A particular apartment complex advertises that the moan square footage of their

Question

A particular apartment complex advertises that the moan square footage of their apartments is 1000 square feet, but you are suspicious and plan to investigate the advertised claim. Your friends in the complex let you measure their apartments, and from the 9 units you measure, you find the average square footage was 929 square feet. Assume these measurements are normally distributed with a known standard deviation of 80 square feet. a. Construct a 90% confidence interval for the mean square footage. SHOW all calculations. b. Calculate the test statistic and p-value to test H_0: mu = 1000 against HA: mu notequalto 1000. Stale your conclusion at the 5% significance level. Show all calculations. To assess the accuracy of a laboratory scale, a standard weight that is known to weight exactly 1 gram is repeatedly weighed a total of n times and the mean x^bar is computed. Suppose the scale readings are Normally distributed with unknown mean mu but known standard deviation sigma = 0.001 grams. Compute the minimum sample size n needed so that a 99% confidence interval for mu has a margin of error no larger than 0.00025 grams. Show all calculations.

Explanation / Answer

a)
z value at 90% CI= 1.645
sd = 80 , mean = 1000 , n = 9

CI = mean +/ - z * ( s/sqrt(n))
= 1000+ /- 1.645 * ( 80 / sqrt(9))
= (956.13,1043.87)

b)
sd = 80 , mean = 1000 , n = 9 , x = 929

t = ( x- mean) / ( s/ sqrt(n))
= ( 929 - 1000) / (80 / sqrt(9))
= -2.6625

p value is calculated using t = -2.6625 , df = 8
p value = .028716.
Here, we reject the null hypothesis

c)
ME = 0.00025 , sd = 0.001
z value at 99% CI = 2.576

ME = +/- z * (s / sqrt(n))
0.00025 = +/- 2.576 *( 0.001/sqrt(n))
n = 106.172

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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