Question

the data given to the right includes data from 46 candies, and 9 of them are red. the company that makes the candy claims that 31% of its candies are red. use the sample data to construct a 90% confidence interval estimate of the percentage of red candies. what do you conclude about the claim of 31%? weights (g) of a sample bag of candy construct a 90% confidence interval estimate of the population percentage of candies that are red. % < p < % (type an integer or decimal rounded to one decimal place as needed.)

Explanation:

Step1: Calculate sample proportion $\hat{p}$

Sample proportion $\hat{p}=\frac{x}{n}$, where $x = 9$ (number of red candies) and $n=46$ (total candies). So, $\hat{p}=\frac{9}{46}\approx0.196$.

Step2: Find $z -$score for $90\%$ confidence interval

For a $90\%$ confidence interval, the significance level $\alpha=1 - 0.90=0.10$, and $\alpha/2=0.05$. The $z -$score $z_{\alpha/2}=z_{0.05}$. From the standard normal table, $z_{0.05} = 1.645$.

Step3: Calculate margin of error $E$

The formula for margin of error $E=z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$. Substitute $\hat{p}=0.196$, $n = 46$, and $z_{\alpha/2}=1.645$.

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Step4: Calculate confidence interval

The confidence interval for proportion $p$ is $\hat{p}-E

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Convert to percentage: $10.0\%

Answer:

$10.0\%

Since the claimed value of $31\%$ does not lie within the $90\%$ confidence interval $(10.0\%,29.2\%)$, at the $90\%$ confidence level, there is sufficient evidence to reject the company's claim that $31\%$ of its candies are red.