please answer all parts (a-c)- this is one giant problem: A random sample of 20

Question

please answer all parts (a-c)- this is one giant problem:

A random sample of 20 recent weddings in a country yielded a mean wedding cost of $26,344.27. Assume that recent wedding costs in this country are normally distributed with a standard deviation of $8000. Complete parts (a) through (c) below. a. Determine a 95% confidence interval for the mean cost, ' of all recent weddings in this country The 95% confidence interval is from $1 to $ (Round to the nearest cent as needed.) b. Interpret your result in part (a). Choose the correct answer below 0 A. We can be 95% confident that the mean cost, , of all recent weddings in this country is somewhere within the confidence interval. B. We can be 95% confident that any randomly selected recent wedding cost in this country is somewhere within the confidence interval ° C. The mean cost, . of all recent weddings in this country is somewhere within the confidence interval. 0 D. We can be 95% confident that the sample cost of recent weddings in this country is somewhere within the confidence interval

Explanation / Answer

a) 95% Interval

Margin of error = (1.960*8000)/Sqrt(20) = 3506.09

95% Interval

(26344.27-3506.09, 26344.27+3506.09)

= (22838.18, 29850.36)

b) Option A is correct

c) Option C is correct

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.