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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

\(\begin{array}{l}\sigma = 0.200\\n = 25\end{array}\)

Given claim: the average weight is \(10kg\).

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:

\({H_0}:\mu = 10kg\)

The alternative hypothesis states the opposite of the null hypothesis (since the null hypothesis states the given claim):

\({H_a}:\mu \ne 10kg\)

Given that, Normal distribution with

\(\begin{array}{l}\sigma = 0.200\\n = 25\end{array}\)

\(\overline x = 9.85\)

\(\alpha = 0.01\)

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

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

\(\begin{array}{l}z = \frac{{\overline x - {\mu _{\overline x }}}}{{{\sigma _{\overline x }}}}\\ = \frac{{9.85 - 10}}{{0.2/\sqrt {25} }}\\ \approx - 3.75\end{array}\)

The P-value is the probability of obtaining a value more extreme or equal to the standardized test statistic z. determine the probability using normal probability table in the appendix.

\(\begin{array}{l}P = P(Z < - 3.75orZ > 3.75)\\ = 2P(Z < - 3.75)\\ = 0\end{array}\)

If the P-value is the smaller than the significance level \(\alpha \), then the null hypothesis is rejected.

\(P < 0.01\),so it鈥檚 reject the \({H_0}\).

It appears that the scale is not calibrated correctly.

Use the value,\(\mu = 10.1\)

Determine \({z_{\alpha /2}} = {z_{0.005}}\) using the normal probability table in the appendix (look up \(0.005\) in the table , the z-score is then the found z-score with the opposite sign);

\({z_{0.005}} = 2.575\)

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:

\(\begin{array}{l}\overline x = \mu - z\frac{\sigma }{{\sqrt n }}\\ = 10 - 2.575\frac{{0.2}}{{\sqrt {25} }}\\ \approx 9.897\end{array}\)

\(\begin{array}{l}\overline x = \mu + z\frac{\sigma }{{\sqrt n }}\\ = 10 + 2.575\frac{{0.2}}{{\sqrt {25} }}\\ \approx 10.103\end{array}\)

We will reject the null hypothesis if the sample mean is smaller than \(9.897\) or larger than \(10.103\).

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

\(\begin{array}{l}z = \frac{{\overline x - {\mu _{\overline x }}}}{{{\sigma _{\overline x }}}}\\ = \frac{{9.897 - 10.1}}{{0.2/\sqrt {25} }}\\ \approx - 5.08\end{array}\)

\(\begin{array}{l}z = \frac{{\overline x + {\mu _{\overline x }}}}{{{\sigma _{\overline x }}}}\\ = \frac{{9.897 + 10.1}}{{0.2/\sqrt {25} }}\\ \approx 0.08\end{array}\)

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

\(\begin{array}{l}\beta = P( - 5.08 < Z < 0.08)\\ = P(Z < 0.08) - P(Z < - 5.08)\\ = 0.5319\\ = 53.19\% \end{array}\)

\(\beta = 37.83\% \)

Use the value,\(\mu = 9.8\)

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

\(\begin{array}{l}z = \frac{{\overline x - {\mu _{\overline x }}}}{{{\sigma _{\overline x }}}}\\ = \frac{{9.897 - 9.8}}{{0.2/\sqrt {25} }}\\ \approx 2.43\end{array}\)

\(\begin{array}{l}z = \frac{{\overline x + {\mu _{\overline x }}}}{{{\sigma _{\overline x }}}}\\ = \frac{{9.897 + 9.8}}{{0.2/\sqrt {25} }}\\ \approx 7.58\end{array}\)

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

\(\begin{array}{l}\beta = P(2.43 < Z < 7.58)\\ = P(Z < 7.58) - P(Z < 2.43)\\ = 1 - 0.9925\\ = 0.0075\end{array}\)

\(\beta = 0.75\% \)