A random sample of 64 observations produced the following summary statistics: \(\bar x = 0.323\) and \({s^2} = 0.034\).

a. Test the null hypothesis that\(\mu = 0.36\)against the alternative hypothesis that\(\mu < 0.36\)using\(\alpha = 0.10\).

b. Test the null hypothesis that \(\mu = 0.36\) against the alternative hypothesis that \(\mu \ne 0.36\) using \(\alpha = 0.10\). Interpret the result.

The summary statistics for the random sample size\(n = 64\)are:

\(\bar x = 0.323\), and\({s^2} = 0.034\).

The two-sided and one-sided hypothesis testing problems \(\mu = 0.36\) need to test at a 10% significance level.

The test statistic is:

\(\begin{aligned}z &= \frac{{\bar x - \mu }}{{\frac{s}{{\sqrt n }}}}\\ &= \frac{{0.323 - 0.36}}{{\frac{{\sqrt {0.034} }}{{\sqrt {64} }}}}\\ &= \frac{{ - 0.037}}{{0.0230}}\\ &= - 1.61\end{aligned}\)

Therefore, the test statistic is \(z = - 1.61\).

a.

The p-value for the left-tailed test is:

\(\begin{aligned}p &= P\left( {Z < - 1.61} \right)\\ &= 0.0537\end{aligned}\).

The probability is the value at the intersection of -1.60 and 0.01 in the z-table.

Since the p-value is less than 0.10, reject the null hypothesis\(\mu = 0.36\).

Therefore, there is sufficient evidence to claim \(\mu < 0.36\).

b.

The p-value for the two-tailed test is:

\(\begin{aligned}p &= 2 \times P\left( {Z < - 1.61} \right)\\ &= 2 \times 0.0537\\ &= 0.1074\end{aligned}\).

The p-value is more than 0.10; therefore, do not reject the null hypothesis\(\mu = 0.36\).

Hence, there is no sufficient evidence to claim \(\mu < 0.36\).