In a random sample of 96 people, 9 of them are left-handed. Let $$ p $$ be the true proportion of lefthanded people in the population. Does the sample indicate that $$ p $$ is higher than 0.03 ? Use a 0.008 level of significance. A) What is the level of significance (comparison level)? B) State the Null and Alternate hypotheses. $$ H_{0}: p=0.03: H_{1}: p0.03 $$ $$ H_{0}: p0.03: H_{1}: p=0.03 $$ $$ H_{0}: p=0.03 ; H_{1}: p

Question

In a random sample of 96 people, 9 of them are left-handed. Let $$ p $$ be the true proportion of lefthanded people in the population. Does the sample indicate that $$ p $$ is higher than 0.03 ? Use a 0.008 level of significance. A) What is the level of significance (comparison level)? B) State the Null and Alternate hypotheses. $$ H_{0}: p=0.03: H_{1}: p>0.03 $$ $$ H_{0}: p>0.03: H_{1}: p=0.03 $$ $$ H_{0}: p=0.03 ; H_{1}: p<0.03 $$ $$ H_{0}: p=0.03 ; H_{1}: p 
eq 0.03 $$ C) Is this a (left-tail, right-tail, one-tail, two-tail) problem? left-tail right-tail one-tail two-tail D) Using 4 decimal places for $$ \sigma_{\hat{P}} $$, what's TRUE about $$ \widehat{P} 96^{?} $$ $$ \widehat{P}_{96} \sim N(0.0174,0.03) $$ $$ \hat{P}_{96} \sim N(0.03,0.0174) $$ $$ \widehat{P}_{96} \sim N(0.0938,0.0174) $$ $$ \widehat{P}_{96} \sim N(0.03,0.0298) $$ $$ \widehat{P}_{96} \sim N(0.0938,0.0298) $$ E) Using 4 decimal places for $$ \hat{p} $$, find the $$ z $$-score of $$ \hat{p} $$ $$ z=\square $$. Round your answer to 2 decimal places. F) What is the p-value of this test? $$ \square $$ . Round your answer to 4 decimal places. G) Does the p-value suggest strong evidence to reject $$ H_{0} $$ ? No. Yes.

UpStudy AI · Accepted Answer

**A)** The level of significance is  
$$
\alpha = 0.008.
$$

**B)** Since we want to test if the true proportion is higher than 0.03, we have  
$$
H_{0} : p = 0.03 \quad \text{versus} \quad H_{1} : p > 0.03.
$$
Thus, the correct option is:  
$$ H_{0}: p=0.03:\, H_{1}: p>0.03 $$.

**C)** Because the alternative hypothesis is $$ p > 0.03 $$, the rejection region will be in the right tail of the normal distribution. Hence, it is a  
**right-tail** test.

**D)** Under $$ H_{0} $$ the sampling distribution of the sample proportion is approximately  
$$
\hat{p} \sim N\left( p, \sqrt{\frac{p(1-p)}{n}} \right).
$$
Here, with $$ p=0.03 $$ and $$ n=96 $$, we have  
$$
\sigma_{\hat{p}} = \sqrt{\frac{0.03(0.97)}{96}}.
$$
Calculating the variance first:  
$$
0.03 \times 0.97 = 0.0291,
$$
$$
\frac{0.0291}{96} \approx 0.000303125,
$$
$$
\sigma_{\hat{p}} \approx \sqrt{0.000303125} \approx 0.0174.
$$
Thus, the correct description is:  
$$
\hat{p}_{96} \sim N(0.03, 0.0174).
$$

**E)** The sample proportion is  
$$
\hat{p} = \frac{9}{96} \approx 0.0938.
$$
The test statistic (z-score) is computed using  
$$
z=\frac{\hat{p}-p}{\sigma_{\hat{p}}}=\frac{0.0938-0.03}{0.0174}.
$$
Calculating the numerator:  
$$
0.0938-0.03=0.0638.
$$
Then,  
$$
z \approx \frac{0.0638}{0.0174} \approx 3.67.
$$
Thus, the z-score is  
$$
z \approx 3.67.
$$

**F)** Since this is a right-tail test, the p-value is  
$$
P(Z>3.67).
$$
Looking this up in a standard normal table or using a calculator gives approximately  
$$
\text{p-value} \approx 0.0001.
$$

**G)** Because the p-value $$ (0.0001) $$ is much smaller than the significance level $$ \alpha = 0.008 $$, there is strong evidence to reject $$ H_{0} $$.  
Thus, the answer is:  
**Yes.**
**A)** The level of significance is 0.008. **B)** $$ H_{0}: p=0.03 $$ versus $$ H_{1}: p>0.03 $$. **C)** This is a right-tail test. **D)** $$ \hat{p}_{96} \sim N(0.03, 0.0174) $$. **E)** The z-score is approximately 3.67. **F)** The p-value is approximately 0.0001. **G)** Yes, the p-value suggests strong evidence to reject $$ H_{0} $$.

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