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The mean cost of a daily newspaper is $1.00USD$

($a$)So, the null hypothesis is: $H_{O:渭鈮�1USD}$

($b$)and the alternative hypothesis is: localid="1650213850458" $H_a:渭鈮�1\mathrm{USD}$.

($c$) The random variable localid="1650214243995" $\stackrel{炉}{X}$is the average cost of a daily newspaper.

($d$) The distribution to use for the test is a normal distribution with parameters localid="1650214278401" $渭=1,蟽/\sqrt{n}=$localid="1650214286464" $0.2/\sqrt{12}$.

we use normal distribution because the population standard deviation is known.

($e$) The test statistic for this test is:

localid="1650214300197" $$\begin{gathered} z=\frac{\stackrel{炉}{x}-渭}{蟽/\sqrt{n}} \\ =\frac{0.95-1}{0.2/\sqrt{12}} \\ =-0.866 \end{gathered}$$

($f$) So, the localid="1650214313924" $p-value$for this test is:

localid="1650214321514" $$\begin{gathered} =2P\{\stackrel{炉}{x}<0.95\} \\ =0.386. \end{gathered}$$

($g$) The graph for this problem is:

($h$): $i$level of confidence: $$\begin{gathered} 伪=1\% \\ =0.01 \end{gathered}$$,

$ii.$Decision: we do not reject localid="1650214495265" $H_0$,

$iii.$Reason for decision.

When the localid="1650214503866" $p-value$is greater than the established alpha value we do not reject the null hypothesis.

Now, we see that:localid="1650214512359" $\text{p-value}=0.386>0.01=伪$.

At a level of significance $1\%$, there is insufficient evidence to conclude that the mean cost of a daily newspaper is not $1.00USD$.

If $\stackrel{炉}{x}$and $s$are the mean and standard deviation of a random sample from a normal distribution with known variance ${蟽}^2,100(1-伪)\%$confidence interval on $渭$is given by

$\stackrel{炉}{x}-z_{\frac{1}{2}}\frac{蟽}{\sqrt{n}}鈮�渭鈮�\stackrel{炉}{x}+z_{\frac{}{2}}\frac{蟽}{\sqrt{n}}$ (Equation 1)

Where $z\frac{鈭�}{2}$is the upper $100\frac{伪}{2}$percentage point of the normal distribution

($i$) We need to find a localid="1650214599220" $95\%CI$on the population mean, then

localid="1650214608627" $\frac{伪}{2}=0.025鈬�z_{\frac{蚁}{2}}=z_{0.025}=1.96$(Equation 2)

We get:

$$\begin{gathered} 0.95-1.96\frac{0.20}{\sqrt{12}}鈮�渭鈮�0.95+1.96\frac{0.20}{\sqrt{12}} \\ 0.95-1.96脳0.113鈮�渭鈮�0.95-1.96脳0.113 \end{gathered}$$

Therefore, a $9.5\%cl$the population means is: $0.84鈮�渭鈮�1.06$