In a random sample of 103 people, 8 of them are left-handed. Let $$ p $$ be the true proportion of lefthanded people in the population. Does the sample indicate that $$ p $$ is other than 0.05 ? Use a 0.008 level of significance. A) What is the level of significance (comparison level)? $$ \square $$ B) State the Null and Alternate hypotheses. $$ H_{0}: p=0.05 ; H_{1}: p0.05 ; H_{1}: p=0.05 $$ $$ H_{0}: p=0.05 ; H_{1}: p0.05 $$ C) Is this a (left-tail, right-tail, one-tail, two-tail) problem? one-tail two-tail left-tail right-tail D) Using 4 decimal places for $$ \sigma_{\hat{P}} $$, what's TRUE about $$ \widehat{P} 103^{?} $$ ? $$ \widehat{P}_{103} \sim N(0-0777,0-0215) $$ $$ \widehat{P} 103 \sim N(0-0215,0-05) $$ $$ \widehat{P}_{103} \sim N(0-0777,0-0264) $$ $$ \widehat{P} 103 \sim N(0-05,0-0215) $$ $$ \widehat{P} 103 \sim N(0-05,0-0264) $$ E) Using 4 decimal places for $$ \hat{p} $$, find the $$ z $$-score of $$ \hat{p} $$ $$ z= $$ $$ \square $$ .Round your answer to 2 decimals. F) What is the p-value of this test? $$ \square $$ . Round your answer to 4 decimals. G) Does the p-value suggest strong evidence to reject $$ H_{0} $$ ? No. Yes.

Question

In a random sample of 103 people, 8 of them are left-handed. Let $$ p $$ be the true proportion of lefthanded people in the population. Does the sample indicate that $$ p $$ is other than 0.05 ? Use a 0.008 level of significance. A) What is the level of significance (comparison level)? $$ \square $$ B) State the Null and Alternate hypotheses. $$ H_{0}: p=0.05 ; H_{1}: p<0.05 $$ $$ H_{0}: p=0.05: H_{1}: p 
eq 0.05 $$ $$ H_{0}: p>0.05 ; H_{1}: p=0.05 $$ $$ H_{0}: p=0.05 ; H_{1}: p>0.05 $$ C) Is this a (left-tail, right-tail, one-tail, two-tail) problem? one-tail two-tail left-tail right-tail D) Using 4 decimal places for $$ \sigma_{\hat{P}} $$, what's TRUE about $$ \widehat{P} 103^{?} $$ ? $$ \widehat{P}_{103} \sim N(0-0777,0-0215) $$ $$ \widehat{P} 103 \sim N(0-0215,0-05) $$ $$ \widehat{P}_{103} \sim N(0-0777,0-0264) $$ $$ \widehat{P} 103 \sim N(0-05,0-0215) $$ $$ \widehat{P} 103 \sim N(0-05,0-0264) $$ E) Using 4 decimal places for $$ \hat{p} $$, find the $$ z $$-score of $$ \hat{p} $$ $$ z= $$ $$ \square $$ .Round your answer to 2 decimals. F) What is the p-value of this test? $$ \square $$ . Round your answer to 4 decimals. G) Does the p-value suggest strong evidence to reject $$ H_{0} $$ ? No. Yes.

UpStudy AI · Accepted Answer

Let's solve the problem step by step.

### A) What is the level of significance (comparison level)?
The level of significance is given as $$ \alpha = 0.008 $$.

### B) State the Null and Alternate hypotheses.
The correct hypotheses for this problem are:
- $$ H_{0}: p = 0.05 $$ (the null hypothesis states that the true proportion of left-handed people is 0.05)
- $$ H_{1}: p \neq 0.05 $$ (the alternative hypothesis states that the true proportion is different from 0.05)

### C) Is this a (left-tail, right-tail, one-tail, two-tail) problem?
Since the alternative hypothesis is $$ H_{1}: p \neq 0.05 $$, this is a two-tail problem.

### D) Using 4 decimal places for $$ \sigma_{\hat{P}} $$, what's TRUE about $$ \widehat{P}_{103} $$?
First, we need to calculate $$ \hat{p} $$ and $$ \sigma_{\hat{P}} $$.

1. Calculate $$ \hat{p} $$:
   $$
   \hat{p} = \frac{x}{n} = \frac{8}{103} \approx 0.0777
   $$

2. Calculate $$ \sigma_{\hat{P}} $$:
   $$
   \sigma_{\hat{P}} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.05(1-0.05)}{103}} = \sqrt{\frac{0.05 \times 0.95}{103}} \approx 0.0215
   $$

Thus, we have:
$$
\widehat{P}_{103} \sim N(0.0777, 0.0215)
$$

### E) Using 4 decimal places for $$ \hat{p} $$, find the $$ z $$-score of $$ \hat{p} $$.
The $$ z $$-score is calculated using the formula:
$$
z = \frac{\hat{p} - p_0}{\sigma_{\hat{P}}}
$$
Where $$ p_0 = 0.05 $$.

Substituting the values:
$$
z = \frac{0.0777 - 0.05}{0.0215} \approx \frac{0.0277}{0.0215} \approx 1.2879
$$

Rounding to two decimals, we get:
$$
z \approx 1.29
$$

### F) What is the p-value of this test?
Since this is a two-tailed test, we need to find the p-value corresponding to the calculated $$ z $$-score. We will use the standard normal distribution to find the p-value.

Using the $$ z $$-score of 1.29, we can find the p-value:
$$
p\text{-value} = 2 \times P(Z > 1.29)
$$

Calculating $$ P(Z > 1.29) $$:
Using a standard normal distribution table or calculator, we find:
$$
P(Z > 1.29) \approx 0.1003
$$

Thus, the p-value is:
$$
p\text{-value} \approx 2 \times 0.1003 \approx 0.2006
$$

Rounding to four decimals, we get:
$$
p\text{-value} \approx 0.2006
$$

### G) Does the p-value suggest strong evidence to reject $$ H_{0} $$?
Since the p-value $$ 0.2006 $$ is greater than the significance level $$ 0.008 $$, we do not have strong evidence to reject $$ H_{0} $$.

**Final Answers:**
A) $$ 0.008 $$  
B) $$ H_{0}: p=0.05 ; H_{1}: p \neq 0.05 $$  
C) two-tail  
D) $$ \widehat{P}_{103} \sim N(0.0777, 0.0215) $$  
E) $$ z = 1.29 $$  
F) $$ 0.2006 $$  
G) No.
A) 0.008 B) $$ H_{0}: p=0.05 ; H_{1}: p \neq 0.05 $$ C) two-tail D) $$ \widehat{P}_{103} \sim N(0.0777, 0.0215) $$ E) $$ z = 1.29 $$ F) 0.2006 G) No.

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