Math / Data & Statistics

Question6. A medical researcher claims that $5 \%$ of children under 18 years of age have asthma. In a random sample of 250 children under 18 years of age, $9.6 \%$ say they have asthma. At $\alpha=0.01$ , is there enough evidence to reject the researcher's claim?

Studdy Solution

STEP 1

What is this asking? Is the asthma rate in kids really higher than the researcher's claim of $5\%$ ? Watch out! Don't mix up sample proportion with population proportion!

STEP 2

1. Define the hypotheses
2. Calculate the test statistic
3. Determine the critical value
4. Make a decision

STEP 3

Alright, let's set up our hypotheses!
The null hypothesis $H_0$ is that the proportion of children with asthma is $p = 0.05$ .
The alternative hypothesis $H_a$ is that the proportion is greater than $0.05$ .
So, we have:
$H_0: p = 0.05$ $H_a: p > 0.05$

STEP 4

Calculate the sample proportion first.
We have $9.6\%$ of $250$ kids, so:
$\hat{p} = \frac{9.6}{100} = 0.096$

STEP 5

Next, compute the test statistic using the formula for a proportion $z$ -test:
$z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$ where $p_0 = 0.05$ and $n = 250$ .

STEP 6

Plug in the numbers:
$z = \frac{0.096 - 0.05}{\sqrt{\frac{0.05 \cdot (1-0.05)}{250}}}$

STEP 7

Calculate the denominator:
$\sqrt{\frac{0.05 \cdot 0.95}{250}} = \sqrt{\frac{0.0475}{250}} = \sqrt{0.00019} \approx 0.0138$

STEP 8

Now, compute the $z$ -value:
$z = \frac{0.096 - 0.05}{0.0138} \approx \frac{0.046}{0.0138} \approx 3.33$

STEP 9

Since we're testing at $\alpha = 0.01$ and it's a one-tailed test, find the critical $z$ -value.
For $\alpha = 0.01$ , the critical $z$ -value is approximately $2.33$ .

STEP 10

Compare the test statistic to the critical value.
Our calculated $z$ -value is $3.33$ , which is greater than $2.33$ .

STEP 11

Since $3.33 > 2.33$ , we reject the null hypothesis.
There's enough evidence to say the asthma rate is higher than $5\%$ .

STEP 12

Yes, there is enough evidence at $\alpha = 0.01$ to reject the researcher's claim that $5\%$ of children under 18 have asthma.

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Question6. A medical researcher claims that $5 \%$ of children under 18 years of age have asthma. In a random sample of 250 children under 18 years of age, $9.6 \%$ say they have asthma. At $\alpha=0.01$ , is there enough evidence to reject the researcher's claim?

Studdy Solution

STEP 1

What is this asking? Is the asthma rate in kids really higher than the researcher's claim of $5\%$ ? Watch out! Don't mix up sample proportion with population proportion!

STEP 2

1. Define the hypotheses
2. Calculate the test statistic
3. Determine the critical value
4. Make a decision

STEP 3

Alright, let's set up our hypotheses!
The null hypothesis $H_0$ is that the proportion of children with asthma is $p = 0.05$ .
The alternative hypothesis $H_a$ is that the proportion is greater than $0.05$ .
So, we have:
$H_0: p = 0.05$ $H_a: p > 0.05$

STEP 4

Calculate the sample proportion first.
We have $9.6\%$ of $250$ kids, so:
$\hat{p} = \frac{9.6}{100} = 0.096$

STEP 5

Next, compute the test statistic using the formula for a proportion $z$ -test:
$z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$ where $p_0 = 0.05$ and $n = 250$ .

STEP 6

Plug in the numbers:
$z = \frac{0.096 - 0.05}{\sqrt{\frac{0.05 \cdot (1-0.05)}{250}}}$

STEP 7

Calculate the denominator:
$\sqrt{\frac{0.05 \cdot 0.95}{250}} = \sqrt{\frac{0.0475}{250}} = \sqrt{0.00019} \approx 0.0138$

STEP 8

Now, compute the $z$ -value:
$z = \frac{0.096 - 0.05}{0.0138} \approx \frac{0.046}{0.0138} \approx 3.33$

STEP 9

Since we're testing at $\alpha = 0.01$ and it's a one-tailed test, find the critical $z$ -value.
For $\alpha = 0.01$ , the critical $z$ -value is approximately $2.33$ .

STEP 10

Compare the test statistic to the critical value.
Our calculated $z$ -value is $3.33$ , which is greater than $2.33$ .

STEP 11

Since $3.33 > 2.33$ , we reject the null hypothesis.
There's enough evidence to say the asthma rate is higher than $5\%$ .

STEP 12

Yes, there is enough evidence at $\alpha = 0.01$ to reject the researcher's claim that $5\%$ of children under 18 have asthma.

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Question6. A medical researcher claims that 5%5 \%5% of children under 18 years of age have asthma. In a random sample of 250 children under 18 years of age, 9.6%9.6 \%9.6% say they have asthma. At α=0.01\alpha=0.01α=0.01, is there enough evidence to reject the researcher's claim?

Studdy Solution

STEP 1

What is this asking? Is the asthma rate in kids really higher than the researcher's claim of 5%5\%5%? Watch out! Don't mix up sample proportion with population proportion!

STEP 2

1. Define the hypotheses2. Calculate the test statistic3. Determine the critical value4. Make a decision

STEP 3

STEP 4

**Calculate the sample proportion** first.We have 9.6%9.6\%9.6% of 250250250 kids, so:p^=9.6100=0.096\hat{p} = \frac{9.6}{100} = 0.096p^​=1009.6​=0.096

STEP 5

Next, **compute the test statistic** using the formula for a proportion zzz-test:z=p^−p0p0(1−p0)nz = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}z=np0​(1−p0​)​​p^​−p0​​where p0=0.05p_0 = 0.05p0​=0.05 and n=250n = 250n=250.

STEP 6

Plug in the numbers:z=0.096−0.050.05⋅(1−0.05)250z = \frac{0.096 - 0.05}{\sqrt{\frac{0.05 \cdot (1-0.05)}{250}}}z=2500.05⋅(1−0.05)​​0.096−0.05​

STEP 7

Calculate the denominator:0.05⋅0.95250=0.0475250=0.00019≈0.0138\sqrt{\frac{0.05 \cdot 0.95}{250}} = \sqrt{\frac{0.0475}{250}} = \sqrt{0.00019} \approx 0.01382500.05⋅0.95​​=2500.0475​​=0.00019​≈0.0138

STEP 8

Now, compute the zzz-value:z=0.096−0.050.0138≈0.0460.0138≈3.33z = \frac{0.096 - 0.05}{0.0138} \approx \frac{0.046}{0.0138} \approx 3.33z=0.01380.096−0.05​≈0.01380.046​≈3.33

STEP 9

Since we're testing at α=0.01\alpha = 0.01α=0.01 and it's a one-tailed test, **find the critical zzz-value**.For α=0.01\alpha = 0.01α=0.01, the critical zzz-value is approximately 2.332.332.33.

STEP 10

**Compare the test statistic** to the critical value.Our calculated zzz-value is 3.333.333.33, which is greater than 2.332.332.33.

STEP 11

Since 3.33>2.333.33 > 2.333.33>2.33, we **reject the null hypothesis**.There's enough evidence to say the asthma rate is higher than 5%5\%5%.

STEP 12

Yes, there is enough evidence at α=0.01\alpha = 0.01α=0.01 to reject the researcher's claim that 5%5\%5% of children under 18 have asthma.

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Question6. A medical researcher claims that $5 \%$ of children under 18 years of age have asthma. In a random sample of 250 children under 18 years of age, $9.6 \%$ say they have asthma. At $\alpha=0.01$ , is there enough evidence to reject the researcher's claim?

What is this asking? Is the asthma rate in kids really higher than the researcher's claim of $5\%$ ? Watch out! Don't mix up sample proportion with population proportion!

1. Define the hypotheses
2. Calculate the test statistic
3. Determine the critical value
4. Make a decision

Calculate the sample proportion first.
We have $9.6\%$ of $250$ kids, so:
$\hat{p} = \frac{9.6}{100} = 0.096$

Next, compute the test statistic using the formula for a proportion $z$ -test:
$z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$ where $p_0 = 0.05$ and $n = 250$ .

Plug in the numbers:
$z = \frac{0.096 - 0.05}{\sqrt{\frac{0.05 \cdot (1-0.05)}{250}}}$

Calculate the denominator:
$\sqrt{\frac{0.05 \cdot 0.95}{250}} = \sqrt{\frac{0.0475}{250}} = \sqrt{0.00019} \approx 0.0138$

Now, compute the $z$ -value:
$z = \frac{0.096 - 0.05}{0.0138} \approx \frac{0.046}{0.0138} \approx 3.33$

Since we're testing at $\alpha = 0.01$ and it's a one-tailed test, find the critical $z$ -value.
For $\alpha = 0.01$ , the critical $z$ -value is approximately $2.33$ .

Compare the test statistic to the critical value.
Our calculated $z$ -value is $3.33$ , which is greater than $2.33$ .

Since $3.33 > 2.33$ , we reject the null hypothesis.
There's enough evidence to say the asthma rate is higher than $5\%$ .

Yes, there is enough evidence at $\alpha = 0.01$ to reject the researcher's claim that $5\%$ of children under 18 have asthma.