Question

deviation of $200. Construct a 95% confidence interval. opulation standard 3. According to a study, 21.1% of 507 female college students were on a of the study. diet at the time a) Construct a 99% confidence interval for the true proportion of all female students who were on a diet at the time of this study. b) Explain what this interval means. c) Is it reasonable to think that only 17 % of college women are on a diet? 4/To assess the accuracy of a laboratory scale, a standard weight that is known to weigh 1 gram is repeatedly weighed 4 times. The resulting measurements (in grams) are: 0.95, 1.02, 1.01, and 0.98. Assume that the weighing by the scale when the true weight is 1 gram are normally distributed with mean u.

Answer 1

4)

solution D Here P = 0.211, N=507, 2=1-0-2115 0-789 at 995. confidence Rapp=258 à the 924 confidence intervals are Þ+ zajes 0211+2-580-211 (0.789) 507 -0.211 to our = (0.164, 0.258) b) we are 99% confident that the true proportion of at the all female Students who were on a diet time of this study. Here p=17% = 0.17 – Here 0.17 es in the stated obtained 99% confidence interval. think that only 17% of - so, it is reasonable to think that only on a diet. college women are

Given,

n = 4

= 3.96/4 = 0.99

s = 0.0316

SE = s/

= 0.0316/ = 0.0158

= 0.05

df = n -1= 4 - 1 = 3

From Table, critical values of t = 3.1824

Confidence interval:

0.99 (3.1824 X 0.0158)

= 0.99 0.0503

= (0.9397 , 1.0403)

So,

Confidence Interval:

0.9397 < < 1.0403