Answer to Question #311383 in Statistics and Probability for Steve

Question

A random sample of 10 students contains the following observations, in hours, for time spent studying in the week before final exams.

28 57 42 35 61 39 55 46 49 38

Assume that the population distribution is normal

(i) Find the sample mean and standard deviation

(ii) Test, at the 5% significance level, the null hypothesis that the population mean is 40 hours against the alternative that it is higher

Answer 1

(i)

$n=10\\\bar{x}=\frac{\sum{x_i}}{n}=45\\s^2=\frac{\sum{\left( x_i-\bar{x} \right) ^2}}{n-1}=111.111\\s=\sqrt{s^2}=10.54093$

(ii)

$H_0:\mu =40\\H_1:\mu >40$

The test statistic is

$T~t_{n-1}=t_9\\T=\sqrt{n}\frac{\bar{x}-\mu}{s}=\sqrt{10}\frac{45-40}{10.54093}=1.5$

The P-value

$P\left( T>1.5 \right) =1-F_{T,9}\left( 1.5 \right) =1-0.916=0.084$

Since the P-value is greater than $\alpha =0.05$, the null hypothesis is not declined.