Under the null hypothesis

H_0

\mu = 9.5

. Since the population standard deviation

\sigma

is unknown, we estimate it with the sample standard deviation

s = 3

. The standardized test statistic is defined as:

t = \frac{\bar{X} - \mu_0}{s/\sqrt{n}} = \frac{\bar{X} - 9.5}{3/\sqrt{16}} = \frac{4(\bar{X} - 9.5)}{3}

which follows a student-t distribution with

n - 1 = 16 - 1 = 15

degrees of freedom. The rejection rule is given in terms of the statistic

T

(defined as

T = \frac{\bar{X} - 9.5}{3/\sqrt{16}}

), rejecting

H_0

T > c

. This is mathematically equivalent to rejecting if the t-statistic is greater than

c

because

T

is the t-statistic itself! Therefore, the significance level

\alpha

(the probability of Type I error) is:

\alpha = P(\text{Reject } H_0 \mid H_0) = P(T > c \mid H_0) = 1 - F_{t_{15}}(c)

. Wait, since

T

is already written as

T = \frac{\bar{X} - 9.5}{3/\sqrt{16}} = \frac{4(\bar{X} - 9.5)}{3}

, the inequality

T > c

is exactly

t > c

. This gives

1 - F_{t_{15}}(c)

. Let's check why the options have

\frac{4(c - 9.5)}{3}

inside the t-distribution function. Ah! If

T = \bar{X}

, then

T > c \implies \bar{X} > c \implies \frac{\bar{X} - 9.5}{3/4} > \frac{c - 9.5}{3/4}

. This is indeed

t > \frac{4(c - 9.5)}{3}

. Thus, the level of significance is

1 - F_{t_{15}}\left(\frac{4(c - 9.5)}{3}\right)

, which matches option A.