Question

Use JMP for most of this: The output voltage of a power supply is assumed to be normally distributed. 16 observations taken at random on voltage are as follows:
10.35, 9.30, 10.00, 9.96, 11.65, 12.00, 11.25, 9.58,
11.54, 9.95, 10.28, 8.37, 10.44, 9.25, 9.38, 10.85

Enter data in JMP. Do Analyze/Distributions in JMP, including Test Mean (use a t-test with a p-value), Test Std Deviation (use the ‘Min PValue’), CI, and a normal probability plot. Print output, and include with homework. Write the answers to a,b, and c, referring to the output, and circling values on output. *Do carry out steps of hypothesis test, as outlined above, and state relevant values from the output in writing, too.

Test the hypothesis that the mean voltage is greater than 9.5, using α = .01.

Do a one-sided 99% confidence interval for the true mean voltage, having JMP calculate the lower

limit. Would this CI lead you to the same conclusion as in part a? How do you know?

Test to see if the true standard deviation of the data is equal to 1.5 or not, using α=.05.

Does the assumption of normality seem reasonable? Explain, using JMP output to advise.

Use the range method to estimate the standard deviation. Is the answer close to the sample standard

deviation?

Answer 1

The sample as

10.35, 9.30, 10.00, 9.96, 11.65, 12.00, 11.25, 9.58,
11.54, 9.95, 10.28, 8.37, 10.44, 9.25, 9.38, 10.85

The t test applies here

The hypotheses are

Reject Ho if p value calculated is more than level of significance (0.01)

test staitistics

t calculated as 3.04039

now p valie is computed using t statistics and Degree of freedm,DF=16-1=15

The P-Value is 0.004132.

Conclusion:

Since the p-value=0.004132 which is lower than 0.01, level of significance hence we reject the null hypothesis and have enough evidence to support the claim.

The formula for estimation is:

μ = M ± t(s_M)

where:

M = sample mean
t = t statistic determined by confidence level
sM = standard error = √(s2/n)

Calculation

M = 10.26
t = 2.95
sM = √(0.9992/16) = 0.25

μ = M ± t(sM)
μ = 10.26 ± 2.95*0.25
μ = 10.26 ± 0.7359

hence

99% CI [9.5241, 10.9959].

yes it follows the result of part a as the Confidence interval shows that the lower limit is above 9.5 hence it foloows the claim of mean>9.5.

The provided sample variance is s2=0.998 and the sample size is given by n=16

Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho:σ2=2.25

Ha:σ2≠2.25

Rejection Region

Based on the information provided, the significance level is α=0.05

and the the rejection region for this two-tailed test is R={χ2:χ2<6.262 or χ2>27.488}

Decision about null hypothesis

Conclusion

Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population variance σ2 is different than 2.25, at the 0.05 significance level.

No this answer is not close to standard deviation.