Math / Data & Statistics

Question\begin{problem} Statistically, $20\%$ of children in California are diagnosed with asthma. A pediatrician believes this proportion is now significantly higher than it used to be. A recently published study reports that, of 357 randomly selected children in California, 80 have a diagnosis of asthma. Does the study support the pediatrician's belief that more than $20\%$ of children in California are now diagnosed with asthma? Use $\alpha=0.01$ . Answer questions $1-8$ .
\begin{enumerate} \item The population of interest is: All children in California with asthma. \item The sample used is: 80 children diagnosed with asthma. \item The variable of interest is: 357 randomly selected children. \item This variable is (circle one): \textbf{Numerical} \quad \textbf{Categorical} \item STEP I: State the null and alternative hypotheses using proper notation. \begin{array}{l} H_{0}= \\ H_{a}= \end{array} \] \item STEP 2: Report the significance level and check the conditions for a valid test. \item STEP 3: Compute the test statistic and p$-value (report the calculator option and inputs). \item STEP 4: Make the decision and write the conclusion (use the class template). \end{enumerate} \end{problem}

Studdy Solution

STEP 1

What is this asking? Is the rate of asthma in Californian children actually higher than 20%, or did a recent study just happen to find a lot of asthmatic kids by chance? Watch out! Don't mix up the sample and the population!
Also, remember that we're looking for evidence of more than 20%, not just a different rate.

STEP 2

1. Set up the hypotheses
2. Check conditions
3. Calculate the test statistic and p-value
4. Make a decision

STEP 3

The null hypothesis ( $H_0$ ) is what we're trying to disprove.
It says "Nope, nothing's changed, the asthma rate is still 20%." In math terms: $H_0: p = 0.20$ , where $p$ is the true proportion of asthmatic children in California.

STEP 4

The alternative hypothesis ( $H_a$ ) is what the pediatrician believes: the asthma rate is greater than 20%.
So, $H_a: p > 0.20$ .

STEP 5

The problem says the 357 kids were randomly selected, so we're good there!
This is important because it lets us generalize our findings to the whole population.

STEP 6

We need to check if our sample is large enough for the math to work out.
We need $n \cdot p_0 \ge 10$ and $n \cdot (1 - p_0) \ge 10$ .
Here, $n = 357$ (our sample size) and $p_0 = 0.20$ (the hypothesized proportion).
Let's check: $357 \cdot 0.20 = 71.4 \ge 10$ $357 \cdot (1 - 0.20) = 357 \cdot 0.80 = 285.6 \ge 10$ Both conditions are met, so we're good to go!

STEP 7

In our sample, 80 out of 357 kids had asthma.
The sample proportion ( $\hat{p}$ ) is: $\hat{p} = \frac{80}{357} \approx 0.224$

STEP 8

The test statistic (a z-score) tells us how far our sample result is from what we'd expect if the null hypothesis were true.
The formula is: $z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$ Plugging in our values: $z = \frac{0.224 - 0.20}{\sqrt{\frac{0.20(1-0.20)}{357}}} = \frac{0.024}{\sqrt{\frac{0.20 \cdot 0.80}{357}}} = \frac{0.024}{\sqrt{\frac{0.16}{357}}} \approx \frac{0.024}{0.021} \approx 1.14$

STEP 9

The p-value is the probability of getting a sample as extreme as ours (or more extreme) if the null hypothesis were actually true.
Since our alternative hypothesis is $p > 0.20$ (a right-tailed test), we want the area to the right of our z-score.
Using a calculator or a z-table, we find that the p-value is approximately 0.127.

STEP 10

Our p-value (0.127) is greater than our significance level ( $\alpha = 0.01$ ).

STEP 11

Since the p-value is greater than $\alpha$ , we fail to reject the null hypothesis.

STEP 12

There is not enough evidence to conclude that the proportion of children in California diagnosed with asthma is greater than 20%.

STEP 13

1. All children in California.
2. 357 randomly selected children in California.
3. Asthma diagnosis (categorical).
4. Categorical
5. $H_0: p = 0.20$ , $H_a: p > 0.20$
6. $\alpha = 0.01$ , Conditions met (random sample, large enough sample size).
7. $z \approx 1.14$ , $p \approx 0.127$ (using 1-PropZTest on a calculator with $p_0 = 0.20$ , $x = 80$ , $n = 357$ , and a right-tailed test).
8. Fail to reject $H_0$ .
There's not enough evidence to say the asthma rate is higher than 20%.

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Studdy Solution

STEP 1

STEP 2

1. Set up the hypotheses
2. Check conditions
3. Calculate the test statistic and p-value
4. Make a decision

STEP 3

The null hypothesis ( $H_0$ ) is what we're trying to disprove.
It says "Nope, nothing's changed, the asthma rate is still 20%." In math terms: $H_0: p = 0.20$ , where $p$ is the true proportion of asthmatic children in California.

STEP 4

The alternative hypothesis ( $H_a$ ) is what the pediatrician believes: the asthma rate is greater than 20%.
So, $H_a: p > 0.20$ .

STEP 5

The problem says the 357 kids were randomly selected, so we're good there!
This is important because it lets us generalize our findings to the whole population.

STEP 6

STEP 7

In our sample, 80 out of 357 kids had asthma.
The sample proportion ( $\hat{p}$ ) is: $\hat{p} = \frac{80}{357} \approx 0.224$

STEP 8

STEP 9

STEP 10

Our p-value (0.127) is greater than our significance level ( $\alpha = 0.01$ ).

STEP 11

Since the p-value is greater than $\alpha$ , we fail to reject the null hypothesis.

STEP 12

There is not enough evidence to conclude that the proportion of children in California diagnosed with asthma is greater than 20%.

STEP 13

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Studdy Solution

STEP 1

STEP 2

1. Set up the hypotheses2. Check conditions3. Calculate the test statistic and p-value4. Make a decision

STEP 3

The **null hypothesis** (H0H_0H0​) is what we're trying to disprove.It says "Nope, nothing's changed, the asthma rate is still 20%." In math terms: H0:p=0.20H_0: p = 0.20H0​:p=0.20, where ppp is the *true* proportion of asthmatic children in California.

STEP 4

The **alternative hypothesis** (HaH_aHa​) is what the pediatrician believes: the asthma rate is *greater* than 20%.So, Ha:p>0.20H_a: p > 0.20Ha​:p>0.20.

STEP 5

The problem says the 357 kids were randomly selected, so we're good there!This is important because it lets us generalize our findings to the whole population.

STEP 6

STEP 7

In our sample, 80 out of 357 kids had asthma.The **sample proportion** (p^\hat{p}p^​) is: p^=80357≈0.224 \hat{p} = \frac{80}{357} \approx 0.224 p^​=35780​≈0.224

STEP 8

STEP 9

STEP 10

Our **p-value (0.127)** is *greater* than our **significance level (α=0.01\alpha = 0.01α=0.01)**.

STEP 11

Since the p-value is greater than α\alphaα, we *fail to reject* the null hypothesis.

STEP 12

There is not enough evidence to conclude that the proportion of children in California diagnosed with asthma is greater than 20%.

STEP 13

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1. Set up the hypotheses
2. Check conditions
3. Calculate the test statistic and p-value
4. Make a decision

The null hypothesis ( $H_0$ ) is what we're trying to disprove.
It says "Nope, nothing's changed, the asthma rate is still 20%." In math terms: $H_0: p = 0.20$ , where $p$ is the true proportion of asthmatic children in California.

The alternative hypothesis ( $H_a$ ) is what the pediatrician believes: the asthma rate is greater than 20%.
So, $H_a: p > 0.20$ .

The problem says the 357 kids were randomly selected, so we're good there!
This is important because it lets us generalize our findings to the whole population.

In our sample, 80 out of 357 kids had asthma.
The sample proportion ( $\hat{p}$ ) is: $\hat{p} = \frac{80}{357} \approx 0.224$

Our p-value (0.127) is greater than our significance level ( $\alpha = 0.01$ ).

Since the p-value is greater than $\alpha$ , we fail to reject the null hypothesis.