Answer:
a) The 95% CI for the mean is [tex]258.8\leq\mu\leq 274.7[/tex].
The value 274 is plausible because it is within the limits of the CI.
b) We can not reject the hypothesis [tex]H_0: \mu=274[/tex]
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.
Step-by-step 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
[tex]M-t_{31}*s/\sqrt{n}\leq \mu\leq M+t_{31}*s/\sqrt{n}[/tex]
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
[tex]M-t_{31}*s/\sqrt{n}\leq \mu\leq M+t_{31}*s/\sqrt{n}\\\\266.8-2.0.395*22/\sqrt{32}\leq\mu\leq 266.8+2.0.395*22/\sqrt{32}\\\\266.8-7.9\leq\mu\leq 266.8+7.9\\\\258.8\leq\mu\leq 274.7[/tex]
b) We have to test the hypothesis
[tex]H_0: \mu=274\\\\H_1: \mu \neq 274[/tex]
The significance level is 0.05.
We calculate the t-statistics:
[tex]t=\frac{M-\mu}{s/\sqrt{n}}=\frac{266.8-274}{22/\sqrt{32}}=\frac{-7.2}{3.9}= 1.84[/tex]
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.
