Question

Do not use MS Excel or statistics software unless stated otherwise (1 -2) (Confidence Interval) The following sample data are measurements of weight of middle school students. 153,148, 151, 163, 114,164, 135, 131,176 (25%) Assume that we do not know the true standard deviation of middle school students Calculate the two-sided 95% confidence intervals on the mean (find both upper and lower limits). I. 2. (25%) Assume now that we know the standard deviation of all middle school students is 19 lb. Calculate the 2-sided 90% confidence interval on the mean. (3-5) (T-test) Using the 10 numbers shown in problems 1 and 2, answer the following questions: 3. (15%) We want to test whether the true mean of middle school students is greater than 140 lb. State the null and alternative hypothesis: а. Но: b. HI: 4. (15%) Based on the .. obtained earlier, what would be the most appropriate conclusion of the test? a. The true mean is exactly 140 lb. b. The mean is different from 140 lb. c. The mean is significantly greater than 140 lb. d. The mean is not significantly greater than 140 lb. 5. (20%) Calculate the test statistics (to) and find the p-value of the test using MS Excel.

Answer 1

	Values ( X )	$\Sigma (X_{i} - \bar{X})^{2}$
	153	21.7781
	148	0.1111
	151	7.1113
	163	215.1121
	114	1178.7755
	164	245.4455
	135	177.7769
	131	300.4433
	176	765.4463
Total	1335	2912.0001

Mean $\bar{X} = \Sigma X_{i} / n$
$\bar{X} = 1335 / 9 = 148.3333$
Standard deviation $S_{X} = \sqrt{\Sigma (X_{i} - \bar{X})^{2}/n-1}$
$S_{X} = \sqrt{ 2912.0001 / 9 -1} = 19.0788$

Part 1

Confidence Interval
$\bar{X} \pm t_{\alpha /2, n-1} S/\sqrt{n}$
$t_{\alpha /2, n-1} = t_{ 0.05 /2, 9- 1 } = 2.306$
$148.3333 \pm t_{\ 0.05/2, 9 -1} * 19.0788/\sqrt{ 9}$
Lower Limit = $148.3333 - t_{\ 0.05/2, 9 -1}19.0788/\sqrt{ 9}$
Lower Limit = 133.668
Upper Limit = $148.3333 + t_{\ 0.05/2, 9 -1}19.0788/\sqrt{ 9}$
Upper Limit = 162.9986
95% Confidence interval is ( 133.668 , 162.9986 )

Part 2

Confidence Interval :-
$\bar{X} \pm Z_{\alpha /2} \sigma /\sqrt{n}$
$Z_{\alpha /2} = Z_{ 0.1 /2} = 1.64$
$148.3333 \pm Z_{ 0.1/2 } * 19/\sqrt{ 9}$
Lower Limit = $148.3333 - Z_{ 0.1/2} \;19/\sqrt{ 9}$
Lower Limit = 137.9159
Upper Limit = $148.3333 + Z_{ 0.1/2} \;19/\sqrt{ 9}$
Upper Limit = 158.7507
90% Confidence interval is ( 137.9159 , 158.7507 )

Part 3

H0 :-   $\mu \leq 140$

H1 :-   $\mu > 140$

Part 4.

95% Confidence interval is ( 133.668 , 162.9986 )

Since   $\mu = 140$ lies in the interval, we can conclude that the true mean is not significantly greater than 140 lb.

Part 5

Test Statistic :-
$t = ( \bar{X} - \mu ) / (S /\sqrt{n})$
$t = ( 148.3333 - 140 ) / ( 19.0788 /\sqrt{ 9 })$
t = 1.3103

Test Criteria :-
Reject null hypothesis if $t \; > \;t_{\alpha, n-1}$
$t_{\alpha, n-1} = t_{0.05 , 9-1} = 1.86$
$t > t_{\alpha, n-1} = 1.3103 < 1.86$
Result :- Fail to reject null hypothesis

Decision based on P value
P - value = P ( t > 1.3103 ) = 0.1132
Reject null hypothesis if P value < $\alpha = 0.05$ level of significance
P - value = 0.1132 > 0.05 ,hence we fail to reject null hypothesis
Conclusion :- Fail to reject null hypothesis