Texts the output voltage of a power supply is assumed to be...

Texts: The output voltage of a power supply is assumed to be normally distributed. Sixteen 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, and 10.85. a) Test the hypothesis that the mean voltage equals 12 V against a two-sided alternative using alpha = 0.05. b) Construct a 95% two-sided confidence interval on μ. c) Test the hypothesis that σ^2 = 1 using alpha = 0.05. d) Construct a 95% two-sided confidence interval on σ.

Transcript

00:03 Part a, we want to test the hypothesis that the mean voltage equals 12 volts against a two -sided alternative using alpha of 0 .05.

00:11 So that would mean the null hypothesis is that the mean is equal to 12 and the alternative is that the mean is not equal to 12.

00:19 That's going to be two -sided.

00:22 Given that alpha is 0 .05, the test statistic is the sample mean minus the hypothesized mean over the sample standard deviation over the square root of the sample size.

00:39 And let's calculate these.

00:45 The sample mean is 10 .259.

00:53 The sample standard deviation is 0 .999.

00:57 Probably could get away with 1 and there are 16 in the sample size.

01:01 And so this calculates to 0 .0 -6 .97.

01:23 And we can compare this against the critical t value with 15 degrees of freedom, 1 less than the sample size with a significance level of 0 .05 as a critical value of negative, we would care about the negative here because of the negative test statistic, but negative 2 .13.

01:56 And since t is even less than our critical, that would mean we would reject the null hypothesis in favor of the alternative that they are different.

02:12 Next we'll construct the 95 % confidence interval using the formula of the sample mean plus or minus the critical t times the sample standard deviation over the square root of the sample size.

02:23 And so that would be 10 .259 plus or minus 2 .13...

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