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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,

\(n = 49\)

The mean is the sum of all values divided by the number of values:

\(\begin{array}{l}\overline x = \frac{{1.38 + 0.44 + 1.09 + .... + 1.44 + 1.28 + 0.51}}{{49}}\\ = \frac{{36.74}}{{49}}\\ \approx 0.749\end{array}\)

The variance is the sum of squares deviations from the mean divided by \(n - 1\). The standard deviation i the square root of the variance:

\(\begin{array}{l}s = \sqrt {\frac{{{{(1.38 - 0.7498)}^2} + .... + {{(0.51 - 0.7498)}^2}}}{{49 - 1}}} \\s \approx 0.3025\end{array}\)

The whiskers of the box plot are at the minimum and maximum value. The box starts at the first quartile, ends at the third quartile and has a vertical line at the median.

The first quartile is at \(25\% \) of the sorted data list, the median at \(50\% \) and the third quartile at \(75\% \).

The figure also contains the most commonly used descriptive measures (such as the mean, median, standard deviation, etc.).

The distribution is skewed to the right (or positively skewed), because the box in the box plot lies to the left between the whiskers and the vertical line of the median in the box of the box plot also lies to the left in the box.

The data values are on the horizontal axis and the standardized normal scores are on the vertical axis.

If the data contains \(n\)data values, then the standardized normal scores are the z-scores in the normal probability table of the appendix corresponding to an area of \(\frac{{j - 0.5}}{n}\)(or the closest area) with \(j \in \{ 1,2,3,...,n\} \).

The smallest standardized score corresponds with the smallest data value, the second smallest standardized score corresponds with the second smallest data value, and so on.

If the pattern in the normal probability plot is roughly linear and does not contain strong curvature, then the distribution is approximately normal.

The normal probability plot contains strong curvature and thus it is not plausible that ALD is at least approximately normal distribution.

It is not necessary to assume normality, because the sample is large and the central limit theorem then tells us that the sampling distribution of the sample mean is approximately normal (thus we are capable of calculating a confidence interval without making assumptions).

Let us assume : \(\alpha = 0.05\)

Given claim: Average is less than \(1.0\).

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

\({H_0}:\mu = 1.0\)

The alternative hypothesis states the claim:

\({H_a}:\mu < 1.0\)

Since the sample is large, we can use the z-test.

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

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

\(z = \frac{{\overline x - \mu }}{{\sigma /\sqrt n }}\)

\(\begin{array}{l}z = \frac{{0.7498 - 1.0}}{{0.3025/\sqrt {49} }}\\z \approx - 5.79\end{array}\)

The P-value is the probability of obtaining a value more extreme or equal to the standard deviation test statistic z, assuming that the null hypothesis is true. Determine the probability table in the appendix.

\(\begin{array}{l}P = P(Z < - 5.79)\\ \approx 0\end{array}\)

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

\(P < 0.05\)\( \Rightarrow \)Reject \({H_0}\)

There is sufficient evidence to support the claim that the true average ALD under these circumstances is less than \(1.0\).

Given that,

\(\begin{array}{l}c = 95\% \\c = 0.95\end{array}\)

Since the sample is large, we can use the confidence interval using the standard normal distribution.

For confidence level \(1 - \alpha = 0.95\) determine \({z_\alpha } = {z_{0.05}}\)using the normal probability table in the appendix(look up \(0.05\) in the table, the z-score is then the found z-score with opposite sign):

\({z_{\alpha /2}} = 1.645\)

Now we take the average of \(1.64\) and \(1.65\), because \(0.05\) lies exactly in the middle between \(0.0495\) and \(0.0505\)

The margin of error is then:

\(\begin{array}{l}E = {z_{\alpha /2}}\frac{\sigma }{{\sqrt n }}\\ = 1.645\left( {\frac{{0.3025}}{{\sqrt {49} }}} \right)\\ \approx 0.0711\end{array}\)

The boundaries of the confidence interval then become:

\(\begin{array}{l}\overline x + E = 0.7498 + 0.0711\\ = 0.8209\end{array}\)

Thus, The upper confidence bound for the true average ALD is \(0.8209\).