Sampling Distribution<\/p>\n
Student\u2019s Name<\/p>\n
Institution Affiliation<\/p>\n
Course Name and Code<\/p>\n
Professor\u2019s Name<\/p>\n
Date<\/p>\n
Sampling Distribution<\/p>\n
The sample proportion for each sample size.<\/p>\n
Sample proportion=X\/n<\/p>\n
Sample Size (n) Number Correctly Scanned (X) Sample Proportion <\/p>\n
25 24 0.96<\/p>\n
50 49 0.98<\/p>\n
100 95 0.95<\/p>\n
200 193 0.965<\/p>\n
The single-proportion sampling error for each sample size.<\/p>\n
Single-proportion sampling error=\u221a ((P(1-P))\/n)<\/p>\n
Sample Size (n) Number Correctly Scanned (X) Sample Proportion Single-proportion sampling error<\/p>\n
25 24 0.96 0.039191836<\/p>\n
50 49 0.98 0.01979899<\/p>\n
100 95 0.95 0.021794495<\/p>\n
200 193 0.965 0.012995191<\/p>\n
The probability of finding 198 correctly scanned packages, for a sample of size n=200.\u00a0<\/p>\n
Since the computer scanner can read a bar code on a package correctly 97%, p=0.97<\/p>\n
Excel formula<\/p>\n
=BINOM.DIST(198, 200, 0.97, FALSE)<\/p>\n
= 0.043042647<\/p>\n
Thus, the probability of finding 198 correctly scanned packages, for a sample of size n=200 is 0.043042647<\/p>\n","protected":false},"excerpt":{"rendered":"
Sampling Distribution Student\u2019s Name Institution Affiliation Course Name and Code Professor\u2019s Name Date Sampling Distribution The sample proportion for each<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[],"tags":[],"class_list":["post-37688","post","type-post","status-publish","format-standard","hentry"],"yoast_head":"\n