidence. If the level is quite low, that is to say, the probability of occurring by chance is tiny, we say the occasion is significant.

Step 2: Explanation

Calculating the null and alternative hypotheses,

$\begin{array}{r}{H}_0:\mu =150\\ H_1:\mu <150\end{array}$

Now, $\stackrel{-}{x}=137{\mathrm{s}}_x=45 n=30$

Using

$t=\frac{\overline{x}-\mu }{s_x/\sqrt{n}}~t_{n-1}$

$t=\frac{137-150}{45/\sqrt{30}}=-1.582$

At the significance level $0.05$ the critical value is$-1.699$

If the critical value is less than the t value the null hypothesis is not rejected.

Hence we can conclude that students in the university an average study $2.5$ hours on an average night.