6) An investment website can tell what devices are used to access the site. The

Question

6) An investment website can tell what devices are used to access the site. The site managers wonder whether they should enhance the facilities for trading via "smartphones" so they want to estimate the proportion of users who access the site that way. They draw a random sample of 200 investors from their customers. Suppose that the population proportion of smart phone users is 36%.

6a) What is the value for the standard error of the sampling distribution of the proportion of smart phone users be?

6b) What is the probability that the sample proportion of smart phone users is greater than 0.36?

6c) What is the probability that sample proportion is between 0.30 and 0.40?

6d) What is the probability that the sample proportion is less than 0.28?

6e) What is the probability that the sample proportion is greater than 0.42?

Explanation / Answer

6a)

p=0.36, n=200

standard error = sqrt(p*(1-p)/n)) = sqrt(0.36*0.64/200) = 0.0339

6b) the probability that the sample proportion of smart phone users is greater than 0.36

z-score corresponding to 0.36 = 0,

probability that the sample proportion of smart phone users is greater than 0.36 = 0.5

6c) What is the probability that sample proportion is between 0.30 and 0.40?

z-scores for 0.30 and 0.40 are,

z1 = (0.30-0.36)/0.0339 = -1.7699

z2 = (0.40-0.36)/0.0339 = 1.1799

P( sample proportion is between 0.30 and 0.40) = P(z1<z<z2) = P(z<z2) - P(z1<z) = 0.8809-0.0383 = 0.8426

6d) What is the probability that the sample proportion is less than 0.28?

z-scores for 0.28, z1 = (0.28-0.36)/0.0339 = -2.3598

P( sample proportion is less than 0;28) = P(z <  -2.3598) = 0.00914

6e) What is the probability that the sample proportion is greater than 0.42?

z-scores for 0.42, z1 = (0.42-0.36)/0.0339 = 1.7699

P( sample proportion is greater than 0.42) = P(z > 1.7699) 1 - P(z< 1.7699) = 1- 0.9616 = 0.0384

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