Question

1. - / 4 points

a usa today snapshot found that 47% of americans associate
ecycling\ with earth day. suppose a random sample of n = 100 adults are polled and the 47% figure is correct.
(a) does the distribution of \\(\hat{p}\\), the sample proportion of adults who list
ecycling\ with earth day, have an approximate normal distribution? if so, what is its mean and standard deviation?
\\(\circ\\) yes, the distribution is approximately normal. the mean is 0.3 and the standard deviation is 0.0458.
\\(\circ\\) no, the distribution is not normal because n(1-p) is less than 5.
\\(\circ\\) yes, the distribution is approximately normal. the mean is 0.47 and the standard deviation is 0.0499.
\\(\circ\\) no, the distribution is not normal because np is less than 5.
\\(\circ\\) yes, the distribution is approximately normal. the mean is 0.23 and the standard deviation is 0.0421.
(b) what is the probability that the sample proportion \\(\hat{p}\\) is less than 44%? (round your answer to four decimal places.)
(c) what is the probability that \\(\hat{p}\\) lies in the interval 0.41 to 0.44? (round your answer to four decimal places.)
(d) what might you conclude about p if the sample proportion were less than 0.30?
\\(\circ\\) the value \\(\hat{p} = 0.30\\) is more than 3 standard deviations from the mean. therefore, it is an unlikely occurrence if p = 0.47. perhaps the sampling was not random, or the 47% figure is not correct.
\\(\circ\\) the value \\(\hat{p} = 0.30\\) is less than 1 standard deviation from the mean. therefore, it is a likely occurrence if p = 0.47. there is no evidence of an inaccurate value of p.
\\(\circ\\) the value \\(\hat{p} = 0.30\\) is more than 3 standard deviations from the mean. therefore, it is a likely occurrence if p = 0.47. there is no evidence of an inaccurate value of p.
\\(\circ\\) the value \\(\hat{p} = 0.30\\) is between 2 and 3 standard deviations from the mean. therefore, it is an unlikely occurrence if p = 0.47. there is no evidence of an inaccurate value of p.
\\(\circ\\) the value \\(\hat{p} = 0.30\\) is between 2 and 3 standard deviations from the mean. therefore, it is an unlikely occurrence if p = 0.47. perhaps the sampling was not random, or the 47% figure is not correct.
you may need to use the appropriate appendix table or technology to answer this question.

Explanation:

Part (a)

Step1: Check Normal Condition

To determine if the sampling distribution of \(\hat{p}\) is approximately normal, we check \(np \geq 5\) and \(n(1 - p) \geq 5\). Here, \(n = 100\) and \(p = 0.47\).

• \(np = 100\times0.47 = 47 \geq 5\)
• \(n(1 - p)=100\times(1 - 0.47)=100\times0.53 = 53 \geq 5\)

So the distribution is approximately normal.

Step2: Find Mean of \(\hat{p}\)

The mean of the sampling distribution of \(\hat{p}\) is equal to the population proportion \(p\). So \(\mu_{\hat{p}}=p = 0.47\).

Step3: Find Standard Deviation of \(\hat{p}\)

The formula for the standard deviation (standard error) of \(\hat{p}\) is \(\sigma_{\hat{p}}=\sqrt{\frac{p(1 - p)}{n}}\).
Substitute \(p = 0.47\) and \(n = 100\):
\[
\sigma_{\hat{p}}=\sqrt{\frac{0.47\times(1 - 0.47)}{100}}=\sqrt{\frac{0.47\times0.53}{100}}=\sqrt{\frac{0.2491}{100}}=\sqrt{0.002491}\approx0.0499
\]

Step1: Standardize the Proportion

We want \(P(\hat{p}<0.44)\). First, standardize \(\hat{p}\) using \(z=\frac{\hat{p}-\mu_{\hat{p}}}{\sigma_{\hat{p}}}\).
We know \(\mu_{\hat{p}} = 0.47\), \(\sigma_{\hat{p}}\approx0.0499\), and \(\hat{p}=0.44\).
\[
z=\frac{0.44 - 0.47}{0.0499}=\frac{- 0.03}{0.0499}\approx - 0.60
\]

Step2: Find the Probability

We need \(P(Z < - 0.60)\). Using the standard normal table or technology, \(P(Z < - 0.60)=0.2743\) (from standard normal table: the area to the left of \(z=-0.60\) is \(0.2743\)).

Step1: Standardize the Lower Bound

For \(\hat{p}=0.41\), calculate \(z_1=\frac{0.41 - 0.47}{0.0499}=\frac{-0.06}{0.0499}\approx - 1.20\).

Step2: Standardize the Upper Bound

For \(\hat{p}=0.44\), calculate \(z_2=\frac{0.44 - 0.47}{0.0499}=\frac{-0.03}{0.0499}\approx - 0.60\) (we already calculated this in part (b)).

Step3: Find the Probability

We want \(P(0.41<\hat{p}<0.44)=P(-1.20 < Z < - 0.60)\).
This is \(P(Z < - 0.60)-P(Z < - 1.20)\).
From standard normal table:

• \(P(Z < - 0.60)=0.2743\)
• \(P(Z < - 1.20)=0.1151\)

So \(P(-1.20 < Z < - 0.60)=0.2743 - 0.1151 = 0.1592\).

Answer:

Yes, the distribution is approximately normal. The mean is \(0.47\) and the standard deviation is \(0.0499\). (Corresponding option: Yes, the distribution is approximately normal. The mean is \(0.47\) and the standard deviation is \(0.0499\).)

Part (b)