Statistics Please show all work. The results of an accelerated test for the life

Question

Statistics Please show all work. The results of an accelerated test for the lifetime of a lithium-ion battery are: time-to-failure (hours): 4.83, 4.60, 4.87, 5.23, 4.72, 5.33, 5.02, 4.85, 4.72, 5.08 Perform a point estimate hypothesis test (P = 5%) to determine if the sample justifies the statement that mu > 4.8 hours. State your conclusion on the time-to-failure. Perform a 95% confidence interval hypothesis test to determine if the sample justifies the statement that n > 4.8 hours. State your conclusion on the time-to-failure. Perform a point estimate hypothesis test (P = 5%) to determine if the sample justifies the statement that a2 = 0.03. State your conclusion on the time-to-failure variance. Perform a 95% confidence interval hypothesis test to determine if the sample justifies the statement that a = 0.1. State your conclusion on the time-to-failure variance.

Explanation / Answer

Part a

H0: µ = 4.8 versus Ha: µ > 4.8

(One tailed test – right tailed test)

t = (Xbar - µ) / [SD/sqrt(n)]

Xbar = 4.925

µ = 4.8

SD = 0.235100451

n = 10

df = n – 1 = 9

t = (4.925 – 4.8) / [0.235100451/sqrt(10)]

t = 1.6813

P-value = 0.0635

Alpha value = 0.05

P-value > Alpha value

So, we do not reject the null hypothesis

We conclude that there is no sufficient evidence to claim that the population mean for time to failure is more than 4.8 hours.

Part b

We are given

Sample Standard Deviation = 0.235100451

Sample Mean = 4.925

Sample Size = 10

df = n – 1 = 9

Confidence Level = 95%

Critical t value = 2.2622

Confidence interval = Xbar -/+ t*SD/sqrt(n)

Confidence interval = 4.925 -/+ 2.2622*0.2351/sqrt(10)

Confidence interval = 4.925 -/+ 0.1682

Lower limit = 4.925 – 0.1682

Lower limit = 4.76

Upper limit = 4.925 + 0.1682

Upper limit = 5.09

The value 4.8 is lies between the confidence interval 4.76 and 8.09, so we do not reject the null hypothesis.

We conclude that there is no sufficient evidence to claim that the population mean for time to failure is more than 4.8 hours.

Part c

H0: 2 = 0.03 versus Ha: 2 0.03

We are given

Level of significance = alpha = 0.05

Sample size = n = 10

Degrees of freedom = n – 1 = 9

Sample standard deviation = 0.055272

Test statistic = Chi square = [(n – 1)*S2]/2

Test statistic = Chi square = [(10 – 1)*0.055272^2]/0.03^2

Test statistic = Chi square = 0.9165

P-value = 0.0004

Alpha value = 0.05

P-value < Alpha value

So, reject the null hypothesis

We conclude that there is insufficient evidence that the population variance is 0.03.

Part d

H0: 2 = 0.03 versus Ha: 2 0.03

Formula for confidence interval is given as below:

[(n – 1)*S2]/ 2alpha/2 < 2 < [(n – 1)*S2]/ 21 - alpha/2

We are given

Sample size = n = 10

S = 0.055272

Lower Chi-Square Value               = 2.7004

Upper Chi-Square Value               = 19.0228

[(10 – 1)*0.055272^2]/ 19.0228 < 2 < [(10 – 1)*0.055272^2]/ 2.7004

0.0014 < 2 < 0.0102

Value 0.03 is not belongs to above confidence interval, so we reject the null hypothesis.

We conclude that there is insufficient evidence that the population variance is 0.03.

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