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Do you drink the cereal milk? A USA Today Poll asked a random sample of $1012$ U.S. adults what they do with the milk in the bowl after they have eaten the cereal. Of the respondents, $67\%$ said that they drink it. Suppose that $70\%$ of U.S. adults actually drink the cereal milk. Let $\hat{p}$ be the proportion of people in the sample who drink the cereal milk.

(a) What is the mean of the sampling distribution of $\hat{p}$ ? Why?

(b) Find the standard deviation of the sampling distribution of $\hat{p}$. Check to see if the $10\%$ condition is met.

(c) Is the sampling distribution of $\hat{p}$ approximately Normal? Check to see if the Normal condition is met.

(d) Find the probability of obtaining a sample of $1012$ adults in which $67\%$ or fewer say they drink the cereal milk. Do you have any doubts about the result of this poll?

Step by step solution

01

Part (a) Step 1: Given Information

Sample proportion $\left(\hat{p}\right)=0.70$,

Population proportion $\left(p\right)=0.67$,

Sample size $\left(n\right)=1012$.

02

Part (a) Step 2: Explanation

The sample distribution's mean can be computed as:

${渭}_{\hat{p}}=p$

$=0.70$

03

Part (b) Step 1: Given Information

Sample proportion $\left(\hat{p}\right)=0.70$,

Population proportion $\left(p\right)=0.67$,

Sample size $\left(n\right)=1012$.

04

Part (b) Step 2: Explanation

The sample proportion's standard deviation is calculated as:

${蟽}_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}}$

$=\sqrt{\frac{0.70(1-0.70)}{1012}}$

$=0.0144052$

05

Part (c) Step 1: Given Information

Sample proportion $\left(\hat{p}\right)=0.70$,

Population proportion $\left(p\right)=0.67$,

Sample size $\left(n\right)=1012$.

06

Part (c) Step 2: Explanation

$np=1010(0.070)$

$=708.4>10$

$n(1-p)=1010(1-0.070)$

$=303.6>10$

Normal approximation could be applied.

07

Part (d) Step 1: Given Information

Sample proportion $\left(\hat{p}\right)=0.70$,

Population proportion $\left(p\right)=0.67$,

Sample size $\left(n\right)=1012$.

08

Part (d) Step 2: Explanation

The probability that fewer or $67\%$are choosing to drink cereal milk is calculated as:

$P(\hat{p}鈮�0.67)=P\left(Z鈮�\frac{0.67-0.70}{0.0144052}\right)$

$=P(Z鈮�-2.08)$

$=0.0188$

The probability is less than $0.05$. Thus, there is doubt on the occurrence of the event.

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