Question

The cost of a daily newspaper varies from city to city. However, the variation among prices remains steady with a standard deviation of $20c$. A study was done to test the claim that the mean cost of a daily newspaper is $1.00$. Twelve costs yield a mean cost of $95\%$with a standard deviation of $18c$. Do the data support the claim at the $1\%$level?

Short answer

($a$) The null hypothesis is localid="1650213643946" $H_0:\mu =1\mathrm{USD}$.

($b$) The alternative hypothesis is: localid="1650213651502" $H_a:\mu \ne 1\mathrm{USD}$.

($c$) A random variable localid="1650213734663" $\overline{X}$is the average cost of a daily newspaper.

($d$) For this test, we use a normal distribution with parameters localid="1650213669545" $\mu =1,\sigma /\sqrt{n}=0.2/\sqrt{12}$.

($e$) The test statistic for this test is: localid="1650213680892" $z=-0.866$.

($f$) localid="1650213689191" $P-value$for this test is: localid="1650213702809" $p=0.386$.

($g$) For this test, we have provided a graph.

($h$) localid="1650213719425" $i$. Level of confidence localid="1650213710887" $\alpha =0.01$. localid="1650213726872" $ii$. Decision: we do not reject localid="1650213743032" $H_0:\mu .$localid="1650213772196" $iii$. Reason for decision: The localid="1650213780425" $p-value$is greater than localid="1650213788509" $0.01.$iv. Conclusion: At a level of significance localid="1650213796781" $1\%$, there is insufficient evidence to conclude that the mean cost of a daily newspaper is not localid="1650213803659" $1.00USD.$(i) A localid="1650213810843" $95\%$localid="1650213818810" $CI$on the population mean is localid="1650213826169" $0.84\le \mu \le 1.06.$

Step-by-step solution

Explanation for Part (a) and (b):

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

($a$)So, the null hypothesis is: $H_{O:\mu \ne 1USD}$

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

Part (c) and (d).

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

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

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

Part (e).

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

localid="1650214300197" $$\begin{gathered} z=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}} \\ =\frac{0.95-1}{0.2/\sqrt{12}} \\ =-0.866 \end{gathered}$$

Part (f).

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

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

Part (g)

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

Part (h) i,ii,iii

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

$ii.$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" $\text{p-value}=0.386>0.01=\alpha$.

Part (h)iv

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 $\overline{x}$and $s$are the mean and standard deviation of a random sample from a normal distribution with known variance ${\sigma }^2,100(1-\alpha )\%$confidence interval on $\mu$is given by

$\overline{x}-z_{\frac{1}{2}}\frac{\sigma }{\sqrt{n}}\le \mu \le \overline{x}+z_{\frac{}{2}}\frac{\sigma }{\sqrt{n}}$ (Equation 1)

Where $z\frac{\propto }{2}$is the upper $100\frac{\alpha }{2}$percentage point of the normal distribution

Part (i):

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

localid="1650214608627" $\frac{\alpha }{2}=0.025\Rightarrow z_{\frac{\rho }{2}}=z_{0.025}=1.96$(Equation 2)

From equation (1) and (2):

We get:

$$\begin{gathered} 0.95-1.96\frac{0.20}{\sqrt{12}}\le \mu \le 0.95+1.96\frac{0.20}{\sqrt{12}} \\ 0.95-1.96\times 0.113\le \mu \le 0.95-1.96\times 0.113 \end{gathered}$$

Therefore, a $9.5\%cl$the population means is: $0.84\le \mu \le 1.06$