Question

USING EXCEL FORMULAS SOLVE THE PROBLEM. MUST USE EXCEL CALCULATIONS AND FORMULAS.!!!

1. Find the data for the problem in the first worksheet named LightbulbLife of the data table down below It gives the data on the lifetime in hours of a sample of 50 lightbulbs. The company manufacturing these bulbs wants to know whether it can claim that its lightbulbs typically last more than 1000 burning hours. So it did a study.
   1. Identify the null and the alternate hypotheses for this study.
   2. Can this lightbulb manufacturer claim at a significance level of 5% that its lightbulbs typically last more than 1000 hours? What about at 1%? Test your hypothesis using both, the critical value approach and the p-value approach. Clearly state your conclusions.
   3. Under what situation would a Type-I error occur? What would be the consequences of a Type-I error?
   4. Under what situation would a Type-II error occur? What would be the consequences of a Type-II error?

lightbulb	Lifetime
1	840.08
2	960
3	953.38
4	981.14
5	938.66
6	1051.14
7	907.84
8	1000.1
9	1073.2
10	1150.66
11	1010.57
12	791.59
13	896.24
14	955.35
15	937.94
16	1113.18
17	1108.81
18	773.62
19	1038.43
20	1126.55
21	950.23
22	1038.19
23	1136.67
24	1031.55
25	1074.28
26	976.9
27	1046.3
28	986.54
29	1014.83
30	920.73
31	1083.41
32	873.59
33	902.92
34	1049.17
35	998.58
36	1010.89
37	1028.71
38	1049.92
39	1080.95
40	1026.41
41	958.95
42	985.17
43	988.49
44	1012.99
45	1070.82
46	1063.13
47	948.57
48	1156.42
49	973.79
50	845.85

Answer 1

The Calculations for the mean and standard deviation are given after the Test.

Right Tailed t test, Single Mean

Given: = 1000 hours, = 997.87 hours, s = 511.49 hours, n = 50, = 0.05

The Hypothesis:

The Null Hypothesis: H0: = 1000: The mean lifetime of a bulb is equal to 1000 burning hours.

The Alternative Hypothesis: Ha: > 1000: The mean lifetime of a bulb is greater than 1000 burning hours..

This is a Right tailed test

The Test Statistic: Since the population standard deviation is unknown, we use the students t test.

The test statistic is given by the equation:

t observed = -0.03

The Excel Calculations

Single Mean - t - Right Tail
	x1	997.87
	μ	1000
	σ	511.49
	n	50
	df = n - 1	49
a	x1-μ	-2.13
b	sqrtn	7.071067812
c	s/sqrtn	72.3356095
z/t	a/c	-0.0294
	tround	-0.03

_________________________________________________________

The P Value Approach

The p Value: The p value (Right tailed) for t = -0.03, for degrees of freedom (df) = n-1 = 49, is; p value = 0.4881

Use Excel Formula TDIST(-0.03,49,1) to get the right tailed p value.

The Decision Rule:  P value is < , Then Reject H0.

The Decision:

At = 0.05: Since P value (0.4881) is > (0.01) , We Fail to Reject H0.

At = 0.05: Since P value (0.4881) is > (0.05) , We Fail To Reject H0.

The Conclusion: There is insufficient evidence at the 95% or the 99% significance level to conclude that the mean lifetime of a bulb is greater than 1000 burning hours..

_______________________________________________

The Critical Value Approach:

The Critical Value: Use the excel formula TINV (Significance level * 2,49). For eg for = 0.05 use TINV(0.1,49)

The critical value (Right Tail) at = 0.05, for df = 49, tcritical= +1.667

The critical value (Right Tail) at = 0.01, for df = 49, tcritical= +2.405

The Decision Rule: If tobserved is > tcritical.

The Decision:

At = 0.05: Since tobserved (-0.03) is < t critical (1.667), we Fail to Reject H0.

At = 0.01: Since tobserved (-0.03) is < t critical (2.405), we Fail to Reject H0.

The Conclusion: There is insufficient evidence at the 95% or the 99% significance level to conclude that the mean lifetime of a bulb is greater than 1000 burning hours..

_________________________________________________

Type I and Type II Errors

A Type I error is the incorrect rejection of a True Null Hypothesis. In this case it would mean that we reject the hypothesis that the mean lifetime of a bulb is equal to 1000 burning hours, when it actually is 1000 burning Hours.

A Type II error is the failure to reject a false null Hypothesis. In this case it would mean that we Fail to rejctt the hypothesis that the mean lifetime of a bulb is equal to 1000 burning hours, when it actually is greater than 1000 burning Hours.

________________________________________________

Calculation for the mean and standard deviation:

Mean = Sum of observation / Total Observations

Standard deviation = SQRT(Variance)

Variance = Sum Of Squares (SS) / n - 1, where

SS = SUM(X - Mean)².

#	X	Mean	(X - Mean)2
1	840.08	498.93	116383.3225
2	960	498.93	212585.5449
3	953.38	498.93	206524.8025
4	981.14	498.93	232526.4841
5	938.66	498.93	193362.4729
6	1051.14	498.93	304935.8841
7	907.84	498.93	167207.3881
8	1000.1	498.93	251171.3689
9	1073.2	498.93	329786.0329
10	1150.66	498.93	424751.9929
11	1010.57	498.93	261775.4896
12	791.59	498.93	85649.8756
13	896.24	498.93	157855.2361
14	955.35	498.93	208319.2164
15	937.94	498.93	192729.7801
16	1113.18	498.93	377303.0625
17	1108.81	498.93	371953.6144
18	773.62	498.93	75454.5961
19	1038.43	498.93	291060.25
20	1126.55	498.93	393906.8644
21	950.23	498.93	203671.69
22	1038.19	498.93	290801.3476
23	1136.67	498.93	406712.3076
24	1031.55	498.93	283684.0644
25	1074.28	498.93	331027.6225
26	976.9	498.93	228455.3209
27	1046.3	498.93	299613.9169
28	986.54	498.93	237763.5121
29	1014.83	498.93	266152.81
30	920.73	498.93	177915.24
31	1083.41	498.93	341616.8704
32	873.59	498.93	140370.1156
33	902.92	498.93	163207.9201
34	1049.17	498.93	302764.0576
35	998.58	498.93	249650.1225
36	1010.89	498.93	262103.0416
37	1028.71	498.93	280666.8484
38	1049.92	498.93	303589.9801
39	1080.95	498.93	338747.2804
40	1026.41	498.93	278235.1504
41	958.95	498.93	211618.4004
42	985.17	498.93	236429.3376
43	988.49	498.93	239668.9936
44	1012.99	498.93	264257.6836
45	1070.82	498.93	327058.1721
46	1063.13	498.93	318321.64
47	948.57	498.93	202176.1296
48	1156.42	498.93	432293.1001
49	973.79	498.93	225492.0196
50	845.85	498.93	120353.4864
Total	49893.43		12819661.46

n	50
Sum	49893.43
Average	997.87
SS	12819661.46
Variance = SS/n-1	261625.74
Std Dev	511.49