Question

(A) A small business ships homemade candies to anywhere in the world. Suppose a random sample of 16 orders is selected and each is weighed. The sample mean was found to be 410 grams and the sample standard deviation was 40 grams. Find the 90% confidence interval for the mean weight of shipped homemade candies. (Round your final answers to the nearest hundredth)
(B) When 500 college students are randomly selected and surveyed; it is found that 155 own a car. Find a 90% confidence interval for the true proportion of all college students who own a car.
(Round your final answers to the nearest hundredth)
(C) Interpret the results (the interval) you got in (A) and (B)

Answer 1

The correct answer to the given question is "$\bold{392.47\ < \mu <\ 427.53}$,$\bold{0.28 \ < P <\ 0.34}$, and for Interpret results go to the C part.

Following are the solution to the given parts:

A)

$\to \bold{(n) = 16}$

$\to \bold{(\bar{X}) = 410}$

$\to \bold{(\sigma) = 40}$

In the given question, we calculate $90\%$ of the confidence interval for the mean weight of shipped homemade candies that can be calculated as follows:

$\to \bold{\bar{X} \pm t_{\frac{\alpha}{2}} \times \frac{S}{\sqrt{n}}}$

$\to \bold{C.I= 0.90}\\\\\to \bold{(\alpha) = 1 - 0.90 = 0.10}\\\\ \to \bold{\frac{\alpha}{2} = \frac{0.10}{2} = 0.05}\\\\ \to \bold{(df) = n-1 = 16-1 = 15}$

Using the t table we calculate $t_{ \frac{\alpha}{2}} = 1.753$  When $90\%$ of the confidence interval:

$\to \bold{410 \pm 1.753 \times \frac{40}{\sqrt{16}}}\\\\ \to \bold{410 \pm 17.53\\\\ \to392.47 < \mu < 427.53}$

So $90\%$ confidence interval for the mean weight of shipped homemade candies is between $392.47\ \ and\ \ 427.53$.

B)

$\to \bold{(n) = 500}$

$\to \bold{(X) = 155}$

$\to \bold{(p') = \frac{X}{n} = \frac{155}{500} = 0.31}$

Here we need to calculate $90\%$ confidence interval for the true proportion of all college students who own a car which can be calculated as

$\to \bold{p' \pm Z_{\frac{\alpha}{2}} \times \sqrt{\frac{p'(1-p')}{n}}}$

$\to \bold{C.I= 0.90}$

$\to\bold{ (\alpha) = 0.10}$

$\to\bold{ \frac{\alpha}{2} = 0.05}$

Using the Z-table we found $\bold{Z_{\frac{\alpha}{2}} = 1.645}$

therefore $90\%$ the confidence interval for the genuine proportion of college students who possess a car is

$\to \bold{0.31 \pm 1.645\times \sqrt{\frac{0.31\times (1-0.31)}{500}}}\\\\ \to \bold{0.31 \pm 0.034}\\\\ \to \bold{0.276 < p < 0.344}$

So $90\%$ the confidence interval for the genuine proportion of college students who possess a car is between $0.28 \ and\ 0.34.$

C)

• In question A, We are  $90\%$ certain that the weight of supplied homemade candies is between 392.47 grams and 427.53 grams.
• In question B, We are  $90\%$ positive that the true percentage of college students who possess a car is between 0.28 and 0.34.

Learn more about confidence intervals:  

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