Math / Data & Statistics

QuestionA hospital readmission is an episode when a patient who has been discharged from a hospital is readmitted again within a certain time period. Nationally the readmission rate for patients with pneumonia is $19 \%$ . A hospital was interested in knowing whether their readmission rate for preumonia was less than the national percentage. They found 9 patients out of 60 treated for pneumonia in a two-month period were readmitted. Complete parts (a) through (d) below. c. Find the value of the test statistic and explain it in context.
The test statistic is $\square$ (Type an integer or a decimal rounded to two decimal places as needed.) The value of the test statistic tells that the observed proportion of readmissions was $\square$ $\square$ $\square$ the null hypothesis proportion of readmissions. (Type an integer or a decimal rounded to two decimal places as needed.) d. The p-value associated with this test statistic is 0.21 . Explain the meaning of the $p$ -value in this context. Based on this result, does the p-value indicate the null hypothesis should be doubted? Select the correct choice below and fill in the answer box within your choice. (Type an integer or a decimal. Do not round.) A. The probability of getting 9 or fewer readmissions for pneumonia of a random sample of 60 patients with preumonia is $\square$ . assuming the population proportion is less than 0.19. B. The probability of getting 9 or fewer readmissions for pneumonia of a random sample of 60 patients with pneumonia is $\qquad$ , assuming the population proportion is 0.19.

Studdy Solution

STEP 1

What is this asking? We want to see if a hospital's pneumonia readmission rate is actually lower than the national average, and we've got some data to help us figure that out!
Specifically, we're looking for the test statistic and what it means, and then we'll dive into the p-value and what it tells us. Watch out! Don't mix up the hospital's readmission rate with the national average.
Also, remember that the p-value isn't the probability that the null hypothesis is true, it's the probability of observing our data (or more extreme data) if the null hypothesis were true.

STEP 2

1. Calculate the observed proportion.
2. Calculate the test statistic.
3. Interpret the test statistic.
4. Interpret the p-value.

STEP 3

First, let's define what we're looking at.
We're dealing with a hospital that wants to know if its readmission rate for pneumonia is less than the national rate of $0.19$ .
Out of $\textbf{60}$ patients treated for pneumonia, $\textbf{9}$ were readmitted.

STEP 4

To calculate the observed proportion of readmissions at this particular hospital, we'll divide the number of readmissions by the total number of pneumonia patients.
So, that's $\frac{9}{60}$ , which simplifies to $0.15$ .
This is our observed proportion, and it's what we'll compare to the national average.

STEP 5

Now, let's calculate the test statistic!
This tells us how far our observed proportion is from the national average, in terms of standard deviations.
The formula for the test statistic is: $z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}$ Where: $\hat{p}$ is the observed proportion ( $0.15$ in our case). $p_0$ is the null hypothesis proportion (the national average, $0.19$ ). $n$ is the sample size ( $60$ patients).

STEP 6

Let's plug in our values: $z = \frac{0.15 - 0.19}{\sqrt{\frac{0.19(1 - 0.19)}{60}}}$ $z = \frac{-0.04}{\sqrt{\frac{0.19 \cdot 0.81}{60}}}$ $z = \frac{-0.04}{\sqrt{\frac{0.1539}{60}}}$ $z = \frac{-0.04}{\sqrt{0.002565}}$ $z = \frac{-0.04}{0.0506}$ $z \approx -0.79$

STEP 7

Our test statistic is approximately $-0.79$ .
This means our observed proportion ( $0.15$ ) is about $\textbf{0.79}$ standard deviations below the national average of $0.19$ .

STEP 8

The p-value given is $0.21$ .
This means that if the true readmission rate at the hospital were actually $0.19$ (the national average), there's a \textbf{21%} chance of observing a readmission rate as low as $0.15$ (or even lower) in a sample of $60$ patients.

STEP 9

Since the p-value is relatively high (greater than common significance levels like $0.05$ ), we don't have enough evidence to reject the null hypothesis.
In other words, we can't confidently say the hospital's readmission rate is significantly lower than the national average.

STEP 10

The test statistic is $-0.79$ .
The value of the test statistic tells us that the observed proportion of readmissions was $0.79$ standard deviations below the null hypothesis proportion of readmissions.
The correct choice is B: The probability of getting 9 or fewer readmissions for pneumonia out of a random sample of 60 patients with pneumonia is $0.21$ , assuming the population proportion is $0.19$ .

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Studdy Solution

STEP 1

STEP 2

1. Calculate the observed proportion.
2. Calculate the test statistic.
3. Interpret the test statistic.
4. Interpret the p-value.

STEP 3

First, let's define what we're looking at.
We're dealing with a hospital that wants to know if its readmission rate for pneumonia is less than the national rate of $0.19$ .
Out of $\textbf{60}$ patients treated for pneumonia, $\textbf{9}$ were readmitted.

STEP 4

STEP 5

STEP 6

Let's plug in our values: $z = \frac{0.15 - 0.19}{\sqrt{\frac{0.19(1 - 0.19)}{60}}}$ $z = \frac{-0.04}{\sqrt{\frac{0.19 \cdot 0.81}{60}}}$ $z = \frac{-0.04}{\sqrt{\frac{0.1539}{60}}}$ $z = \frac{-0.04}{\sqrt{0.002565}}$ $z = \frac{-0.04}{0.0506}$ $z \approx -0.79$

STEP 7

Our test statistic is approximately $-0.79$ .
This means our observed proportion ( $0.15$ ) is about $\textbf{0.79}$ standard deviations below the national average of $0.19$ .

STEP 8

The p-value given is $0.21$ .
This means that if the true readmission rate at the hospital were actually $0.19$ (the national average), there's a \textbf{21%} chance of observing a readmission rate as low as $0.15$ (or even lower) in a sample of $60$ patients.

STEP 9

Since the p-value is relatively high (greater than common significance levels like $0.05$ ), we don't have enough evidence to reject the null hypothesis.
In other words, we can't confidently say the hospital's readmission rate is significantly lower than the national average.

STEP 10

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Studdy Solution

STEP 1

STEP 2

1. Calculate the observed proportion.2. Calculate the test statistic.3. Interpret the test statistic.4. Interpret the *p*-value.

STEP 3

First, let's **define** what we're looking at.We're dealing with a hospital that wants to know if its readmission rate for pneumonia is less than the national rate of 0.190.190.19.Out of 60\textbf{60}60 patients treated for pneumonia, 9\textbf{9}9 were readmitted.

STEP 4

STEP 5

STEP 6

STEP 7

Our test statistic is approximately −0.79-0.79−0.79.This means our observed proportion (0.150.150.15) is about 0.79\textbf{0.79}0.79 standard deviations *below* the national average of 0.190.190.19.

STEP 8

The *p*-value given is 0.210.210.21.This means that if the *true* readmission rate at the hospital were actually 0.190.190.19 (the national average), there's a \textbf{21%} chance of observing a readmission rate as low as 0.150.150.15 (or even lower) in a sample of 606060 patients.

STEP 9

Since the *p*-value is relatively high (greater than common significance levels like 0.050.050.05), we don't have enough evidence to reject the null hypothesis.In other words, we can't confidently say the hospital's readmission rate is *significantly* lower than the national average.

STEP 10

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1. Calculate the observed proportion.
2. Calculate the test statistic.
3. Interpret the test statistic.
4. Interpret the p-value.

First, let's define what we're looking at.
We're dealing with a hospital that wants to know if its readmission rate for pneumonia is less than the national rate of $0.19$ .
Out of $\textbf{60}$ patients treated for pneumonia, $\textbf{9}$ were readmitted.

Our test statistic is approximately $-0.79$ .
This means our observed proportion ( $0.15$ ) is about $\textbf{0.79}$ standard deviations below the national average of $0.19$ .

The p-value given is $0.21$ .
This means that if the true readmission rate at the hospital were actually $0.19$ (the national average), there's a \textbf{21%} chance of observing a readmission rate as low as $0.15$ (or even lower) in a sample of $60$ patients.

Since the p-value is relatively high (greater than common significance levels like $0.05$ ), we don't have enough evidence to reject the null hypothesis.
In other words, we can't confidently say the hospital's readmission rate is significantly lower than the national average.