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The hypothesis\({G_0}:{\mu _D} = 0\) is not sensible in this context, because the mean equal to zero would mean that the diagnosis occurs at the start of the first symptom, which is not possible (a diagnosis occurs always after the symptoms started occurring)

Let us assume:

\(\alpha = 0.05\)

Given claim: more than \(25\)months

The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis and the alternative hypothesis state the opposite of each other. The null hypothesis needs to contain the value mentioned in the claim.

\(\begin{array}{l}{H_0}:{\mu _d} = - 25\\{H_a}:{\mu _d} < - 25\end{array}\)

Determine the sample mean of the differences. The mean is the sum of all values divided by the number of values.

\(\bar d = \frac{{ - 24 - 12 - 55 + \ldots + - 61 - 69 - 80}}{{15}} \approx - 38.6\)

Determine the sample standard deviation of the differences:

\({s_d} = \sqrt {\frac{{{{( - 24 - ( - 38.6))}^2} + \ldots + {{( - 80 - ( - 38.6))}^2}}}{{15 - 1}}} \approx 23.1788\)

Determine the value of the test statistic:

\(t = \frac{{\bar d - d}}{{{s_d}/\sqrt n }} = \frac{{ - 38.6 - 0}}{{23.1788/\sqrt {15} }} \approx - 6.450\)

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme, assuming that the null hypothesis is true. The Pvalue is the number (or interval) in the column title of the Student's T distribution in the appendix containing the \(t\)-value in the row\(df = n - 1 = \$ \$ 15 - 1 = 14\) :

\(P < 0.0005\)

If the P-value is less than the significance level, reject the null hypothesis.

\(P < 0.05 \Rightarrow {\mathop{\rm Reject}\nolimits} {H_0}\)

There is sufficient evidence to support the claim that on average diagnosis occurs more than \(25\)months after the onset of symptoms.