Formulate an alternative hypothesis for each of the following null hypotheses. (
ID: 3221797 • Letter: F
Question
Formulate an alternative hypothesis for each of the following null hypotheses. (a) H_0: support for a presidential candidate is unchanged after the start use of the use of TV commercials. (b) H_0: The proportion of viewers watching a particular local news channel is less than 30%. (c) H_0: The median grade point average of undergraduate mathematics majors is 2.9 It is suspected that a coin is not balanced (not fair). Let p be the probability of tossing a head. To test H_0:p = 0.5 against the alternative hypothesis H_a: p > 0.5, a coin is tossed 15 times. Let Y equal the number of times a head is observed in the 15 tosses of this coin. Assume the rejection region to be {Y210}. (a) Find alpha. (b) Find beta for p = 0.7. (c) Find beta for p = 0.6. (d) Find the rejection region for {Y greaterthanorequalto K) for alpha = 0.01, and alpha = 0.03 (e) For the alternative H_a: p = 0.7, find beta for the values of alpha given in (d). In Exercise 6.1.3: (a) Assume that the rejection region is {Y greaterthanorequalto 28}. Calculate alpha and beta if p = 06. Compare the results with the corresponding values obtained in Exercise 6.1.3. (This gives the effect of enlarging the rejection region on alpha and beta.) (b) Assume that the rejection region is {Y greaterthanorequalto 28}. Calculate alpha and beta if p = 06 and (i) the coin is tossed 20 times, or (ii) the coin is tossed 25 times (This shows the effect of increasing the sample size on alpha and beta for a fixed rejection region.) Suppose we have a random sample of size 25 from a normal population with an unknown mean mu and a standard deviation of 4. We wish to test the hypothesis H_0: mu = 10 versus H_a: mu >10. Let the rejection region be defined by: reject H_0 if the sample mean X > 11.2. (a) Find alpha (b) Find beta for H_a: mu = 11. (c) What should the sample size be if alpha = 0.01 and beta = 0.2? A process for making steel pipe is under control if the diameter of the pipe has mean 3.0 in. with standard deviation of no more than 0.250in.Explanation / Answer
We need to solve 6.1.3 to get to 6.1.4 -
6.1.3 -
p0 <- 0.5 # expected value under H0
sigma <- sqrt(15*p0*(1-p0)) # theoretical standard deviation
p1 <- 0.7 # expected value under H1
alpha <- 0.03 # probability of type I error
crit <- qbinom(1-alpha, 15, p0)
beta <- pbinom(crit-1, 15,p1) = 0.484
Repeat the above steps for 6.1.4. Sorry I couldn't complete that part!