Question

i Question Hep he dean of a ur ersity esim atestat the mean number of classroom hours per week faire faculty暷110Aa.member ofthe studert mnal you want to test his dam. Anndom sampieofthe n mber efciassoom hon bro AMme fatty for one wokis shown itetable below Ata·001 oan you re ect the dean's dam? Complete parts (a) through (d) below Assume the population is no alydshhhd 10.5 7.1 13.8 84 4.8 11.3 13.5 Which of the folowing comedty sta Ho and H OR He' μ< 11.0 p2110 u110 11.0 μ< 11.0 (b)use ๒deology to fird to P alue Which of the folowing is coet? OA Fal to repect Ho because he pwah" is ๒stwn to sigroance ievel ○B. Rgect Ho because the Pvalue is less than teydoarce level OC. Reject Ho because the Pivalue is grealer than the signficance level O D. Fal to reject Hp because the Pvalue is greater than the signficance evel Interpret the decision in the conteat of the niginal clam hours per wfoumuity is grealer than 110 Cick to select your anewers

Answer 1

	Values ( X )	$\Sigma (X_{i} - \bar{X})^{2}$
	10.5	0.5625
	7.1	7.0225
	13.8	16.4025
	8.4	1.8225
	4.8	24.5025
	11.3	2.4025
	13.5	14.0625
	8.6	1.3225
Total	78	68.1

$Mean \bar{X} = \Sigma X_{i} / n$
$\bar{X} = 78 / 8 = 9.75$
$Standard deviation S_{X} = \sqrt{\Sigma (X_{i} - \bar{X})^{2}/n-1}$
$S_{X} = \sqrt{ 68.1 / 8 -1} = 3.1191$

To Test :-

H0 :-   $\mu = 11.0$

H1 :-   $\mu\neq 11.0$

Test Statistic :-

$t = ( \bar{X} - \mu ) / (S /\sqrt{n})$

$t = ( 9.75 - 11 ) / ( 3.1191 /\sqrt{ 8 })$

t = -1.1335

Test Criteria :-

Reject null hypothesis if $| t |\; > \;t_{\alpha /2, n-1}$

$t_{\alpha /2, n-1} = t_{\0.01 /2, 8-1} = 3.499$

$| t |\; > \;t_{\alpha /2, n-1} = 1.1335 < 3.499$

Result :- Fail to reject null hypothesis

P value

Looking for the value | t | = 1.1335 in t table across n - 1 = 8 - 1 = 7 degree of freedom.

| t | = 1.1335 lies between the values 1.119 and 1.415 with respective P value are 0.30 and 0.20.

Using excel to calculate exact P value = 0.294

Decision based on P value

Reject null hypothesis if P value < $\alpha=0.01$ level of significance

0.294 > 0.01, we fail to reject null hypothesis

Conclusion :- Accept Null Hypothesis

$\mu = 11.0$

Fail to reject H0, since P value is greater than level of significance.

At 15 level of significance, there are not sufficient evidence to reject the claim that the mean number of classroom hours per week for full time faculty is 11.0