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Textbook (Online ision) QUESTION 1
A random sample of 13 mathematics dictionaries had an average price of $$ 46.52$ with a sample standard deviation of $$ 6.83$ . Find the $90 %$ confidence for $μ$ , the population mean price of all mathematics dictionaries.
A. $(43.14, 49.90)$
B. $(42.39, 50.65)$
C. $(42.81, 50.23)$
D. $(43.40, 49.64)$
QUESTION 2
The lengths of Atlantic croaker fish are normally distributed, with a mean of 10.4 inches and a standard deviation of 2.3 inches. Let $x$ be a random vari represents the length of an Atlantic croaker fish. Suppose an Atlantic croaker fish is randomly selected. Find the probability that length of the fish is be inches and 12 inches
A. 0.397
B. 0.153
C. 0.603
D. 0.138
QUESTION 3
In a survey of 1200 adults, 562 said that they plan to watch the next Olympics on television. Using these sample statistics, calculate the margin of erro $95 %$ confidence interval for the proportion of all adults that plan to watch the next Olympics on television
A. 0.039
B. 0.024
C. 0.014
D. $\cap \cap 28$
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CHAPTER 15 SU ....docx
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$4 \frac{13}{13}$
is 10
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(17 $d +$

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Answer

Margin of Error = z_c * $\sqrt{\frac{p (1 - p)}{n}}$ = 0.0244

Steps

Step 1 ：QUESTION 1: n=13, $\bar{x} = 46.52$ , s=6.83, $α = 0.10$ , t_c=1.782

Step 2 ：CI = $(\bar{x} - t_{c} \frac{s}{\sqrt{n}}, \bar{x} + t_{c} \frac{s}{\sqrt{n}})$ = $(43.14, 49.90)$

Step 3 ：QUESTION 2: $μ = 10.4$ , $σ = 2.3$ , P(9

Step 4 ：QUESTION 3: n=1200, x=562, p= $\frac{x}{n}$ =0.4683, $α = 0.05$ , z_c=1.96

Step 5 ：CI = $(p - z_{c} \sqrt{\frac{p (1 - p)}{n}}, p + z_{c} \sqrt{\frac{p (1 - p)}{n}})$ = $(0.4439, 0.4927)$

Step 6 ：Margin of Error = z_c * $\sqrt{\frac{p (1 - p)}{n}}$ = 0.0244