Question

113 The margin of error for the 95% confidence interval of the mean fill is. 2 A 0.48 3 B 0.41 4 C 0.36 5 D 0.32 714 As a class project, each of 280 students taking E270 is required to obtain a sample of n = 100 students and build a B 95% confidence interval for the distance travelled to the campus. The instructor thus receives 280 different interval e estimates. The instructor would expect_ ofthese intervals to capture the population mean distance travelled. DA 280 266 100 а в 95 5 Questions 15-19 are related to the following 5 To build an interval estimate for the proportion (percentage) of American households who live paycheck-to-paycheck, a random sample 700 full-time workers showed 546 live paycheck-to-paycheck. 3.15 The point estimate of the population proportion is А 0.74 B 0.76 0.78 D 0.80 16 The standard error of the sample proportion is Η Α 0.0231 Тв 0.0208 C 0.0196 D 0.0157 А 17 The interval estimate for the population proportion with a 90% confidence level is 0.761 0.799 0.754 0.806 0.744 0.816 0.734 0.826 B 18 In the previous question, the margin of error for the intervals such that in 19 out of 20 cases such intervals would capture the actual population proportion of households who live paycheck-to-paycheck is A 0.036 B 0.031

Answer 1

Solution-:

.Ans :15-17 Let, Number of Americans households who live paycheck- to- paycheck

Given :

(15) The point estimate of population proportion is, $p=\frac{x}{n}=\frac{546}{700}=0.78$

Therefore, Option (C) 0.78 is correct.

(16) The standard error of sample proportin is,

  $SE_{p}=\sqrt{\frac{p*q}{n}}=\sqrt{\frac{0.78*0.22}{700}}=0.0157$

Therefore, Option (D) 0.0157 is correct.

(17) The 90% C.I. for population proportion is ,

$(p-Z_{\alpha/2}*SE_{p},p+Z_{\alpha/2}*SE_{p})$

Where,

  

The required C.I. is

Therefore, Option (B) is correct.

For (Q. 18)

Here, $X=19,n=20$

$\therefore p=\frac{x}{n}=\frac{19}{20}=0.95$ ,

  $SE_{p}=\sqrt{\frac{p*q}{n}}=\sqrt{\frac{0.95*0.05}{200}}=0.049$