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A resident of Seattle, Washington, claims that it rains there nearly every day, meaning [tex]99\%[/tex] of the days have at least some rain. To investigate this claim, a skeptic selects a random sample of 50 days and finds that it rained on 46 of the 50 days. Is this convincing evidence that it rains less often than "every day" in Seattle?

[tex]\[
\begin{array}{l}
H_0: p = 0.99 \\
H_a: p \ \textless \ 0.99
\end{array}
\][/tex]

where [tex]p[/tex] is the true proportion of days that it rains in Seattle, Washington.

The [tex]P[/tex]-value of this test is approximately 0. What decision should be made at the [tex]\alpha = 0.05[/tex] level?

A. Because [tex]0 \ \textless \ 0.05[/tex], we reject [tex]H_0[/tex].

B. Because [tex]0 \ \textless \ 0.05[/tex], we fail to reject [tex]H_0[/tex].

C. Because [tex]0 \ \textless \ 0.05[/tex], we reject [tex]H_a[/tex].

D. Because [tex]0 \ \textless \ 0.05[/tex], we fail to reject [tex]H_a[/tex].

Sagot :

GinnyAnswer
To determine whether there is convincing evidence that it rains less often than "every day" (99% of the days) in Seattle, we perform a hypothesis test for the proportion of days that it rains.

### Step-by-Step Solution

#### Step 1: State the Hypotheses
We need to set up our null and alternative hypotheses:
- [tex]\( H_0: p = 0.99 \)[/tex] (The true proportion of days it rains is 99%)
- [tex]\( H_a: p < 0.99 \)[/tex] (The true proportion of days it rains is less than 99%)

#### Step 2: Collect Data and Compute the Sample Proportion
In a sample of 50 days, it rained on 46 days. Therefore, the sample proportion [tex]\( \hat{p} \)[/tex] is:
[tex]\[ \hat{p} = \frac{46}{50} = 0.92 \][/tex]

#### Step 3: Calculate the Standard Error
The standard error (SE) of the sample proportion under the null hypothesis is calculated as:
[tex]\[ \text{SE} = \sqrt{\frac{p_0(1 - p_0)}{n}} \][/tex]
Plugging in the values:
[tex]\[ \text{SE} = \sqrt{\frac{0.99 \cdot (1 - 0.99)}{50}} = 0.014071247279470294 \][/tex]

#### Step 4: Compute the Test Statistic (Z-score)
The Z-score for the sample proportion is calculated as:
[tex]\[ Z = \frac{\hat{p} - p_0}{\text{SE}} \][/tex]
Substituting the values:
[tex]\[ Z = \frac{0.92 - 0.99}{0.014071247279470294} = -4.974683381630904 \][/tex]

#### Step 5: Obtain the P-value
The P-value is the probability of observing a test statistic as extreme as, or more extreme than, the one observed under the null hypothesis. For a left-tailed test, the P-value is:
[tex]\[ \text{P-value} = P(Z \leq -4.974683381630904) = 3.2677185178473833 \times 10^{-7} \][/tex]

#### Step 6: Make the Decision
To determine whether to reject the null hypothesis, we compare the P-value to the significance level [tex]\( \alpha = 0.05 \)[/tex].

Given:
[tex]\[ \text{P-value} = 3.2677185178473833 \times 10^{-7} \][/tex]
[tex]\[ \alpha = 0.05 \][/tex]

Since the P-value is much smaller than [tex]\( \alpha \)[/tex] (0.05), we reject the null hypothesis.

#### Conclusion
Because [tex]\( 0 < 0.05 \)[/tex], we reject [tex]\( H_0 \)[/tex]. This means we have convincing evidence at the [tex]\( \alpha = 0.05 \)[/tex] level to conclude that it rains less often than "every day" (99% of the days) in Seattle.

Therefore, the decision is:
Because [tex]\( 0 < 0.05 \)[/tex], we reject [tex]\( H_0 \)[/tex].
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