Here are the problems written out, in case you don't have the text:

• 6.14: Let the observed value of the mean X-bar of a random sample of size 20 from a distribution that is N(mu, 80) be 81.2. Find a 95 percent confidence interval for mu.
• 6.16: Let a random sample of size 17 from the normal distribution N(mu, sigma^2) yield x-bar = 4.7 and s^2 = 5.76. Determine a 90 percent confidence interval for mu.
• 6.26: It is known that a random variable X has a Poisson distribution with parameter mu. A sample of 200 observations from this population has a mean equal to 3.4. Construct an approximate 90 percent confidence interval for mu.
• 6.28: Let X_1, X_2, ..., X_n be a random sample from N(mu, sigma^2) where both parameters mu and sigma^2 are unknown. A confidence interval for sigma^2 can be found as follows. We know that (n * S^2)/(sigma^2) is chi-squared with n-1 df. Thus we can find constants a and b so that P ( n S^2/sigma^2 < b ) = 0.975 and P( a < n S^2/sigma^2 < b ) = 0.95.
  (a) Show that this second probability statement can be written as P( n S^2/b < sigma^2 < n S^2/a ) = 0.95.
  (b) If n = 9 and s^2 = 7.63, find a 95 percent confidence interval for sigma^2.
  (c) If mu is known, how would you modify the preceding procedure for finding a confidence interval for sigma^2?
• 6.30: Let two independent random samples, each of size 10, from two normal distributions N(mu_1, sigma^2) and N(mu_2, sigma^2) yield x-bar = 4.8, (s_1)^2 = 8.64, y-bar = 5.6, (s_2)^2 = 7.88. Find a 95 percent confidence interval for mu_1 - mu_2.
• 6.42: Let us assume that the life of a tire in miles, say X, is normally distributed with mean theta and standard deviation 5000. Past experience indicates that theta = 30,000. The manufacturer claims that the tires made by a new process have mean theta > 30,000, and it is very possible that theta = 35,000. let us check his claim by testing H_0: theta = 30,000 against H_1: theta > 30,000. We shall observe n independent values of X, say x_1, ..., x_n, and we shall reject H_0 (thus accept H_1) if and only if x-bar >= c. Determine n and c so that the power function K(theta) of the test has the values K(30,000) = 0.01 and K(35,000) = 0.98.
• 6.45: Let Y_1 < Y_2 < Y_3 < Y_4 be the order statistics of a random sample of size n=4 from a distribution with pdf f(x; theta) = 1/theta, 0 < x < theta, zero elsewhere, where 0 < theta. The hypothesis H_0: theta = 1 is rejected and H_1: theta > 1 is accepted if the observed Y_4 > c.
  (a) Find the constant c so that the significance level is alpha = 0.05.
  (b) Find the power function of the test.
• 6.46: Assume that the weight of cereal in a "10-ounce box" is N ( mu, sigma^2). To test H_0: mu = 10.1 against H_1: mu > 10.1, we take a random sample of size n = 16 and observe that x-bar = 10.4 and s = 0.4.
  (a) Do we accept or reject H_0 at the 5 percent significance level?
  (b) What is the approximate p-value of this test?
• 6.48: Among the data collected for the World Health Organization air quality monitoring project is a measure of suspended particles in mcg/(m^3). let X and Y equal the concentration in mcg/(m^3) of suspended particles in the city center (commercial district) for Melbourne and Houston, respectively. Using n=13 observations of X and m=16 observations of Y, we shall test H_0: mu_X = mu_Y against H_1: mu_X < mu_Y.
  (a) Define the test statistic and critical region, assuming the variances are equal. Let alpha = 0.05.
  (b) If x-bar = 72.9, s_x = 25.6, y-bar = 81.7, and s_y = 28.3, calculate the value of the test statistic and state your conclusion.
• 6.61: It is proposed to fit the Poisson distribution to the following data:
  

  x	|	0	1	2	3	x > 3
  Frequency	|	20	40	16	18	6

  (a) Compute the corresponding chi-square goodness-of-fit statistic. Hint: In computing the mean, treat x > 3 as x = 4.
  (b) How many degrees of freedom are associated with this chi-square?
  (c) Do these data result in the rejection of the Poisson model at the alpha = 0.05 significance level?