Question

(1 point) Here we consider the sleep habits of med students versus non-med students. The study consists of the hours of sleep per day obtained from 27 med students and 30 non-med students. The summarized data is given in the table. Here, degrees of freedom (d.f.) that you must use in your calculations are given below. is the mean hours of sleep per day from each sample. The Student Type n2s Med (x) 27 5.6 0.7744 0.88 Non-Med (x2) 30 6.2 1.9881 1.41 degrees of freedom: d.f.49 Test the claim that the mean hours of sleep for med and non-med students is different. Use a 0.05 significance level (a) Find the test statistic. (b) Find the P-value. (c) ls there sufficient data to support the claim? Yes No Test the claim that, on average, med students get less sleep than non-med students. Use a 0.05 significance level (d) Is there sufficient data to support the claim? Yes No

Answer 1

Hypotheses are:

(a)

= (( 5.6-6.2)-0)/(sqrt((0.7744/27)+( 1.988 1/30))) n, n' =-1.947

(b)

The p-value using excel function "=TDIST(1.947,49,2)" is 0.0573.

(c)

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

(d)

Now hypotheses are:

$H_{0}:mu_{1}-mu_{2}=0$

$H_{a}:mu_{1}-mu_{2}<0$

The p-value using excel function "=TDIST(1.947,49,1)" is 0.0286.

Since  p-value is lesser than 0.05 so we reject the null hypothesis.