MBALN 603 Statistics Examination for Hypothesis

Part 1

Keywords:Confidence Intervals for Means, Normal distribution, t-student distribution

Exercise 1. (5 points) For the following data set

114	112	123	111	109	103
112	113	107	129	108	126
131	129	126	115	123	108
131	115	134	133	132	114
104	127	128	137	114	139
128	117	121	110	141	121
138	140	122	130	115	102

Table 1: Data set Exercise 1

Construct a

90%confidence interval for the population mean;
95%confidence interval for the population mean;
99%confidence interval for the population mean;
Whatassumption do you need to make about the population of interest to construct the confidence intervals ?

Exercise 2. (5 points) For the following data set

136	141	137	145
131	135	138	136
130	132	136	134
137	133	141	139

Table 2: Data set Exercise 2

Construct a

1. 90%confidence interval for the population mean;

2. 95%confidence interval for the population mean;

3. 99%confidence interval for the population mean;

4.Whatassumption do you need to make about the population of interest to construct the confidence intervals

Exercise 3. (5 points) Two boxes, A and B contain equal numbers of balls (black and white). A sample of 100 balls selected with replacement from each of the boxes revealed 52 white balls form box A and 44 white balls from box B.

Testthe hypothesis that the two boxes have equal ratio of white balls, using a significance level of 0.05
Testthe hypothesis that box A has greater ratio of white balls than box B, using a significance level of 0.05

Exercise 4. (5 points) Consider the sample data for grades A, B, C & D of four different classes, which are tabulated in table 3. Test whether the four means are equal at the 0.05 level of significance.

Perform a test that will compare all 4 means at once. We are not interested in pair-wise comparisons at this stage.

Please show all the steps in your calculations.

Table 3: Data set Exercise 4

80	81	83	78
83	79	72	79
80	83	80	80
65	76	81	98
92	65	87	57
81	74	68	81
65	34	68	71
34	45	78	91
57	61	51	46
71	69	66	72
71	59	63	82
81	67	61	52
72	75	64
81	80	74
36	55	78
83	67
91	94
73
51
98

Answer

Part 1. Exercise 1:

Data
114	112	123	111	109	103
112	113	107	129	108	126
131	129	126	115	123	108
131	115	134	133	132	114
104	127	128	137	114	139
128	117	121	110	141	121
138	140	122	130	115	102

Population Mean =		121.2381	Confidence intervals	90%
			Confidence intervals	95%
Number of Sample =		42	Confidence intervals	99%
Standard deviation =		11.08022
Standard error =		1.709715	t(90%)	1.644854
			t(95%)	1.959964
			t(99%)	2.575829
1)	Upper 90% confidence limit =		124.0503
	Lower 90% confidence limit =		118.4259
2)	Upper 95% confidence limit =		124.5891
	Lower 95% confidence limit =		117.8871
3)	Upper 99% confidence limit =		125.642
	Lower 99% confidence limit =		116.8342

The population mean is 121.2381.

90% confidence interval of the population mean =

Upper 90% confidence limit = 124.0503.

Lower 90% confidence limit = 118.4259.

95% confidence interval of the population mean =

Upper 95% confidence limit = 124.5891 (Efron and Tibshirani 1986).

Lower 95% confidence limit = 117.8871.

99% confidence interval of the population mean =

Upper 99% confidence limit = 125.642.

Lower 99% confidence limit = 116.8342.

The assumptions needed to construct the confidence intervals are-

Each sample is independent to each other.
All the samples are normally distributed (Payton, Greenstone and Schenker 2003).
There exists no significant outlier in the data set.

Exercise 2:

Data
136	141	137	145
131	135	138	136
130	132	136	134
137	133	141	139

Population Mean =		136.3125	Confidence intervals	90%
			Confidence intervals	95%
Number of Sample =		16	Confidence intervals	99%
Standard deviation =		3.961797
Standard error =		0.990449	t(90%)	1.644854
			t(95%)	1.959964
			t(99%)	2.575829
1)	Upper 90% confidence limit =		137.9416
	Lower 90% confidence limit =		134.6834
2)	Upper 95% confidence limit =		138.2537
	Lower 95% confidence limit =		134.3713
3)	Upper 99% confidence limit =		138.8637
	Lower 99% confidence limit =		133.7613

(Source: Altman and Gardner 1988)

The population mean is 136.3125.

90% confidence interval of the population mean =

Upper 90% confidence limit = 137.9416.

Lower 90% confidence limit = 134.6834.

95% confidence interval of the population mean =

Upper 95% confidence limit = 138.2537.

Lower 95% confidence limit = 134.3713.

99% confidence interval of the population mean =

Upper 99% confidence limit = 138.8637.

Lower 99% confidence limit = 133.7613.

The assumptions needed to construct the confidence intervals are-

Each sample are independent to each other (Ci 1987).
All the samples are normally distributed.
There exists no significant outlier in the data set.

Exercise 3:

Null hypothesis:	The proportions of white balls in two boxes are equal.
Alternative hypothesis:	The proportions of white balls in box A is greater than box B.
Box A			Box B
Total balls (n1) =	100		Total balls (n2) =	100
White balls =	52		White balls =	44
Proportion (p1) =	0.52		Proportion (p2) =	0.44
Difference in proportions (p1 - p2) =		0.08
Total samples (n1+n2) =		200
Total white balls =		96
Total proportion (p-bar) =		0.48
Z-statistic =		1.132277
Level of significance =		5%
Z-critical =		1.959964
Sig. =		Cannot reject null hypothesis

Hypothesis:

The hypotheses are-

Null hypothesis (H₀): Box A has equal proportion of white ball such as proportion of white ball in Box B.

Alternative hypothesis (H_A): Box A has higher proportion of white ball than the proportion of white ball in Box B.

Level of significance:

The level of significance is assumed to be 5%.

Test applied:

Two sample proportional Z-test.

Calculated Z-statistic:

Z = = 1.132277 (Sahai and Khurshid 1996)

(p1= proportion of white ball in box A; p2= proportion of white ball in box B; p=proportion of white ball together in box A and box B, n1= total number of balls in box A; n2= total number of balls in box B).

Decision-making:

1.959964>1.132277, so, Z_sig.>Z_crit. Therefore, the null hypothesis could not be rejected with 95% probability (Levine et al. 1999).

Interpretation:

Therefore, it could be interpreted that Box A has equal proportion of white ball of proportion of white ball in Box B.

Exercise 4:

Anova: Single Factor
Summary
Groups	Count	Sum	Average	Variance
Grade A	20	1445	72.25	292.4079
Grade B	17	1164	68.47059	217.2647
Grade C	15	1074	71.6	95.68571
Grade D	12	887	73.91667	238.6288
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	236.9355	3	78.97851	0.364614	0.778776	2.758078
Within Groups	12996.5	60	216.6084

Total	13233.44	63

Hypothesis:

The hypotheses are-

Null hypothesis (H₀): The averages of four different grades A, B, C and D of four different classes are equal to each other (Heiberger and Neuwirth 2009).

Alternative hypothesis (H₀): There exists at least one inequality in the averages of four different grades A, B, C and D of four different classes.

Observations:

The average of grade A is 72.25, grade B is 68.47, grade C is 71.6 and grade D is 73.92.

Test applied:

One-way ANOVA.

Level of Significance:

The level of significance is found to be 5%.

Test Statistic:

Calculated F-statistic = 0.364614 with p-value = 0.778776.

Decision-making:

As, calculated p-value is greater than the 5% level of significance, therefore, null hypothesis could not be rejected with 95% probability (Christensen 1987). On the other hand, as F_crit.is greater than F_cal.(2.758078>0.364614), therefore, null hypothesis is accepted with 95% possibility (Ross and Willson 2017).

Interpretation:

The averages of four different grades A, B, C and D of four different classes are equal to each other.

References:

Altman, D.G. and Gardner, M.J., 1988. Statistics in Medicine: Calculating confidence intervals for regression and correlation. British medical journal (Clinical research ed.), 296(6631), p.1238.

Christensen, R., 1987. One-Way ANOVA. In Plane Answers to Complex Questions (pp. 57-69). Springer, New York, NY.

Ci, B., 1987. Confidence intervals. Lancet, 1, pp.494-497.

Efron, B. and Tibshirani, R., 1986. Bootstrap methods for standard errors, confidence intervals, and other measures of statistical accuracy. Statistical science, pp.54-75.

Heiberger, R.M. and Neuwirth, E., 2009. One-way anova. In R through excel (pp. 165-191). Springer, New York, NY.

Levine, D.M., Berenson, M.L., Stephan, D. and Lysell, D., 1999. Statistics for managers using Microsoft Excel (Vol. 660). Upper Saddle River, NJ: Prentice Hall.

Payton, M.E., Greenstone, M.H. and Schenker, N., 2003. Overlapping confidence intervals or standard error intervals: what do they mean in terms of statistical significance?. Journal of Insect Science, 3(1).

Ross, A. and Willson, V.L., 2017. One-Way Anova. In Basic and Advanced Statistical Tests (pp. 21-24). SensePublishers, Rotterdam.

Sahai, H. and Khurshid, A., 1996. Formulae and tables for the determination of sample sizes and power in clinical trials for testing differences in proportions for the two?sample design: a review. Statistics in medicine, 15(1), pp.1-21.

Buy MBALN 603 Statistics Examination for Hypothesis Answers Online

Talk to our expert to get the help with MBALN 603 Statistics Examination for Hypothesis Answers to complete your assessment on time and boost your grades now

The main aim/motive of the management assignment help services is to get connect with a greater number of students, and effectively help, and support them in getting completing their assignments the students also get find this a wonderful opportunity where they could effectively learn more about their topics, as the experts also have the best team members with them in which all the members effectively support each other to get complete their diploma assignments. They complete the assessments of the students in an appropriate manner and deliver them back to the students before the due date of the assignment so that the students could timely submit this, and can score higher marks.Â The experts of the assignment help services at urgenthomework.com are so much skilled, capable, talented, and experienced in their field of programming homework help writing assignments, so, for this, they can effectively write the best economics assignment help services.

Get Online Support for MBALN 603 Statistics Examination for Hypothesis Assignment Help Online

); }

Not the Exact Question you were looking for ? Post your question for assignment help and get instant help on your homework and assignment questions from our experts

114	112	123	111	109	103
112	113	107	129	108	126
131	129	126	115	123	108
131	115	134	133	132	114
104	127	128	137	114	139
128	117	121	110	141	121
138	140	122	130	115	102

80	81	83	78
83	79	72	79
80	83	80	80
65	76	81	98
92	65	87	57
81	74	68	81
65	34	68	71
34	45	78	91
57	61	51	46
71	69	66	72
71	59	63	82
81	67	61	52
72	75	64
81	80	74
36	55	78
83	67
91	94
73
51
98

114	112	123	111	109	103
112	113	107	129	108	126
131	129	126	115	123	108
131	115	134	133	132	114
104	127	128	137	114	139
128	117	121	110	141	121
138	140	122	130	115	102

80	81	83	78
83	79	72	79
80	83	80	80
65	76	81	98
92	65	87	57
81	74	68	81
65	34	68	71
34	45	78	91
57	61	51	46
71	69	66	72
71	59	63	82
81	67	61	52
72	75	64
81	80	74
36	55	78
83	67
91	94
73
51
98

MBALN 603 Statistics Examination for Hypothesis

Part 1

Answer

Part 1.

Exercise 1:

Data

Exercise 2:

Data

Exercise 3:

Box A

Box B

Level of significance:

Test applied:

Calculated Z-statistic:

Decision-making:

Interpretation:

Exercise 4:

Summary

ANOVA

Hypothesis:

Observations:

Test applied:

Level of Significance:

Test Statistic:

Decision-making:

Interpretation:

References:

Buy MBALN 603 Statistics Examination for Hypothesis Answers Online

Get Online Support for MBALN 603 Statistics Examination for Hypothesis Assignment Help Online

114	112	123	111	109	103
112	113	107	129	108	126
131	129	126	115	123	108
131	115	134	133	132	114
104	127	128	137	114	139
128	117	121	110	141	121
138	140	122	130	115	102

80	81	83	78
83	79	72	79
80	83	80	80
65	76	81	98
92	65	87	57
81	74	68	81
65	34	68	71
34	45	78	91
57	61	51	46
71	69	66	72
71	59	63	82
81	67	61	52
72	75	64
81	80	74
36	55	78
83	67
91	94
73
51
98