Final Answer:
a) The standard error of the mean is approximately \(0.2276\) (rounded to four decimal places).
Step-by-step explanation:
a) The standard error of the mean (SE) is calculated using the formula \(SE = \frac{s}{\sqrt{n}}\), where \(s\) is the sample standard deviation and \(n\) is the sample size. In this case, the sample standard deviation is given as \(1.6\) and the sample size is \(51\). Substituting these values into the formula, we get \(SE = \frac{1.6}{\sqrt{51}} \approx 0.2276\), rounded to four decimal places.
b) To find the critical value of \(t^*\) for a \(95\%\) confidence interval with \(50\) degrees of freedom (sample size \(51 - 1 = 50\)), we consult a t-distribution table or use statistical software. For a two-tailed test at \(0.025\) significance level, the critical value is approximately \(\pm 2.0096\) (assuming a symmetric distribution).
c) Constructing a \(95\%\) confidence interval involves using the formula \(\bar{x} \pm t^*\frac{s}{\sqrt{n}}\). Substituting the given values, we get \(80 \pm 2.0096 \times \frac{1.6}{\sqrt{51}}\).
The interval is calculated to be approximately \((79.4917, 80.5083)\). Since the null hypothesis value (\(\mu = 85\)) is not within this interval, at \(\alpha = 0.050\), we can reject the null hypothesis. The evidence suggests that the population mean is significantly different from \(85\).