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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.

Recall that

\[
\bar{x} = 21.4 \quad \text{and} \quad s^2 = 0.64.
\] **(a)** Calculate the test statistic for testing \(H_0: \ \mu = 22\), where \(\mu\) is the true weight of a “22 ounce” Charles Mulligan’s porterhouse. Report your answer rounded to three decimal places.

**(b)** Calculate the p-value for the test \(H_0: \ \mu = 22\) versus \(H_1: \ \mu < 22\). Report your decision when \(\alpha = 0.05\).

Some useful code:

- Excel:
`T.DIST(x, df, 1)`

gives the area to the**left**of`x`

. - Excel:
`TDIST(x, df, 1)`

gives the area to the**right**of`x`

. (An older command. Only accepts positive values for`x`

.) `R`

:`pt(x, df)`

gives the area to the**left**of`x`

.

**(c)** Calculate the test statistic for testing \(H_0: \ \sigma \leq 0.50\), where \(\sigma\) is the true standard deviation of the weight of a “22 ounce” Charles Mulligan’s porterhouse.

**(d)** Calculate the p-value for the test \(H_0: \ \sigma \leq 0.50\) versus \(H_1: \ \sigma > 0.50\). Report your decision when \(\alpha = 0.01\).

Some useful code:

- Excel:
`CHISQ.DIST(x, df, 1)`

gives the area to the**left**of`x`

. - Excel:
`CHIDIST(x, df, 1)`

gives the area to the**right**of`x`

. (An older command.) `R`

:`pchisq(x, df)`

gives the area to the**left**of`x`

.

**(e)** Suppose the true value of \(\sigma\) is indeed 0.50. Was your decision in part **(d)** correct? If not, what Type of error was made? Justify your answer.

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)** Calculate the \(z\) test statistic for the test \(H_0: \ p = 0.70\) versus \(H_1: \ p > 0.70\) where \(p\) is the true proportion of Champaign-Urbanians that support marijuana legalization. Report your answer rounded to two decimal places.

**(b)** Calculate the (approximate, using \(z\)) p-value for the test \(H_0: \ p = 0.70\) versus \(H_1: \ p > 0.70\).

**(c)** Calculate the **exact** p-value for the test \(H_0: \ p = 0.70\) versus \(H_1: \ p > 0.70\). That is, calculate the probability of seeing as many supporters as observed, or more, assuming the null hypothesis is true.

Some useful code:

- Excel:
`BINOM.DIST(x, n, p, 1)`

gives the probability of less than or equal to`x`

. `R`

:`pbinom(x, n, p)`

gives the probability of less than or equal to`x`

.

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 we obtain a random sample of 80 Champaign voters, of which 55 support marijuana legalization. We also obtain a random sample of 100 Urbana voters, of which 75 support marijuana legalization. Let \(p_C\) be the true proportion of Champaign voters who support marijuana legalization and let \(p_U\) be the true proportion of Urbana voters who support marijuana legalization.

**(a)** Calculate a 99% confidence interval for \(p_U - p_C\).

**(b)** Calculate the p-value for the test \(H_0: \ p_U = p_C\) versus \(H_1: \ p_U \neq p_C\).