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The null hypothesis states that the population mean is equal to the value mentioned in the claim. If the null hypothesis is the claim, then the alternative hypothesis states the opposite of the null hypothesis.

\(\begin{array}{l}{H_0}:\mu = 0\\{H_a}:\mu \ne 0\end{array}\)

The formula for the value of the test statistic is given by, \(t = \frac{{\bar x - {\mu _0}}}{{s/\sqrt n }}\).

(a)

NORMAL PROBABILITY PLOT:

The horizontal axis shows the data values, while the vertical axis shows the standardised normal scores.

The standardised normal scores are the z-scores in the normal probability table in the appendix corresponding to an area of \(\frac{{j - 0.5}}{n}\) (or the closest area) with \(j \in \{ 1,2,3,...,n\} \) if the data comprises n data values. The least standardised score correlates to the smallest data value, and so on.

The population distribution is essentially normal if the pattern in the normal probability plot is roughly linear and does not contain strong curvature.

The population distribution is about normal, and known that the conditions for testing the necessary hypotheses are met because the normal probability plot does not have severe curvature and is essentially linear.

(b)

The mean is the ration of sum of all values and the total number of values.

\(\begin{aligned}{c}\bar x &= \frac{{0.53 + 0.65 + 0.46 + 0.50 + 0.37}}{5}\\ &= \frac{{2.51}}{5}\\ &\approx 0.502\end{aligned}\)

The square of the variance is the standard deviation.

\(\begin{aligned}{c}s &= \sqrt {\frac{{{{(0.53 - 0.502)}^2} + \ldots . + {{(0.37 - 0.502)}^2}}}{{5 - 1}}} \\ &\approx 0.1023\end{aligned}\)

Let the given be:

\(\begin{array}{l}n = 5\\\alpha = 0.05/0.01\end{array}\)

Claim that the true average amount left is less than\(10\% \)of\(0.6oz\), that corresponds with\(0.6oz\).

The value of the test statistic:

\(\begin{aligned}{c}t &= \frac{{\bar x - {\mu _0}}}{{s/\sqrt n }}\\ &= \frac{{0.502 - 0.6}}{{0.1023/\sqrt 5 }}\\ &\approx - 2.142\end{aligned}\)

The P-value is the chance of getting the test statistic's result, or a number that is more severe. The P-value is the number (or interval) in the column header of the T table in the appendix that contains the t-value in the row\(\begin{array}{c}df = n - 1\\ = 5 - 1\\ = 4\end{array}\)for the student.

\(0.025 < P < 0.05\)

As the P-value is smaller than the significance level, so the null hypothesis is rejected.

\(\begin{array}{l}P < 0.05 \Rightarrow Reject {H_0}\\P > 0.01 \Rightarrow Fail to Reject {H_0}\end{array}\)

The true average amount left is less than \(10\% \) of \(0.6oz\), that corresponds with \(0.6oz\) is supported by appropriate evidence at significance level \(0.05\). The true average amount left is less than \(10\% \) of \(0.6oz\), that corresponds with \(0.6oz\) is not supported by appropriate evidence at significance level \(0.01\).

(c)

Let the given be:

\(\begin{array}{l}n = 5\\\alpha = 0.05/0.01\end{array}\)

Claim that the true average amount left is less than\(10\% \)of\(0.6oz\), that corresponds with\(0.6oz\).

The null hypothesis states that the population mean is equal to the value mentioned in the claim. If the null hypothesis is the claim, then the alternative hypothesis states the opposite of the null hypothesis.

\(\begin{array}{l}{H_0}:\mu = 0.6\\{H_a}:\mu < 0.6\end{array}\)

Type I error: When\({H_0}\)is true, reject the null hypothesis\({H_0}\).

Interpretation: The true average amount remaining is less than\(10\% \)of the advertised net contents, which is incorrect.

Type Il error: When\({H_0}\)is false, fail to reject the null hypothesis\({H_0}\).

Interpretation: The true average quantity remaining is at least \(10\% \) of the reported net contents, which is incorrect.