Math / Data & Statistics

QuestionQ1(6). Parents were asked if sports are equally important for boys and girls. Suppose that the proportion of parents who agree the equal importance in the population is actually 0.65 . In a survey, of the 400 parents interviewed, $70 \%$ agreed that boys and girls should have equal opportunities to participate in sports.
1(2). Describe the sampling distribution of the sample proportion of parents who agree that boys and girls should have equal opportunities.
2(4). What is the probability of observing a sample proportion as large as or larger than the observed value $\hat{p}$ $=0.70 ?$

Studdy Solution

STEP 1

What is this asking? We're checking how likely it is to see a sample where 70% of 400 parents agree about equal sports opportunities, if we actually know that 65% of all parents agree. Watch out! Don't mix up the true proportion (from everyone) with the sample proportion (from the survey).

STEP 2

1. Describe the sampling distribution
2. Calculate the probability

STEP 3

Alright, so we're diving into the world of sampling distributions!
We're given that the true proportion of parents who agree is $\text{p} = 0.65$ .
This is our golden number, the truth we hold dear!

STEP 4

Now, we've got a sample of 400 parents, which is a pretty decent size.
We denote our sample size as $n = 400$ .
This is important because the bigger our sample, the closer our sample proportion should get to the true proportion.
It's like magic!

STEP 5

The sampling distribution of the sample proportion ( $\hat{p}$ ) is approximately normal if $n \cdot p \geq 10$ and $n \cdot (1-p) \geq 10$ .
Let's check if this holds true for our case.
We have $400 \cdot 0.65 = 260$ and $400 \cdot (1 - 0.65) = 400 \cdot 0.35 = 140$ .
Both are way bigger than 10, so we're good to go!
The distribution is approximately normal!
Woohoo!

STEP 6

The mean of the sampling distribution is the true proportion itself, which is $\mu_{\hat{p}} = p = 0.65$ .
This makes sense, right?
On average, our sample proportion should be close to the real deal.

STEP 7

The standard deviation of the sampling distribution is given by the formula: $\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$ Let's plug in our values: $\sigma_{\hat{p}} = \sqrt{\frac{0.65 \cdot (1-0.65)}{400}} = \sqrt{\frac{0.65 \cdot 0.35}{400}} = \sqrt{\frac{0.2275}{400}} = \sqrt{0.00056875} \approx 0.0238$ So, our standard deviation is approximately 0.0238.
This tells us how spread out our sample proportions are likely to be.

STEP 8

Now for the grand finale!
We want to find the probability of getting a sample proportion of 0.70 or greater, given our known true proportion and standard deviation.
We're looking for $P(\hat{p} \geq 0.70)$ .

STEP 9

To do this, we'll use the z-score!
The z-score tells us how many standard deviations away our sample proportion is from the mean.
The formula is: $z = \frac{\hat{p} - p}{\sigma_{\hat{p}}}$ Plugging in our values: $z = \frac{0.70 - 0.65}{0.0238} = \frac{0.05}{0.0238} \approx 2.10$ So our z-score is approximately 2.10.

STEP 10

Now, we look up this z-score in a standard normal table (or use a calculator) to find the probability of getting a z-score less than 2.10.
This value is approximately 0.9821.
However, we want the probability of getting a z-score greater than 2.10.
So we subtract our value from 1: $1 - 0.9821 = 0.0179$ .

STEP 11

The sampling distribution of the sample proportion is approximately normal with a mean of 0.65 and a standard deviation of approximately 0.0238.
The probability of observing a sample proportion as large as or larger than 0.70 is approximately 0.0179.

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Studdy Solution

STEP 1

What is this asking? We're checking how likely it is to see a sample where 70% of 400 parents agree about equal sports opportunities, if we actually know that 65% of all parents agree. Watch out! Don't mix up the true proportion (from everyone) with the sample proportion (from the survey).

STEP 2

1. Describe the sampling distribution
2. Calculate the probability

STEP 3

Alright, so we're diving into the world of sampling distributions!
We're given that the true proportion of parents who agree is $\text{p} = 0.65$ .
This is our golden number, the truth we hold dear!

STEP 4

Now, we've got a sample of 400 parents, which is a pretty decent size.
We denote our sample size as $n = 400$ .
This is important because the bigger our sample, the closer our sample proportion should get to the true proportion.
It's like magic!

STEP 5

STEP 6

The mean of the sampling distribution is the true proportion itself, which is $\mu_{\hat{p}} = p = 0.65$ .
This makes sense, right?
On average, our sample proportion should be close to the real deal.

STEP 7

STEP 8

Now for the grand finale!
We want to find the probability of getting a sample proportion of 0.70 or greater, given our known true proportion and standard deviation.
We're looking for $P(\hat{p} \geq 0.70)$ .

STEP 9

STEP 10

STEP 11

The sampling distribution of the sample proportion is approximately normal with a mean of 0.65 and a standard deviation of approximately 0.0238.
The probability of observing a sample proportion as large as or larger than 0.70 is approximately 0.0179.

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Studdy Solution

STEP 1

What is this asking? We're checking how likely it is to see a sample where 70% of 400 parents agree about equal sports opportunities, if we *actually* know that 65% of *all* parents agree. Watch out! Don't mix up the *true* proportion (from everyone) with the sample proportion (from the survey).

STEP 2

1. Describe the sampling distribution2. Calculate the probability

STEP 3

Alright, so we're diving into the world of sampling distributions!We're given that the *true* proportion of parents who agree is p=0.65\text{p} = 0.65p=0.65.This is our **golden number**, the truth we hold dear!

STEP 4

Now, we've got a sample of 400 parents, which is a pretty decent size.We denote our sample size as n=400n = 400n=400.This is important because the bigger our sample, the closer our sample proportion *should* get to the true proportion.It's like magic!

STEP 5

STEP 6

The mean of the sampling distribution is the **true proportion** itself, which is μp^=p=0.65\mu_{\hat{p}} = p = 0.65μp^​​=p=0.65.This makes sense, right?On average, our sample proportion should be close to the real deal.

STEP 7

STEP 8

Now for the grand finale!We want to find the probability of getting a sample proportion of 0.70 or greater, given our known true proportion and standard deviation.We're looking for P(p^≥0.70)P(\hat{p} \geq 0.70)P(p^​≥0.70).

STEP 9

STEP 10

STEP 11

The sampling distribution of the sample proportion is approximately normal with a mean of 0.65 and a standard deviation of approximately 0.0238.The probability of observing a sample proportion as large as or larger than 0.70 is approximately 0.0179.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

What is this asking? We're checking how likely it is to see a sample where 70% of 400 parents agree about equal sports opportunities, if we actually know that 65% of all parents agree. Watch out! Don't mix up the true proportion (from everyone) with the sample proportion (from the survey).

1. Describe the sampling distribution
2. Calculate the probability

Alright, so we're diving into the world of sampling distributions!
We're given that the true proportion of parents who agree is $\text{p} = 0.65$ .
This is our golden number, the truth we hold dear!

Now, we've got a sample of 400 parents, which is a pretty decent size.
We denote our sample size as $n = 400$ .
This is important because the bigger our sample, the closer our sample proportion should get to the true proportion.
It's like magic!

The mean of the sampling distribution is the true proportion itself, which is $\mu_{\hat{p}} = p = 0.65$ .
This makes sense, right?
On average, our sample proportion should be close to the real deal.

Now for the grand finale!
We want to find the probability of getting a sample proportion of 0.70 or greater, given our known true proportion and standard deviation.
We're looking for $P(\hat{p} \geq 0.70)$ .

The sampling distribution of the sample proportion is approximately normal with a mean of 0.65 and a standard deviation of approximately 0.0238.
The probability of observing a sample proportion as large as or larger than 0.70 is approximately 0.0179.