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The null hypothesis, denoted by H₀, is the claim that is initially assumed to be true (the 鈥減rior belief鈥� claim). The alternative hypothesis, denoted by H_a, is the assertion that is contradictory to H₀.

The null hypothesis will be rejected in favour of the alternative hypothesis only if sample evidence suggests that H₀ is false. If the sample does not strongly contradict H₀, we will continue to believe in the plausibility of the null hypothesis. The two possible conclusions from a hypothesis-testing analysis are then reject H0 or fail to reject H₀.

Given that, Normal distribution with

\(\sigma = 60\)

Given claim: the average compressive strength is more than \(1300KN/{m^2}\).

The average is represented by the population mean \(\mu \).

The null hypothesis states that the population mean is equal to the value mentioned in the claim:

\({H_0}:\mu = 1300KN/{m^2}\)

The alternative hypothesis states the given claim:

\({H_a}:\mu > 1300KN/{m^2}\)

Given that, Normal distribution with

\(\sigma = 60\)

\(\begin{array}{l}n = 10\\\overline x = 1340\end{array}\)

\(\begin{array}{l}\mu = 1300\\\alpha = 0.01\end{array}\)

Determine \({z_\alpha } = {z_{0.01}}\) using the normal probability table in the appendix (look up \(0.01\) in the table , the z-score is then the found z-score with the opposite sign);

\({z_{0.01}} = 2.33\)

The sampling distribution of the sample mean \(\overline x \) has mean \(\mu \)and standard deviation \(\frac{\sigma }{{\sqrt n }}\).

The corresponding sample mean is the population mean (of the null hypothesis) increased by the product of the z-score and the standard deviation:

\(\begin{array}{l}\overline x = \mu + z\frac{\sigma }{{\sqrt n }}\\ = 1300 + 2.33\frac{{60}}{{\sqrt {10} }}\\ \approx 1344.2086\end{array}\)

We will reject the null hypothesis if the sample mean is larger than \(1344.2086\).

Since \(\overline x = 1340\) is less than \(1344.2086\), we fail to reject the null hypothesis \({H_0}\).

Given that, Normal distribution with

\(\sigma = 60\)

\(n = 10\)

\(\begin{array}{l}\mu = 1350\\\alpha = 0.01\end{array}\)

Probability distribution:

Since the distribution of \(X\) is normal, the sampling distribution of the sample mean \(\overline x \) is also normal.

The sampling distribution of the sample mean \(\overline x \) has mean \(\mu \)and standard deviation \(\frac{\sigma }{{\sqrt n }}\).

\(\begin{array}{l}{\mu _{\overline x }} = \mu = 1350\\{\sigma _{\overline x }} = \frac{\sigma }{{\sqrt n }}\\{\sigma _{\overline x }} = \frac{{60}}{{\sqrt {10} }}\\{\sigma _{\overline x }} \approx 18.9737\end{array}\)

Thus the probability distribution of the test statistic is a normal distribution with mean \(1350\) and standard deviation \(18.9737\).

Determine \({z_\alpha } = {z_{0.01}}\) using the normal probability table in the appendix (look up \(0.01\) in the table , the z-score is then the found z-score with the opposite sign);

\({z_{0.01}} = 2.33\)

The sampling distribution of the sample mean \(\overline x \) has mean \(\mu \)and standard deviation \(\frac{\sigma }{{\sqrt n }}\).

The corresponding sample mean is the population mean (of the null hypothesis) increased by the product of the z-score and the standard deviation:

\(\begin{array}{l}\overline x = \mu + z\frac{\sigma }{{\sqrt n }}\\ = 1300 + 2.33\frac{{60}}{{\sqrt {10} }}\\ \approx 1344.2086\end{array}\)

We will reject the null hypothesis if the sample mean is larger than \(1344.2086\).

The z-score is the value decreased by the mean, divided by the standard deviation:

\(\begin{array}{l}z = \frac{{\overline x - {\mu _{\overline x }}}}{{{\sigma _{\overline x }}}}\\ = \frac{{1344.2086 - 1350}}{{18.9737}}\\ \approx - 0.31\end{array}\)

Determine the probability that we fail to reject the null hypothesis using the normal probability table in the appendix,

\(\begin{array}{l}\beta = P(\overline X \le 1344.2086)\\ = P(Z < - 0.31)\\ = 0.3783\end{array}\)

\(\beta = 37.83\% \)