Question

A random sample of 32 days for the Three Gorges Dam in China provides an average of 266.8 gigawatts of power produced per day with a standard deviation of 22 gigawatts. (a) Compute a 95% confidence interval for the true mean power per day. Is 274 gigawatts a plausible value for µ? Explain. (b) Complete a hypothesis test at α = 0.05 to test µ = 274 versus µ 6= 274. (c) Are the results from (a) and (b) consistent (i.e., do you get the same conclusion from both)?

Answer 1

a) The 95% CI for the mean is
`258.8\leq\mu\leq 274.7`.

The value 274 is plausible because it is within the limits of the CI.

b) We can not reject the hypothesis
`H_0: \mu=274`

c) They are consistent. The conclusion from both results is that 274 could be the real mean. In both cases we couldn't prove 274 it is not the mean.

Explanation:

In this problem we have a sample of n=32 with mean M=266.8 and standard deviation s=22.

a) To compute a 95% confidence interval (CI), we calculate

`M-t_(31)*s/√(n)\leq \mu\leq M+t_(31)*s/√(n)`

First we have to estimate t, with 32-1=31 degrees of freedom for a two-tailed 95% CI.

By looking up in the t-table, we get t=2.0395.

Then the confidence interval is

`M-t_(31)*s/√(n)\leq \mu\leq M+t_(31)*s/√(n)\\\\266.8-2.0.395*22/√(32)\leq\mu\leq 266.8+2.0.395*22/√(32)\\\\266.8-7.9\leq\mu\leq 266.8+7.9\\\\258.8\leq\mu\leq 274.7`

b) We have to test the hypothesis

`H_0: \mu=274\\\\H_1: \mu \\eq 274`

The significance level is 0.05.

We calculate the t-statistics:

`t=(M-\mu)/(s/√(n))=(266.8-274)/(22/√(32))=(-7.2)/(3.9)= 1.84`

The p-value for t=1.84 and df=31 is P=0.07536.

As P=0.07 is greater than the significance level (0.05), we can not reject the null hypothesis. The interpretation of this is we can not claim that the mean is not 274.