It was determined that the percentage of a's in the English language in the 1800 s was $12 \%$. A random sample of 600 letters from a current newspaper contained 60 a's. Test the hypothesis that the proportion of a's has changed in modern times, using the 0.05 level of significance. Treat the newspaper as a random sample of all letters used.
State the null and alternative hypotheses.
A.
\[
\begin{array}{l}
H_{0}: p \neq 0.12 \\
H_{a}: p=0.12
\end{array}
\]
B.
\[
\begin{array}{l}
H_{0}: p> 0.12 \\
H_{a}: p< 0.12
\end{array}
\]
C.
\[
\begin{array}{l}
H_{0}: p=0.12 \\
H_{a}: p \neq 0.12
\end{array}
\]
D.
\[
\begin{array}{l}
H_{0}: p=0.12 \\
H_{a}: p> 0.12
\end{array}
\]
E.
\[
\begin{array}{l}
H_{0}: p< 0.12 \\
H_{a}: p> 0.12
\end{array}
\]
Determine the z-test statistic.
\[
z=\square \text { (Round to two decimal places as needed.) }
\]
Find the $p$-value.
\[
\text { p-value }=\square \text { (Round to three decimal places as needed. })
\]
Choose the correct conclusion.
A. Do not reject $\mathrm{H}_{0}$. The proportion of a's is significantly different from 0.12 .
B. Do not reject $\mathrm{H}_{0}$. The proportion of a's is not significantly different from 0.12 .
C. Reject $\mathrm{H}_{0}$. The proportion of a's is not significantly different from 0.12 .
D. Reject $\mathrm{H}_{0}$. The proportion of a's is significantly different from 0.12 .
(1) Time Remaining: 01:22:24
Next
Answer
Finally, we compare the p-value to the level of significance (0.05). If the p-value is less than the level of significance, we reject the null hypothesis. In this case, 0.041 < 0.05, so we reject H0. This means we believe the proportion of 'a's has changed in modern times. So, the correct conclusion is: \[\boxed{\text{Reject } H_{0}. \text{The proportion of 'a's is significantly different from 0.12.}}\]
Step 1 ：The null hypothesis (H0) is that the proportion of 'a's has not changed, i.e., it is still 0.12. The alternative hypothesis (Ha) is that the proportion of 'a's has changed, i.e., it is not 0.12. So, the correct hypotheses are: \[H_{0}: p=0.12\] \[H_{a}: p \neq 0.12\]
Step 2 ：Next, we calculate the z-test statistic using the formula: \[z = \frac{{\hat{p} - p_{0}}}{{\sqrt{\frac{{p_{0} * (1 - p_{0})}}{{n}}}}}\] where \(\hat{p}\) is the sample proportion, \(p_{0}\) is the assumed population proportion under the null hypothesis, and n is the sample size. In this case, \(\hat{p} = \frac{{60}}{{600}} = 0.10\), \(p_{0} = 0.12\), and n = 600. Plugging these values into the formula gives: \[z = \frac{{0.10 - 0.12}}{{\sqrt{\frac{{0.12 * (1 - 0.12)}}{{600}}}}} = -2.04\]
Step 3 ：The p-value is the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true. For a two-tailed test (because Ha: p ≠ 0.12), we find the probability of observing a z-score less than -2.04 or greater than 2.04. This gives a p-value of 0.041.
Step 4 ：Finally, we compare the p-value to the level of significance (0.05). If the p-value is less than the level of significance, we reject the null hypothesis. In this case, 0.041 < 0.05, so we reject H0. This means we believe the proportion of 'a's has changed in modern times. So, the correct conclusion is: \[\boxed{\text{Reject } H_{0}. \text{The proportion of 'a's is significantly different from 0.12.}}\]
