A) find the mean u^p
11 answers
To find the mean \( \mu_{p} \) of the sampling distribution of the sample proportion \( p \), you can use the formula:
\[ \mu_{p} = p \]
where \( p \) is the population proportion.
In this case, you are given that 37% of teenagers aged 12-17 own smartphones, which can be expressed as:
\[ p = 0.37 \]
Thus, the mean \( \mu_{p} \) is:
\[ \mu_{p} = 0.37 \]
Therefore, the answer is:
\[ \mu_{p} = 0.3700 \] (rounded to four decimal places).
To find the standard deviation \( \sigma_{p} \) of the sampling distribution of the sample proportion \( p \), you can use the formula:
\[ \sigma_{p} = \sqrt{\frac{p(1 - p)}{n}} \]
where:
Given:
We can calculate \( \sigma_{p} \) as follows:
Calculate \( 1 - p \): \[ 1 - p = 1 - 0.37 = 0.63 \]
Now plug in the values into the formula: \[ \sigma_{p} = \sqrt{\frac{0.37 \times 0.63}{101}} \]
Calculate the numerator: \[ 0.37 \times 0.63 = 0.2331 \]
Divide by the sample size \( n = 101 \): \[ \frac{0.2331}{101} \approx 0.002308 \]
Take the square root: \[ \sigma_{p} \approx \sqrt{0.002308} \approx 0.0480 \]
Thus, the standard deviation \( \sigma_{p} \) is approximately:
\[ \sigma_{p} \approx 0.0480 \] (rounded to four decimal places).
To find the probability that more than 40% of the sampled teenagers own a smartphone, we can use the normal approximation to the binomial distribution. We will convert the sample proportion to a z-score and then use the standard normal distribution to find the probability.
We already calculated these in previous steps:
To calculate the z-score for the proportion of 0.40:
\[ z = \frac{p' - \mu_{p}}{\sigma_{p}} \]
Where:
Substituting the values:
\[ z = \frac{0.40 - 0.37}{0.0480} \approx \frac{0.03}{0.0480} \approx 0.625 \]
Next, we need to find the probability that corresponds to a z-score of 0.625. You can use a standard normal distribution table or a calculator.
Using a z-table or calculator, you can find that:
\[ P(Z < 0.625) \approx 0.7357 \]
However, since we want the probability that more than 40% own a smartphone, we need:
\[ P(Z > 0.625) = 1 - P(Z < 0.625) = 1 - 0.7357 \approx 0.2643 \]
Thus, the probability that more than 40% of the sampled teenagers own a smartphone is:
\[ P(p' > 0.40) \approx 0.2643 \] (rounded to four decimal places).
To find the probability that the proportion of the sampled teenagers who own a smartphone is between 0.34 and 0.48, we will again use the normal approximation to the binomial distribution and calculate the z-scores for both proportions.
As calculated previously:
For \( p_1 = 0.34 \): \[ z_1 = \frac{p_1 - \mu_p}{\sigma_p} = \frac{0.34 - 0.37}{0.0480} = \frac{-0.03}{0.0480} \approx -0.625 \]
For \( p_2 = 0.48 \): \[ z_2 = \frac{p_2 - \mu_p}{\sigma_p} = \frac{0.48 - 0.37}{0.0480} = \frac{0.11}{0.0480} \approx 2.2917 \]
For \( z_1 = -0.625 \): Using the standard normal distribution table or calculator, we find: \[ P(Z < -0.625) \approx 0.2659 \]
For \( z_2 = 2.2917 \): Using the standard normal distribution table or calculator, we find: \[ P(Z < 2.2917) \approx 0.9890 \]
To find the probability that the proportion of the sampled teenagers who own a smartphone is between 0.34 and 0.48:
\[ P(0.34 < p < 0.48) = P(Z < 2.2917) - P(Z < -0.625) \] \[ P(0.34 < p < 0.48) \approx 0.9890 - 0.2659 \approx 0.7231 \]
Thus, the probability that the proportion of sampled teenagers who own a smartphone is between 0.34 and 0.48 is:
\[ P(0.34 < p < 0.48) \approx 0.7231 \] (rounded to four decimal places).
To find the probability that less than 48% of sampled teenagers own smartphones, we will again use the normal approximation to the binomial distribution and calculate the z-score for the proportion of 0.48.
We already have:
For \( p = 0.48 \): \[ z = \frac{p - \mu_{p}}{\sigma_{p}} = \frac{0.48 - 0.37}{0.0480} = \frac{0.11}{0.0480} \approx 2.2917 \]
Using the standard normal distribution table or a calculator: \[ P(Z < 2.2917) \approx 0.9890 \]
Thus, the probability that less than 48% of sampled teenagers own smartphones is:
\[ P(p < 0.48) \approx 0.9890 \] (rounded to four decimal places).
To determine whether it would be unusual if less than 30% of the sampled teenagers owned smartphones, we need to calculate the probability of that event happening.
We'll use the same mean \( \mu_{p} \) and standard deviation \( \sigma_{p} \) calculated earlier:
For \( p = 0.30 \): \[ z = \frac{p - \mu_{p}}{\sigma_{p}} = \frac{0.30 - 0.37}{0.0480} = \frac{-0.07}{0.0480} \approx -1.4583 \]
Using the standard normal distribution table or a calculator, we find: \[ P(Z < -1.4583) \approx 0.0726 \]
Since the probability \( P(p < 0.30) \approx 0.0726 \) is less than 0.05 (or 5%), this would be considered an unusual event.
Thus, we can conclude:
It would be unusual if less than 30% of the sampled teenagers owned smartphones since the probability is approximately 0.0726.
