Statistics

[1] Mars states that the population proportion of blue M&Ms is 0.240. When approximating the binomial distribution as NORMAL, what is the probability of calculating a sample proportion of 0.237 or less blue M&Ms from a sample size of n = 70?

mu = np= 0.24*70 = 16.8 sd = sqrt(npq) = sqrt(70*0.24*(1-0.24)) = 3.5732 x = 0.237*70 = 16.59 Need to find P(x <= 16.59) =? z = (16.59 - 16.8)/3.5732 = -0.0587 P(x < = 16.59) = P(z < = -0.0587) = 0.4765 (Ans.)

[2] On average, Americans watch 10 complete baseball games per season, with a population standard deviation of 15 games. What is the probability that a sample of 100 Americans watch an average of 12 games or more?

standard error (SE) = σ/sqrt(n) = 15/sqrt 100 = 1.5  z-score = (12-15)/1.5 = -2  we want to find P(z> - 2)

So looking the table we get 97.725%

[3] Assume that it is true that people cannot tell the difference between Coke and Pepsi. If a sample of 25 people participated in a blind taste test, can we assume that the sampling distribution of the proportion of people that correctly chose Pepsi is approximately normal? (Why or why not?)

No, because there are only 2 possible outcomes, Pepsi or Coke.  This means that the distribution is discrete. Normal distribution only applies to continuous scenarios, where there is a range between 2 extremes.  In this case there is no range between 2 extremes, only 2 extremes.

[4] The average length of stay at a local hospital is 4.9 days. Assuming a population standard deviation of 3.5 days, what is the probability a random sample of 50 patients has an average stay of 5.2 days or less?

P[ average < 5.2}  =P[ (X - 4.9)/3.5/sqrt 50)< (5.2 - 4.9)/3.5/sqrt 50)]  =P[ Z< (5.2 - 4.9)/3.5/sqrt 50)]  =P[ Z< 0.6061]  =0.7291

[5] In an application to estimate the mean number of miles that employees commute to work roundtrip each day, the following information is given: n = 20; sample mean = 4.33; sample standard deviation of s = 3.50 miles. Assuming the parent population is normal, what is the upper limit for a 95 percent confidence interval?

 n = 20 x-bar = 4.33 standard deviation, σ = 3.5 Margin of error, E = s.e.× Zα  s.e. = σ / √n = 3.5/√20 = 0.7826 Zα = 1.96 at 95% confidence So, E = 0.7826*1.96 = 1.5339 = 1.53 (approx.) A 95% confidence interval for the mean amount spent daily per person at the park is (4.33 - 1.53, 4.33+1.53) = (2.8, 5.86)  So the upper limit for a 95 percent confidence interval is 5.86 miles.

[6] Suppose you want to estimate the population proportion of statistics nerds at UMUC. To do so, you take a sample of 100 students and find that 20 of those students were bona-fide, genuine statistics nerds. What is the lower limit of the 90% confidence interval around your sample proportion?

[7] Assume the GPA of STAT 200 students is distributed normally with a population standard deviation of .10. To test whether the average GPA of Stats II students is GREATER THAN 3.12, a sample of size 49 was drawn and a sample mean of 3.15 was calculated. What is the test statistic?

[8] Using the information from Question [7], what is the p-value?

[9] What is your conclusion to the test in [7] at the 0.05 significance level?

[10] When conducting a hypothesis test, is it possible to keep the probability of a Type I error constant while decreasing the probability of a Type II error?

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