please show all steps The Engineering Department at Sims Software Inc. recently

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The Engineering Department at Sims Software Inc. recently developed two chemical solutions designed to increase the usable life of computer disks. A sample of disks treated with the first solution lasted 86, 78, 66 83, 84, 81, 84. 109. 65, and 102 hours. Those treated with the second solution lasted 91, 71, 75, 76, 87, 79, 73, 76, 79, 78, 87, 90, 76, and 72 hours. Assume the population standard deviations are not the same. At the 10 significance level, can we conclude that there is a difference in the length of time the two types of treatment lasted? Click here for the Excel Data File 1. The degrees of freedom is 2. Reject Ho if t Round down your answer to nearest whole number.) or t> (Negative amounts should be indicated by a minus sign. Round your answers to 3 decimal places.). 3. The test statistic is 4. Decision: Donot rejec H041-a Round your answer to 2 decimal places.)

Explanation / Answer

Since there is difference in population standard deviation, separate variances will be used under t test.

sample 1

Sample 2

86

91

78

71

66

75

83

76

84

87

81

79

84

73

109

76

65

79

102

78

87

90

76

72

Null Hypothesis: there is no significant difference => H0:   µ1 - µ2 = 0

Alternate Hypothesis: there is a significant difference => H1:       µ1   µ2

Level of significance = .1

It is a two tiled test.

Mean of sample 1 (x1) = 83.8

Mean of sample 2 (x2) = 79.29

Standard deviation of sample 1 (S1) = 12.97

Standard deviation of sample 2 (S2) = 6.46

n1 = 10                  n2 = 14

Degree of freedom = smaller of (n1-1) and (n2-1)

Thus, degree of freedom = n1-1 = 10-1 = 9

t = (x1-x2)/((S1^2/n1) + (S2^2/n2))^.5

t = (83.8 – 79.29)/(12.97^2/10   + 6.46^2/14)^.5

t = 1.013

at .1 level of significance and two tail with degree of freedom of 9,

Critical value of t = 1.833

Since, value of t calculated is less than the critical value, Alternative hypothesis is rejected add null hypothesis is accepted.

1 D.F = 9

2. Reject Null hypothesis, if t < -1.833   or t > 1.833

3. Test statistics t = 1.013

4. Decision: do not reject ( accept the null hypothesis H0)

sample 1

Sample 2

86

91

78

71

66

75

83

76

84

87

81

79

84

73

109

76

65

79

102

78

87

90

76

72

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