`[1] -3.333333`

STAT 120

Bastola

You want to…

… compare the proportion of students who use a Windows-based PC to the proportion who use a Mac.

Inference for one proportion

Inference for two proportions

You want to…

… compare the proportion of students who study abroad between those attending public universities and those at private universities.

Inference for one proportion

Inference for two proportions

You want to…

… compare the proportion of in-state students at a university to the proportion from outside the state.

Inference for one proportion

Inference for two proportions

The correct answer is a. This is one categorical variable: in-state or out-of-state

You want to…

… compare the proportion of in-state students who get financial aid to the proportion of out-of-state students who get financial aid

This is…

Inference for one proportion

Inference for two proportions

Are metal tags detrimental to penguins? A study looked at the 10 year survival rate of penguins tagged either with a metal tag or an electronic tag. 20% of the 167 metal tagged penguins survived, compared to 36% of the 189 electronic tagged penguins.

Is there a statistically discernible difference in survival rates? \[\begin{align*} H_0: p_M=p_E \quad H_A: p_M \neq p_E \end{align*}\] \(p=\) true survival rate

Source: Saraux, et. al. (2011). “Reliability of flipper-banded penguins as indicators of climate change,” Nature, 469, 203-206.

20% of the 167 metal tagged penguins survived, compared to 36% of the 189 electronic tagged penguins.

Survived | Died | Total | |
---|---|---|---|

Metal Tag | 33 | 134 | 167 |

Electronic Tag | 68 | 121 | 189 |

Total | 101 | 255 | 356 |

Are the conditions met for using the normal distribution for inference?

a). Yes

b). No

**Pooling is done to combine the data from both groups and come up with a combined proportion, assuming the groups are the same**

We don’t know \(\mathrm{p}_{\mathrm{M}}\) or \(\mathrm{p}_{\mathrm{E}}\), so how do we compute the SE for our hypothesis test?

- Assume the two proportions are equal and use .b[one proportion for both groups.
- Our best guess of this one proportion comes from combining data from both groups and computing the overall proportion, called the pooled proportion p.
- Hint: the pooled proportion will always be somewhere in between the two sample proportions.

If the distribution of the sample statistic is normal: A confidence interval can be calculated by \[\begin{align*} \text { sample statistic } \pm z^* \times S E \end{align*}\] where \(z^*\) is a \(N(0,1)\) percentile depending on the level of confidence. A p-value is the area in the tail(s) of a \(N(0,1)\) beyond \[\begin{align*} z=\frac{\text { sample statistic }- \text { null value }}{\text { SE }} \end{align*}\]

\[\begin{align*} H_0: p_1&=p_2\\ H_a: p_1&\neq p_2 \end{align*}\]

\[\begin{align*} \hat{p}_{pooled}=\frac{33+68}{167+189}=0.2837 \end{align*}\]

\[z=\frac{\left(\hat{p}_1-\hat{p}_2\right)-0}{\sqrt{\frac{\hat{p}_{pooled}(1-\hat{p}_{pooled})}{n_1}+\frac{\hat{p}_{pooled}(1-\hat{p}_{pooled})}{n_2}}}\]

If observed counts in the two-way table are at least 10, then the \(p\)-value can be computed as the area in the tail(s) of a standard normal beyond \(z\). .out-t[Always use pooled proportion for the SE.

20% of the 167 metal tagged penguins survived, compared to 36% of the 189 electronic tagged penguins.

33 survived with metal tags and 68 with electronic The pooled proportion is: \[\begin{align*} \hat{p}_{pooled}=\frac{33+68}{167+189}=0.2837 \end{align*}\]

SE for our test: \[\begin{align*} S E=\sqrt{\frac{0.284(1-.284)}{167}+\frac{0.284(1-.284)}{189}}=0.048 \end{align*}\]

20% of the 167 metal tagged penguins survived, compared to 36% of the 189 electronic tagged penguins. The pooled SE is 0.048.

\[\begin{align*} \text { Standardized test stat: } \quad z=\frac{(0.2-0.36)-0}{0.048}=-3.34 \end{align*}\]

Reject the null

A difference in survival rates as extreme, or more extreme, than \(16 \%\) would occur by chance only about \(0.08 \%\) of the time. There is a statistically discernible difference \((\mathrm{z}=-3.34, \mathrm{p}=0.0008)\)

How much do the rates differ? - Compute a \(95 \%\) CI for the difference…

How do we compute the SE? - We can’t use the pooled version since we’ve concluded the proportions differ!

For large enough \(\mathrm{n}_1\) and \(\mathrm{n}_2\) : \(\quad\) statistic \(\pm \quad z^* \times S E\)

\[\left(\hat{p}_1-\hat{p}_2\right) \pm z^* \sqrt{\frac{\hat{p}_1\left(1-\hat{p}_1\right)}{n_1}+\frac{\hat{p}_2\left(1-\hat{p}_2\right)}{n_2}}\]

\(20 \%\) of the 167 metal tagged penguins survived, compared to \(36 \%\) of the 189 electronic tagged penguins. Give a \(90 \%\) confidence interval for the difference in proportions (metal - electronic).

What is \(\mathrm{z}^*\) for the confidence interval?

1.280

1.645

1.960

2.575

0.90

\(20 \%\) of the 167 metal tagged penguins survived, compared to \(36 \%\) of the 189 electronic tagged penguins. Give a \(90 \%\) confidence interval for the difference in proportions (metal - electronic).

\(90 \% C I\) for \(\mathrm{p}_M-\mathrm{p}_E:\)

\[\begin{align*} (0.2-0.36) &\pm 1.645 \cdot \sqrt{\frac{0.2(1-0.2)}{167}+\frac{0.36(1-0.36)}{189}}\\ &=-0.16 \pm 1.645 \times 0.047\\ &=(-0.237,-0.09) \end{align*}\]

We are \(90\%\) confident that the survival rate is between \(9%\) and \(23.7%\) lower for metal tagged penguins, as opposed to electronically tagged.

\(20 \%\) of the 167 metal tagged penguins survived, compared to \(36 \%\) of the 189 electronic tagged penguins. Give a \(95 \%\) confidence interval for the difference in proportions (metal - electronic).

What is \(\mathrm{z}^*\) for the confidence interval?

1.280

1.645

1.960

2.575

0.90

\(20 \%\) of the 167 metal tagged penguins survived, compared to \(36 \%\) of the 189 electronic tagged penguins. Give a \(95 \%\) confidence interval for the difference in proportions (metal - electronic).

\(95 \% C I\) for \(\mathrm{p}_M-\mathrm{p}_E:\)

\[\begin{align*} (0.20-0.36) &\pm 1.96 \sqrt{\frac{0.20 \times 0.80}{167}+\frac{0.36 \times 0.64}{189}}\\ &=-0.16 \pm 1.96 \times 0.047\\ &=(-0.251,-0.069) \end{align*}\]

We are \(95 \%\) confident that between \(6.9 \%\) to \(25.1 \%\) fewer penguins survive when metal tags are used compared to electronic tags.

- Please download the Class-Activity-18 template from moodle and go to class helper web page

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