Before it closed, Ron Swanson was a frequent patron of Charles Mulligan’s Steakhouse in Indianapolis, Indiana. Ron enjoyed the experience so much, during each visit he took a picture with his steak.

Ron also weighed each steak he consumed. He has a record of eating six “22 ounce” Charles Mulligan’s porterhouse steaks. Ron found that these six steaks weighed

\[ 22.4 \text{ oz}, \ 20.8 \text{ oz}, \ 21.6 \text{ oz}, \ 20.2 \text{ oz}, \ 21.4 \text{ oz}, \ 22.0 \text{ oz} \]

Suppose that “22 ounce” Charles Mulligan’s porterhouse steaks follow a \(N(\mu, \sigma^2)\) distribution and that Ron’s six steaks were a random sample.

**(a)** Compute the sample standard deviation, \(s\), of these six steaks. Do not use a computer. You may only use \(+\), \(-\), \(\times\), \(\div\), and \(\sqrt{}\) on a calculator. Show **all** work.

**Solution:**

\(x_i\) | \(x_i - \bar{x}\) | \((x_i - \bar{x})^2\) |
---|---|---|

22.4 | 1.0 | 1.00 |

20.8 | -0.6 | 0.36 |

21.6 | 0.2 | 0.04 |

20.2 | -1.2 | 1.44 |

21.4 | 0 | 0.00 |

22.0 | 0.6 | 0.36 |

So, we have

\[ \sum_{i = n}^{n} x_i = 128.4 \]

Thus

\[ \bar{x} = \frac{1}{n} \sum_{i = n}^{n} x_i = \frac{1}{6}(128.4) = 21.4 \]

Note that,

\[ \sum_{i = n}^{n} (x_i - \bar{x}) = 0. \]

More importantly,

\[ \sum_{i = n}^{n} (x_i - \bar{x})^2 = 3.2 \]

Thus,

\[ s^2 = \frac{1}{n - 1}\sum_{i = n}^{n} (x_i - \bar{x})^2 = \frac{1}{5}(3.2) = 0.64 \]

Then finally,

\[ s = \sqrt{s^2}= \sqrt{0.64} = \boxed{0.8} \]

```
x = c(22.4, 20.8, 21.6, 20.2, 21.4, 22.0)
mean(x)
```

`## [1] 21.4`

`var(x)`

`## [1] 0.64`

`sd(x)`

`## [1] 0.8`

**(b)** Construct a \(95\%\) two-sided confidence interval for the true mean weight of a “22 ounce” Charles Mulligan’s porterhouse steak, \(\mu\).

**Solution:**

Here \(\sigma\) is unknown and \(n\) is small, so we use \(t\).

\[ \bar{x} \pm t_{\alpha / 2}(n - 1)\frac{s}{\sqrt{n}} \]

We have,

- \(n = 6\),
- \(\bar{x} = 21.4\),
- \(s = 0.8\),
- \(1 - \alpha = 0.95\), so \(\alpha / 2 = 0.025\),
- \(t_{\alpha / 2}(n - 1) = t_{0.025}(5) = 2.571\).

\[ 21.4 \pm 2.571 \frac{0.8}{\sqrt{6}} \]

\[ \boxed{\bf 21.4 \pm 0.8397} \]

\[ \boxed{\bf (20.5603, 22.2397)} \]

```
n = length(x)
est = mean(x)
s = sd(x)
alpha = 0.05
crit = qt(alpha / 2, df = n - 1, lower.tail = FALSE)
se = s / sqrt(n)
margin = crit * se
lower = est - margin
upper = est + margin
c(lower, upper) # above answer contains some rounding, hence the difference
```

`## [1] 20.56045 22.23955`

`t.test(x, level = 0.95)$conf.int`

```
## [1] 20.56045 22.23955
## attr(,"conf.level")
## [1] 0.95
```

**(c)** Construct a \(95\%\) confidence lower bound for the true mean weight of a “22 ounce” Charles Mulligan’s porterhouse steak, \(\mu\).

**Solution:**

\[ \left[ \bar{x} - t_{\alpha}(n - 1)\frac{s}{\sqrt{n}}, \infty \right) \]

We have,

- \(n = 6\),
- \(\bar{x} = 21.4\),
- \(s = 0.8\),
- \(1 - \alpha = 0.95\), so \(\alpha = 0.05\),
- \(t_{\alpha}(n - 1) = t_{0.05}(5) = 2.0150\).

\[ 21.4 - 2.0150 \frac{0.8}{\sqrt{6}} = 20.7419 \]

\[ \boxed{[20.7419, \infty)} \]

```
n = length(x)
est = mean(x)
s = sd(x)
alpha = 0.05
crit = qt(alpha, df = n - 1, lower.tail = FALSE)
se = s / sqrt(n)
margin = crit * se
(lower = est - margin)
```

`## [1] 20.74189`

**(d)** Construct a \(90\%\) two-sided confidence interval for the true standard deviation of the weight of a “22 ounce” Charles Mulligan’s porterhouse steak, \(\sigma\).

**Solution:**

\[ \left(\sqrt{\frac{(n - 1) \cdot s^2}{\chi_{\alpha / 2}^2(n-1)}}, \sqrt{\frac{(n - 1) \cdot s^2}{\chi_{1 - \alpha / 2}^2(n-1)}} \right) \]

- \(n = 6\),
- \(s = 0.8\),
- \(1 - \alpha = 0.90\), so \(\alpha/2 = 0.05\), and \(1 - \alpha/2 = 0.95\)
- \(\chi_{\alpha / 2}^2(n - 1) = \chi_{0.05}^2(5) = 11.07\).
- \(\chi_{1 - \alpha / 2}^2(n-1) = \chi_{0.95}^2(5) = 1.145\).

\[ \left( \sqrt{\frac{(6 - 1) \cdot 0.8^2}{11.07}} , \sqrt{\frac{(6 - 1) \cdot 0.8^2}{1.145}} \right) \]

\[ \boxed{(0.5377, 1.6718)} \]

```
n = 6
s = 0.8
alpha = 0.10
crit_lower = qchisq(alpha / 2, df = n - 1, lower.tail = FALSE)
crit_upper = qchisq(1 - alpha / 2, df = n - 1, lower.tail = FALSE)
lower = sqrt((n - 1) * s ^ 2 / crit_lower)
upper = sqrt((n - 1) * s ^ 2 / crit_upper)
c(lower, upper)
```

`## [1] 0.5376398 1.6714060`

**(e)** Construct a \(90\%\) confidence upper bound for the true standard deviation of the weight of a “22 ounce” Charles Mulligan’s porterhouse steak, \(\sigma\).

**Solution:**

\[ \left(0, \sqrt{\frac{(n - 1) \cdot s^2}{\chi_{1 - \alpha}^2(n-1)}} \right) \]

- \(n = 6\),
- \(s = 0.04\),
- \(1 - \alpha = 0.90\)
- \(\chi_{1 - \alpha}^2(n-1) = \chi_{0.90}^2(5) = 1.610\).

\[ \left( 0, \sqrt{\frac{(6 - 1) \cdot 0.8^2}{1.610}} \right) \]

\[ \boxed{(0, 1.4098)} \]

```
n = 6
s = 0.8
alpha = 0.10
crit_upper = qchisq(1 - alpha, df = n - 1, lower.tail = FALSE)
upper = sqrt((n - 1) * s ^ 2 / crit_upper)
c(0, upper)
```

`## [1] 0.00000 1.40968`

Last year, ballots in Champaign-Urbana contained the following question to assess public opinion on an issue:

“Should the State of Illinois legalize and regulate the sale and use of marijuana in a similar fashion as the State of Colorado?”

Suppose that we would like to understand Champaign-Urbana’s 2017 opinion on marijuana legalization. To satisfy our curiosity, we obtain a random sample of 120 Champaign-Urbanians and find that 87 support marijuana legalization.

**(a)** Construct a \(99\%\) confidence interval for \(p\), the true proportion of Champaign-Urbanians that support marijuana legalization.

**Solution:**

- \(x = 87\)
- \(n = 120\)
- \(\alpha = 0.01\), \(\alpha/2 = 0.005\)
- \(z_{\alpha / 2} = z_{0.005} = 2.576\)

\[ \hat{p} = \frac{x}{n} = \frac{87}{120} = 0.725 \]

\[ \hat{p} \pm z_{\alpha / 2} \sqrt{\frac{\hat{p} \cdot (1 - \hat{p})}{n}} \]

\[ 0.725 \pm 2.576 \sqrt{\frac{0.725 \cdot 0.275}{120}} \]

\[ \boxed{\bf 0.725 \pm 0.1050} \]

\[ \boxed{\bf (0.62, 0.83)} \]

**(b)** Suppose that a pollster wants to estimate the true proportion of Champaign-Urbanians that support marijuana legalization to within 0.04, with \(95\%\) confidence. How many Champaign-Urbanians should this pollster poll? Assume the pollster has no prior knowledge about the proportion.

**Solution:**

- \(\hat{p} = 0.50\), since we have no prior knowledge. (Worst case scenario.)
- \(\epsilon = 0.04\)
- \(\alpha = 0.05\), \(\alpha/2 = 0.025\)
- \(z_{\alpha / 2} = z_{0.025} = 1.960\)

\[ n = \left\lceil \left( \frac{z_{\alpha / 2} }{\epsilon} \right) ^ 2 \cdot \hat{p}(1 - \hat{p}) \right\rceil = \left\lceil \left( \frac{1.960}{0.04} \right) ^ 2 \cdot 0.50 \cdot 0.50 \right\rceil = \boxed{601} \]

**(c)** Now assume the pollster believes that support for legalization is somewhere between \(65\%\) and \(85\%\) and they would like to estimate the true proportion of Champaign-Urbanians that support marijuana legalization to within 0.04, with \(90\%\) confidence. How many Champaign-Urbanians should this pollster poll?

**Solution:**

- \(\tilde{p} = 0.65\), since it is closest to 0.50. (Worst case scenario.)
- \(\epsilon = 0.04\)
- \(\alpha = 0.10\), \(\alpha/2 = 0.05\)
- \(z_{\alpha / 2} = z_{0.05} = 1.645\)

\[ n = \left\lceil \left( \frac{z_{\alpha / 2} }{\epsilon} \right) ^ 2 \cdot \tilde{p}(1 - \tilde{p}) \right\rceil = \left\lceil \left( \frac{1.645}{0.04} \right) ^ 2 \cdot 0.65 \cdot 0.35 \right\rceil = \boxed{385} \]

Suppose students in a Statistics class are interested in the average score of an exam, but the instructor has only graded (a random sample of) 13 of the (many) exams. The instructor states that a \(90\%\) confidence interval for the true mean is given by \((79.14, 82.86)\) and that you can assume the grades follow a normal distribution.

Using only this information, calculate \(\bar{x}\), \(s\), and finally, a \(95\%\) confidence interval for \(\mu\), the true mean of the exam.

**Solution:**

First, since the interval is symmetric about the sample mean

\[ \bar{x} = \frac{79.14 + 82.86}{2} = 81 \]

We can also obtain the margin of error, which is half the length of the interval,

\[ \epsilon = \frac{82.86 - 79.14}{2} = 1.86 \]

The critical value of an \(90\%\) confidence interval with 13 observations is

\[ t_{0.05}(12) = 1.782 \]

Then solving

\[ 1.86 = 1.782 \cdot \frac{s}{\sqrt{13}} \]

we find that

\[ s = 3.7634. \]

The critical value of an \(95\%\) confidence interval with 13 observations is

\[ t_{0.025}(12) = 2.1788 \]

\[ \bar{x} \pm t_{\alpha / 2}(n - 1)\frac{s}{\sqrt{n}} \]

Then a \(95\%\) confidence interval is given by

\[ 81 \pm 2.1788 \frac{3.7634}{\sqrt{13}} \]

\[ \boxed{\bf 81 \pm 2.2742} \]

\[ \boxed{\bf (78.7258, 83.2742)} \]

Suppose that 10 students visit the Stars Hollow Apple Orchard and each pick (a random sample of) 15 Fuji apples, weigh them, then create a \(90\%\) confidence interval for the true mean weight of a Fuji apple at the Stars Hollow Apple Orchard. What is the probably that at most 2 of these intervals do not contain the true mean weight of a Fuji apple at the Stars Hollow Apple Orchard?

Since we haven’t actually seen any of the intervals, and assuming they are created with the correct procedure, each has a \(90\%\) of containing the true mean, thus they have a \(10\%\) chance to not contain the true mean.

Define \(X\) to be the number of intervals that do not contain the true mean. Then,

\[ X \sim \text{binom}(n = 10, p = 0.10) \]

So finally we want

\[ \begin{aligned} P(X \leq 2) &= P(X = 0) + P(X = 1) + P(X = 2) \\ &= 0.3486784 + 0.3874205 + 0.1937102 \\ &= \boxed{0.9298092} \end{aligned} \]

`dbinom(c(0, 1, 2), size = 10, prob = 0.10)`

`## [1] 0.3486784 0.3874205 0.1937102`

`sum(dbinom(c(0, 1, 2), size = 10, prob = 0.10))`

`## [1] 0.9298092`

`pbinom(2, size = 10, prob = 0.10)`

`## [1] 0.9298092`