Math / Data & Statistics

Question11 12 13
Coffee: The National Coffee Association reported that $65 \%$ of U.S. adults drink coffee daily. A random sample of 300 U.S. adults is selected. Round your answers to at least four decimal places as needed.
Part 1 of 6 (a) Find the mean $\mu_{p}$ .
The mean $\mu_{\hat{p}}$ is 0.65 .
Part 2 of 6 (b) Find the standard deviation $\sigma_{\hat{p}}$ .
The standard deviation $\sigma_{\hat{p}}$ is 0.0275 .
Part 3 of 6 (c) Find the probability that more than $66 \%$ of the sampled adults drink coffee daily.
The probability that more than $66 \%$ of the sampled adults drink coffee daily is 0.3564 .
Part 4 of 6 (d) Find the probability that the proportion of the sampled adults who drink coffee daily is between 0.57 and 0.71 .
The probability that the proportion of the sampled adults who drink coffee daily is between 0.57 and 0.71 is 0.9836 . Part: $5 / 6$
Part 6 of 6 (f) Using a cutoff of 0.05 , would it be unusual if less than $63 \%$ of the sampled adults drink coffee daily? it $\square$ (Choose one) $\mathbf{V}$ be unusual if less than $63 \%$ of the sampled adults drink coffee daily, since the probability is $\square$ would would not

Studdy Solution

STEP 1

1. The sample size is $n = 300$ .
2. The population proportion of U.S. adults who drink coffee daily is $p = 0.65$ .
3. The sampling distribution of the sample proportion $\hat{p}$ is approximately normal because $np$ and $n(1-p)$ are both greater than 5.

STEP 2

1. Calculate the mean $\mu_{\hat{p}}$ .
2. Calculate the standard deviation $\sigma_{\hat{p}}$ .
3. Calculate the probability that more than 66% of the sampled adults drink coffee daily.
4. Calculate the probability that the proportion of the sampled adults who drink coffee daily is between 0.57 and 0.71.
5. Determine if it would be unusual if less than 63% of the sampled adults drink coffee daily.

STEP 3

Calculate the mean $\mu_{\hat{p}}$ :
The mean of the sampling distribution of the sample proportion is given by the population proportion $p$ .
$\mu_{\hat{p}} = p = 0.65$

STEP 4

Calculate the standard deviation $\sigma_{\hat{p}}$ :
The standard deviation of the sampling distribution of the sample proportion is calculated using the formula:
$\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.65 \times 0.35}{300}}$
$\sigma_{\hat{p}} \approx 0.0275$

STEP 5

Calculate the probability that more than 66% of the sampled adults drink coffee daily:
Using the standard normal distribution, find the z-score for $\hat{p} = 0.66$ :
$z = \frac{0.66 - 0.65}{0.0275} \approx 0.3636$
Find the probability $P(\hat{p} > 0.66)$ using the standard normal distribution table or calculator:
$P(\hat{p} > 0.66) \approx 0.3564$

STEP 6

Calculate the probability that the proportion of the sampled adults who drink coffee daily is between 0.57 and 0.71:
Find the z-scores for $\hat{p} = 0.57$ and $\hat{p} = 0.71$ :
$z_{0.57} = \frac{0.57 - 0.65}{0.0275} \approx -2.9091$
$z_{0.71} = \frac{0.71 - 0.65}{0.0275} \approx 2.1818$
Find the probability $P(0.57 < \hat{p} < 0.71)$ using the standard normal distribution table or calculator:
$P(0.57 < \hat{p} < 0.71) \approx 0.9836$

STEP 7

Determine if it would be unusual if less than 63% of the sampled adults drink coffee daily:
Find the z-score for $\hat{p} = 0.63$ :
$z = \frac{0.63 - 0.65}{0.0275} \approx -0.7273$
Find the probability $P(\hat{p} < 0.63)$ using the standard normal distribution table or calculator:
$P(\hat{p} < 0.63) \approx 0.2335$
Since $0.2335 > 0.05$ , it would not be unusual if less than 63% of the sampled adults drink coffee daily.
The conclusion is: it would not be unusual if less than 63% of the sampled adults drink coffee daily, since the probability is greater than 0.05.

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