Question

A survey of 150 elementary school girls found that 94 saw the latest Disney movie. Construct a 90% confidence interval for the proportion of girls who saw that movie. Can we claim that 70% saw it?

Answer 1

The 90% confidence interval for the proportion of girls who saw the latest Disney movie is approximatel
`y \( \left(0.872, 0.973\right) \).`Therefore, we cannot claim that 70% saw it, as this value is outside the confidence interval.

Step-by-step explanation:

In constructing the confidence interval, we use the formula for the confidence interval of a proportion:

Confidence Interval =
`\hat{p} \pm z * \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \]`

-
`\( \hat{p} \)` is the sample proportion (94/150 in this case),

- z is the z-score corresponding to the confidence level (1.645 for 90% confidence),

-n is the sample size (150 in this case).

Plugging in these values, we get the confidence interval. In this context, the interval
`\(\left(0.872, 0.973\right)\)` means we are 90% confident that the true proportion of girls who saw the movie is within this range.

To address whether 70% saw it, we note that 70% is outside the confidence interval. Therefore, based on this sample, there is not enough evidence to claim that 70% of elementary school girls saw the latest Disney movie.

This conclusion is subject to the assumptions of random sampling and independence. If these conditions are met, the interval provides a range of plausible values for the true proportion of girls who saw the movie.

Answer 2

The 90% confidence interval for the proportion of elementary school girls who saw the latest Disney movie is approximately [0.898, 0.987]. Given that this interval does not include 70%, we cannot claim with 90% confidence that 70% of the girls saw the movie.

Step-by-step explanation:

To construct a confidence interval for the proportion of girls who saw the latest Disney movie, we can use the formula:

`\[ \hat{p} \pm Z * \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \]`

Where:

`\(\hat{p}\)` is the sample proportion (94/150 = 0.6267),

n is the sample size (150),

Z is the Z-score corresponding to a 90% confidence level.

The Z-score for a 90% confidence level is approximately 1.645. Plugging in the values:

`\[ 0.6267 \pm 1.645 * \sqrt{(0.6267 * (1 - 0.6267))/(150)} \]`

This calculation yields the interval [0.898, 0.987].

The confidence interval represents the range within which we can be 90% confident the true population proportion lies. Since 70% falls outside this interval, we cannot claim, with 90% confidence, that 70% of the girls saw the movie.

It's important to note that the width of the confidence interval indicates the precision of our estimate. In this case, the narrow interval suggests a relatively precise estimate of the proportion of girls who watched the Disney movie.