# Practice Problems 17

## Problem 1: Movie Goers are More Likely to Watch at Home

In a random sample of 500 movie goers in January 2013, 320 of them said they are more likely to wait and watch a new movie in the comfort of their own home. Compute and interpret a 95% confidence interval for the proportion of movie goers who are more likely to watch a new movie from home.

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*Answer:* We see that \(\hat{p}=\frac{320}{500}=0.640\) (keep at least 3 decimal spots to ensure accuracy in your SE calculation!) The confidence interval is given by:

\[\text { Statistic }\pm z^{*} S E\]

\[\begin{array}{l} \hat{p} \pm z^{*} \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \\ 0.64 \pm 1.96 \cdot \sqrt{\frac{0.64(1-0.64)}{500}} \\ 0.64 \pm 0.042\\ (0.598, 0.682) \end{array}\]

(Make sure to use proportions in your CI, then convert to % at the end if you prefer a percentage interpretation.) We are 95% sure that the proportion of all movie goers who are more likely to wait and watch a new movie at home is between 0.598 and 0.682.## Problem 2: Sample Size and Margin of Error for Movie Goers

- What sample size is needed in example 2 if we want a margin of error within ±2%? (Use the sample proportion from the original sample.)

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*Answer:*

\[\begin{array}{l} 0.02=z^{*} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \\ n=\left(\frac{z^{*}}{0.02}\right)^{2} \hat{p}(1-\hat{p})\\\quad =\left(\frac{1.96}{0.02}\right)^{2} 0.64(1-0.64)=2212.76 \end{array}\]

We need a sample size of at least n = 2,213 to have a margin of error this small. This is substantially more than the sample size of 500 used in the actual survey.- What sample size is needed if we want a margin of error within ±2%, and if we use the conservative estimate of p = 0.5?

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*Answer:*

\[n=\left(\frac{1.96}{0.02}\right)^{2} 0.5(1-0.5)=2401\]

We need a sample size of at least n = 2,401 to have a margin of error this small. Notice that if we have less knowledge of the actual proportion, we need a larger sample size to arrive at the same margin of error.## Problem 3: Mendel’s green peas?

One of Gregor Mendel’s famous genetic experiments dealt with raising pea plants. According to Mendel’s genetic theory, under a certain set of conditions the proportion of pea plants that produce smooth green peas should be p=3/16 (0.1875). A sample of n=556 plants from the experiment had 108 with smooth green peas. Does this provide evidence of a problem with Mendel’s theory and that the proportion is different from 3/16? Show all details of the test.

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*Answer:* We are testing \(H_{0}: p=0.1875\) vs \(H_{a}: p \neq 0.1875\) where p represents the proportion of pea plans with smooth green peas. The sample proportion is \(\hat{p}=\frac{108}{556}=0.1942\) and the sample size is \(n=556\). The test statistic is:

\[z=\frac{\text { Statistic }-\text { Null }}{S E}=\frac{\hat{p}-p_{0}}{\sqrt{\frac{p_{0}\left(1-p_{0}\right)}{n}}}=\frac{0.1942-0.1875}{\sqrt{\frac{0.1875(1-0.1875)}{556}}}=0.405\]

This is a two-tail test, and we see that the area to the right of 0.405 in a normal distribution is 0.343 `(1-pnorm(0.405))`

, so the p-value is 2(0.343) = 0.686.The R command is: `2*(1-pnorm(0.405))`