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Math  /  Data & Statistics

Question✓3\checkmark 3 14 ✓5\checkmark 5 6 ✓7\checkmark 7 8 9 10 =11=11 12 13 14
WOW: In the computer game World of Warcraft, some of the strikes are critical strikes, which do more damage. Assume that the probability of a critical strike is the same for every attack, and that attacks are independent. During a particular fight, a character has 241 critical strikes out of 596 attacks.
Part: 0/30 / 3 □\square
Part 1 of 3 (a) Construct an 80%80 \% confidence interval for the proportion of strikes that are critical strikes. Round the answer to at least three decimal places.
An 80%80 \% confidence interval for the proportion of strikes that are critical strikes is □\square <p<<p< □\square .

Studdy Solution

STEP 1

What is this asking? We need to figure out a range, with 80% confidence, for the real percentage of critical strikes, given that 241 out of 596 attacks were critical. Watch out! Don't mix up the sample proportion with the true population proportion.
We're estimating the *true* proportion using the *sample*!

STEP 2

1. Calculate the sample proportion.
2. Find the critical z-value.
3. Calculate the margin of error.
4. Construct the confidence interval.

STEP 3

Let's **dive in**!
We're given that there were **241** critical strikes out of a total of **596** attacks.
The *sample proportion* (p^\hat{p}) is simply the number of critical strikes divided by the total number of attacks.

STEP 4

So, we **calculate** p^\hat{p} as: p^=Number of critical strikesTotal number of attacks=241596≈0.40436 \hat{p} = \frac{\text{Number of critical strikes}}{\text{Total number of attacks}} = \frac{241}{596} \approx 0.40436 We'll keep more decimal places for now to avoid rounding errors later, but remember the problem asked us to round to at least three decimal places in the final answer.

STEP 5

Since we want an **80%** confidence interval, this means there's 100%−80%2=10%\frac{100\% - 80\%}{2} = 10\% of area in each tail of the normal distribution.
We need to find the *z-score* (zα/2z_{\alpha/2}) that corresponds to the area to the *left* of it being 1−10%=90%1 - 10\% = 90\% or 0.900.90.

STEP 6

Looking up a *z-table* or using a calculator, we find that the *z-score* for 0.900.90 is approximately **1.28**.
So, zα/2≈1.28z_{\alpha/2} \approx 1.28.

STEP 7

The *margin of error* (ME) tells us how much our sample proportion might differ from the true population proportion.
The formula for the margin of error is: ME=zα/2⋅p^(1−p^)n ME = z_{\alpha/2} \cdot \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} Where nn is the sample size.

STEP 8

Let's **plug in** our values: zα/2≈1.28z_{\alpha/2} \approx 1.28, p^≈0.40436\hat{p} \approx 0.40436, and n=596n = 596. ME=1.28⋅0.40436⋅(1−0.40436)596 ME = 1.28 \cdot \sqrt{\frac{0.40436 \cdot (1 - 0.40436)}{596}} ME≈1.28⋅0.40436⋅0.59564596 ME \approx 1.28 \cdot \sqrt{\frac{0.40436 \cdot 0.59564}{596}} ME≈1.28⋅0.24088596 ME \approx 1.28 \cdot \sqrt{\frac{0.24088}{596}} ME≈1.28⋅0.00040416 ME \approx 1.28 \cdot \sqrt{0.00040416} ME≈1.28⋅0.02010 ME \approx 1.28 \cdot 0.02010 ME≈0.02573 ME \approx 0.02573

STEP 9

Finally, we can **build** our confidence interval!
The formula is: p^−ME<p<p^+ME \hat{p} - ME < p < \hat{p} + ME

STEP 10

Substituting our values: 0.40436−0.02573<p<0.40436+0.02573 0.40436 - 0.02573 < p < 0.40436 + 0.02573 0.37863<p<0.43009 0.37863 < p < 0.43009 Rounding to three decimal places, we get: 0.379<p<0.430 0.379 < p < 0.430

STEP 11

An 80% confidence interval for the proportion of strikes that are critical strikes is 0.379<p<0.4300.379 < p < 0.430.

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