Suppose that a researcher is interested in the effect of caffeine on typing speed. A group of nine individuals are administered a typing test. The following day, they repeat the typing test, this time after taking 400 mg of caffeine. (Note: This is not recommended.) The data gathered, measured in words per minute, is

```
decaf = c(98, 124, 107, 105, 80, 43, 73, 68, 69)
caff = c(104, 128, 110, 108, 86, 53, 72, 73, 72)
```

decaf | caff |
---|---|

98 | 104 |

124 | 128 |

107 | 110 |

105 | 108 |

80 | 86 |

43 | 53 |

73 | 72 |

68 | 73 |

69 | 72 |

Note that these are paired observations.

Use the **sign test** with a significance level of 0.05 to assess whether or not caffeine has an effect on typing speed. That is, test

\[ H_0\colon \ m_D = m_C - m_N = 0 \quad \text{vs} \quad H_A\colon \ m_D = m_C - m_N \neq 0 \]

where

- \(m_C\) is the median typing speed in words per minute of individuals using caffeine
- \(m_N\) is the median typing speed in words per minute of individuals not using caffeine

Since it is possible that the caffeine makes typing speed worse, use a two-sided test.

You may use the following probabilities calculated in `R`

.

`round(dbinom(x = 0:9, size = 9, prob = 0.5), 3)`

`## [1] 0.002 0.018 0.070 0.164 0.246 0.246 0.164 0.070 0.018 0.002`

Report:

- The
**p-value**of the test - A
**decision**when \(\alpha = 0.05\).

`# use this chunk to complete any necessary calculations in R`

**P-Value:**Your p-value here.**Decision:**Your decision here.

Does meditation have an effect on blood pressure. A group of six college aged individuals were given a routine physical examination including a measurement of their systolic blood pressure. (Measured in millimeters of mercury.) A week after their physicals, the same six individuals returned for a guided meditation session. Immediately afterwords there (systolic) blood pressure was measured. The data gathered is

```
physical = c(125, 108, 185, 135, 112, 133)
meditation = c(120, 114, 160, 131, 124, 125)
```

physical | meditation |
---|---|

125 | 120 |

108 | 114 |

185 | 160 |

135 | 131 |

112 | 124 |

133 | 125 |

Note that these are paired observations.

Use the **sign test** with a significance level of 0.10 to assess whether or not meditation has an effect on blood pressure. That is, test

\[ H_0\colon \ m_D = m_M - m_P = 0 \quad \text{vs} \quad H_A\colon \ m_D = m_M - m_P \neq 0 \]

where

- \(m_P\) is the median systolic blood pressure in millimeters of mercury measured without meditation
- \(m_M\) is the median systolic blood pressure in millimeters of mercury measured with meditation

Since it is possible that the meditation makes blood pressure worse, use a two-sided test.

You may use the following probabilities calculated in `R`

.

`round(dbinom(x = 0:6, size = 6, prob = 0.5), 3)`

`## [1] 0.016 0.094 0.234 0.312 0.234 0.094 0.016`

Report:

- The
**p-value**of the test - A
**decision**when \(\alpha = 0.10\).

`# use this chunk to complete any necessary calculations in R`

**P-Value:**Your p-value here.**Decision:**Your decision here.

Return to the sleep data in Exercise 2. This time test

- \(H_0\): The distribution of systolic blood pressure is
**the same**with and without meditation - \(H_A\): The distribution of systolic blood pressure is
**different**with and without meditation

To do so, use a **permutation test** that permutes the *statistic*

\[ \bar{x}_D \]

where \(\bar{x}_D\) is the sample mean difference. Assume that the distribution of blood pressure with and without meditation has the same shape, but may have different locations. Use at least 10000 permutations.

```
physical = c(125, 108, 185, 135, 112, 133)
meditation = c(120, 114, 160, 131, 124, 125)
```

- Create a histogram that illustrates the distribution of the statistic used.
- Report the p-value of the test.

```
# use this chunk to complete any necessary permutation calculations
# also calculate statistic on observed data
```

`# use this chunk to create the histogram`

`# use this chunk to calculate the p-value of the test`

Which profession pays more? Data Scientist of Actuary? A (far too small) survey of junior (less than three years experience) data scientist and actuaries resulted in the following data:

```
data_sci = c(88000, 121000, 91000, 50000, 78000, 95000)
actuary = c(63000, 75000, 81000, 75000, 85000)
```

Use a **permutation test** that permutes the *statistic*

\[ t = \frac{(\bar{x} - \bar{y}) - 0}{s_p\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} \]

to test

- \(H_0\): The distribution of salaries is
**the same**for junior data scientists and actuaries - \(H_A\): The distribution of salaries is
**different**for junior data scientists and actuaries

Assume that the distribution of salaries for both has the same shape, but may have different locations. Use at least 10000 permutations.

- Create a histogram that illustrates the distribution of the statistic used.
- Report the p-value of the test.

```
# use this chunk to complete any necessary permutation calculations
# also calculate statistic on observed data
```

`# use this chunk to create the histogram`

`# use this chunk to calculate the p-value of the test`

Repeat exercise 3, but use an appropriate test available in the `R`

function `wilcox.test()`

.

Report:

- The
**p-value**of the test - A
**decision**when \(\alpha = 0.05\).

`# use this chunk to complete any necessary calculations in R`

**P-Value:**Your p-value here.**Decision:**Your decision here.