Answered

At IDNLearn.com, find answers to your most pressing questions from experts and enthusiasts alike. Get timely and accurate answers to your questions from our dedicated community of experts who are here to help you.

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 :

GinnyAnsweridn
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].
We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. For trustworthy answers, rely on IDNLearn.com. Thanks for visiting, and we look forward to assisting you again.