Homework 4

Table of Contents

• Problem 1
• Problem 2
• Problem 3
• Problem 4
• Problem 5

Problem 1

To determine whether the bacteria count was lower in the west basin of Lake Macatawa than in the east basin, $n = 37$ samples of water were taken from the west basin and the number of bacteria colonies in $100$ milliliters of water was counted. The sample characteristics were $\mean{x} = 11.95$ and $s = 11.80$, measured in hundreds of colonies. Find an approximate $95\%$ confidence interval for the mean number of colonies (say, $\mu_W$) in $100$ milliliters of water in the west basin.

Solution.

While the true standard deviation is unknown, the sample size is large enough to allow us to use the central limit theorem, so the confidence interval will be

$$11.95 \pm z_{0.025} \frac{11.80}{\sqrt{37}} = \boxed{\br{8.15, 15.75}}.$$

Problem 2

A leakage test was conducted to determine the effectiveness of a seal designedto keep the inside of a plug airtight. An air needle was inserted into the plug, and the plug and needlewere placed under water. The pressure was then increased until leakage was observed. Let $X$ equal the pressure in pounds per square inch. Assume that the distribution of $X$ is $\mathcal{N}\p{\mu, \sigma^2}$. The following $n = 10$ observations of $X$ were obtained:

Use the observations to:

1. Find a point estimate of $\mu$.
2. Find a point estimate of $\sigma$.
3. Find a $95\%$ one-sided confidence interval for $\mu$ that provides an upper bound for $\mu$.

Solution.

1. $$\boxed{\mean{x} \approx 3.72}$$
2. $$\boxed{s \approx 0.46}$$

3. Since the sample is small and the variance is unknown, we need to use a $t$-distribution with $10 - 1 = 9$ degrees of freedom:

   $$T = \frac{\mean{X} - \mu}{S/\sqrt{10}} \sim t\p{9}$$

   For a $95\%$ one-sided confidence interval, we use

   $$\P\p{T \leq t_{0.05}\p{9}} = 0.95.$$

   This gives the one-sided interval

   $$\mu \leq \mean{x} + t_{0.05}\p{9} \frac{s}{\sqrt{10}} = \boxed{\poc{-\infty, 3.99}}.$$

Problem 3

Students in a semester-long health-fitness program have their percentage of body fat measured at the beginning of the semester and at the end of the semester. The following measurements give these percentages for $9$ men and for $8$ women:

1. Let $X$ be the change in percentage of body fat for females before and after the program. Assume $X \sim \mathcal{N}\p{\mu_X, \sigma_X^2}$ for some unknown $\sigma_X^2$. Construct a $90\%$ confidence interval for $\mu_X$.

2. Let $Y$ be the change in percentage of body fat for males before and after the program. Assume $Y \sim \mathcal{N}\p{\mu_Y, \sigma_Y^2}$ for some unknown $\sigma_Y^2$. Construct a $90\%$ confidence interval for $\mu_Y$.

3. Assume $\sigma_X^2 = \sigma_Y^2 = \sigma^2$ for some unknown $\sigma^2$. Construct a $90\%$ confidence interval for $\mu_X - \mu_Y$.

4. Is the program effective in reducing percentage of body fat? Is it more effective for males or for females?

Solution.

1. Since the sample size is small and the variance is unknown, we use a $t$-distribution for our confidence interval. Our sample for $X$ is

   $$\begin{array}{rrrr} -0.01 & 1.87 & -1.95 & -1.90 \\ -1.36 & -2.70 & -1.64 & -0.85 \end{array}$$

   Thus, our confidence interval is

   $$\mean{y} \pm t_{0.05}\p{7} \frac{s_Y}{\sqrt{8}} = \boxed{\br{-2.03, -0.11}}.$$

2. Similarly, our sample for $Y$ is

   $$\begin{array}{rrrrr} -1.13 & -3.70 & -0.98 & 0.98 & -0.89 \\ -0.49 & -1.40 & 2.10 & -0.91 \end{array}$$

   We get the confidence interval

   $$\mean{x} \pm t_{0.05}\p{8} \frac{s_X}{\sqrt{9}} = \boxed{\br{-1.70, 0.28}}.$$

3. Since the variances are the same, we will use a $t$-distribution with degrees of freedom $9 + 8 - 2 = 15$. Recall the pooled sample variance:

   $$S_p^2 = \frac{\p{n_X-1} S_X^2 + \p{n_Y-1} S_Y^2}{n_X + n_Y - 2}.$$

   The final confidence interval will be

   $$\mean{x} - \mean{y} \pm t_{0.05}\p{15} s_p \sqrt{\frac{1}{9} + \frac{1}{8}} = \boxed{\br{-0.94, 1.65}}.$$

4. Based on the samples and confidence intervals, the program seems to be effective for females, but inconclusive for males (since it contains $0$). Similarly, based on the confidence interval for the difference of the means, there is no evidence that the program is more effective for a particular sex.

Problem 4

Let $X_1, \ldots, X_5$ be a random sample of SAT mathematics scores, assumed to be $\mathcal{N}\p{\mu_X, \sigma^2}$, and let $Y_1, \ldots Y_5$ be an independent random sample of SAT verbal scores, assumed to be $\mathcal{N}\p{\mu_Y , \sigma^2}$. If the following data are observed:

find a $90\%$ confidence interval for $\mu_X - \mu_Y$.

Solution.

Because the sample size is small and the variances are unknown, we use a $t$-distribution with degrees of freedom $5 + 5 - 2 = 8$. The confidence interval is

$$\mean{x} - \mean{y} \pm t_{0.05}\p{8} s_p \sqrt{\frac{1}{5} + \frac{1}{5}} = \boxed{\br{-111.27, 70.87}}.$$

Alternatively, you can also view the problem as $n = 5$ samples from

$$X - Y = \mathcal{N}\p{\mu_X - \mu_Y, 2\sigma^2},$$

in which case, you use a $t$-distribution with degrees of freedom $5 - 1 = 4$, which gives

$$\mean{x} - \mean{y} \pm t_{0.05}\p{4} \frac{s}{\sqrt{5}} = \boxed{\br{-115.98, 75.58}}.$$

Problem 5

Let $X$ and $Y$ be the life time (in hours) of two types of light bulbs, respectively. Assume $X \sim \mathcal{N}\p{\mu_X, 689}$ and $Y \sim \mathcal{N}\p{\mu_Y , 735}$. Suppose a random sample of $23$ type $X$ light bulbs yields an average life time of $956.2$ hours, and a random sample of $28$ type $Y$ light bulbs yields an average life time of $978.6$ hours. Construct a $95\%$ confidence interval for $\mu_X - \mu_Y$.

Solution.

The variances are known, so we can use a normal distribution.

$$\mean{x} - \mean{y} \pm z_{0.025} \sqrt{\frac{\sigma_X^2}{23} + \frac{\sigma_Y^2}{28}} = \boxed{\br{-37.09, -7.71}}.$$