Answer

(a) Single Student Probability:

For an individual student, standard deviation $\sigma = 16,000$ and mean $\mu = 6,000$.

$$Z = \frac{X - \mu}{\sigma} = \frac{0 - 6000}{16000} = -0.375$$

Using a standard normal distribution table, $P(Z \le -0.375) \approx 0.3538$ (or approximately 35.4%).

(b) Sample Mean Probability:

For a sample size of $n = 64$, the standard error is:

$$ \frac{\sigma}{\sqrt{n}} = \frac{16000}{\sqrt{64}} = \frac{16000}{8} = 2000$$

Assuming true mean $\mu = 0$:

$$Z = \frac{\bar{X} - \mu}{{\sigma}/\sqrt{n}} = \frac{6000 - 0}{2000} = 3.00$$

$$P(Z \ge 3.00) = 1 - 0.9987 = 0.0013 \text{ (or } 0.13\%\text{)}$$

(c) Explanation:

The results support the school's claim. Part (a) shows that finding a single student with no gain is quite common (about a 35% chance), so a single anecdotal testimonial does not disprove the program's success. On the other hand, part (b) shows that if the true effect were zero, observing a sample average gain of $6,000$ across 64 students would be extremely improbable (0.13%).

Answer

(a)

For an individual student, standard deviation $\sigma = 120$ and mean $\mu = 80$.

$$Z = \frac{X - \mu}{\sigma} = \frac{0 - 80}{120} = -\frac{2}{3} \approx -0.67$$

$$P(Z \le -0.67) \approx 0.2514 \text{ (or approximately } \mathbf{25.1\%}\text{)}$$

(b)

For a sample size of $n = 36$, the standard error is:

$$SE = \frac{\sigma}{\sqrt{n}} = \frac{120}{\sqrt{36}} = \frac{120}{6} = 20$$

Assuming true mean $\mu = 0$:

$$Z = \frac{\bar{X} - \mu}{SE} = \frac{80 - 0}{20} = 4.00$$

$$P(Z \ge 4.00) \approx 0.00003 \text{ (or } \mathbf{0.003\%}\text{)}$$

(c)

The single student's outcome does not undermine the school's claim. Given the large variation in individual test scores ($\sigma = 120$), roughly 1 in 4 students will see no improvement even when the course works as advertised. However, obtaining a sample mean gain of 80 points across 36 students purely by chance (if the course were useless) is practically impossible ($Z = 4.00$).

Answer

We have that the sampling distribution of the mean is $\bar{X} \sim N\left(0.69, \frac{0.69 \times 0.31}{1500}\right)$.

(a) We want to compute $P[0.67 < \bar{X} < 0.71]$, we have

$$P[0.67 < \bar{X} < 0.71] = P[0.67 - 0.69 < \bar{X} - 0.69 < 0.71 - 0.69]$$ $$= P\left[ \frac{0.67 - 0.69}{\sqrt{0.69 \times 0.31 / 1500}} < \frac{\bar{X} - 0.69}{\sqrt{0.69 \times 0.31 / 1500}} < \frac{0.71 - 0.69}{\sqrt{0.69 \times 0.31 / 1500}} \right]$$ $$\approx P[-1.676 < Z < 1.676]$$ $$= 90.62\%$$

(b) The chance of seeing 71% or more is about 4.69%, so this is borderline but somewhat unlikely.

(c) We can compute

$$P[\bar{X} > 0.71] = P[\bar{X} - 0.69 > 0.71 - 0.69]$$ $$= P\left[ \frac{\bar{X} - 0.69}{\sqrt{0.69 \times 0.31 / 3000}} > \frac{0.71 - 0.69}{\sqrt{0.69 \times 0.31 / 3000}} \right]$$ $$\approx P[Z > 2.3708]$$ $$= 0.89\%$$

Now the probability is quite small, at less than 1 in one hundred it is unlikely that the true mean is 0.69.

Answer

We have that the sampling distribution of the mean is $\bar{X} \sim N\left(0.42, \frac{0.42 \times 0.58}{1000}\right)$.

(a) We want to compute $P[0.39 < \bar{X} < 0.45]$, we have

$$P[0.39 < \bar{X} < 0.45] = P[0.39 - 0.42 < \bar{X} - 0.42 < 0.45 - 0.42]$$ $$= P\left[ \frac{0.39 - 0.42}{\sqrt{0.42 \times 0.58 / 1000}} < \frac{\bar{X} - 0.42}{\sqrt{0.42 \times 0.58 / 1000}} < \frac{0.45 - 0.42}{\sqrt{0.42 \times 0.58 / 1000}} \right]$$ $$\approx P[-1.921 < Z < 1.921]$$ $$= 94.52\%$$

(b) The chance of seeing 45% or more is about 2.74%, so this is fairly small. By chance we would see such numbers less than three times in a hundred, so it is hard to tell if this is too small or large enough to expect. Same as the previous question. Basically these calculations are either clear (really large or small) or borderline. We discuss how to think about this in Chapter 8.

(c) We can compute

$$P[\bar{X} > 0.45] = P[\bar{X} - 0.42 > 0.45 - 0.42]$$ $$= P\left[ \frac{\bar{X} - 0.42}{\sqrt{0.42 \times 0.58 / 2000}} > \frac{0.45 - 0.42}{\sqrt{0.42 \times 0.58 / 2000}} \right]$$ $$\approx P[Z > 2.7165]$$ $$= 0.33\%$$

Now the probability is quite small, at about 3 in one thousand it is unlikely that the true mean is 0.42.

Answer

We have $n = 1000$ and $\pi = 0.33$ from our theory. The central limit theorem is clearly appropriate since the sample size is very large.

(a) We want to compute $P[\bar{X} \ge 0.37]$, we have

$$P[\bar{X} \ge 0.37] = P[\bar{X} - 0.33 \ge 0.37 - 0.33]$$ $$= P\left[\frac{\bar{X} - 0.33}{\sqrt{0.33 \times 0.67 / 1000}} \ge \frac{0.37 - 0.33}{\sqrt{0.33 \times 0.67 / 1000}}\right]$$ $$\approx P[Z \ge \frac{0.04}{0.01487}]$$ $$\approx P[Z \ge 2.69]$$ $$= 0.0036$$

or about 0.36%.

(b) Not really, we would not see a sample average of 0.37 very often if the truth was 0.33, it is more likely that the true value is larger.

(c) For this we want to know the sample size such that $P[\bar{X} \ge 0.37] = 0.01$. So we want

$$P[\bar{X} \ge 0.37] = P[\bar{X} - 0.33 \ge 0.37 - 0.33]$$ $$= P\left[\frac{\bar{X} - 0.33}{\sqrt{0.33 \times 0.67 / n}} \ge \frac{0.37 - 0.33}{\sqrt{0.33 \times 0.67 / n}}\right]$$ $$= P\left[Z \ge \frac{0.37 - 0.33}{\sqrt{0.33 \times 0.67 / n}}\right]$$ $$= 0.01$$

so from the normal tables we have that $z_0$ for this is 2.33. So $\frac{0.37 - 0.33}{\sqrt{0.33 \times 0.67 / n}} = 2.33$

$$\sqrt{\frac{0.2211}{n}} = \frac{0.04}{2.33}$$ $$\frac{0.2211}{n} = 0.0002947$$

solves to 750.3, so we need about 751 observations for this.

Answer

We have $n = 1200$ and $\pi = 0.52$ from our theory. The central limit theorem is clearly appropriate since the sample is very large.

(a) We want to compute $P[\bar{X} \ge 0.56]$, we have

$$P[\bar{X} \ge 0.56] = P[\bar{X} - 0.52 \ge 0.56 - 0.52]$$ $$= P\left[\frac{\bar{X} - 0.52}{\sqrt{0.52 \times 0.48 / 1200}} \ge \frac{0.56 - 0.52}{\sqrt{0.52 \times 0.48 / 1200}}\right]$$ $$\approx P[Z \ge \frac{0.04}{0.01442}]$$ $$\approx P[Z \ge 2.77]$$ $$= 0.0028$$

or about 0.28%.

(b) Not really, we would not see a sample average of 0.56 very often if the truth was 0.52, it is much more likely that the true value is higher.

(c) For this we want to know the sample size such that $P[\bar{X} \ge 0.56] = 0.05$. So we want

$$P[\bar{X} \ge 0.56] = P[\bar{X} - 0.52 \ge 0.56 - 0.52]$$ $$= P\left[\frac{\bar{X} - 0.52}{\sqrt{0.52 \times 0.48 / n}} \ge \frac{0.56 - 0.52}{\sqrt{0.52 \times 0.48 / n}}\right]$$ $$= P\left[Z \ge \frac{0.56 - 0.52}{\sqrt{0.52 \times 0.48 / n}}\right]$$ $$= 0.05$$

so from the normal tables we have that $z_0$ for this is 1.645. So $\frac{0.56 - 0.52}{\sqrt{0.52 \times 0.48 / n}} = 1.645$

solves to 422.3, so we need about 423 observations for this.