The following observations resulted when sampling from a normal population:77; 69; 65; 75; 76; 63; 71; 74; 70; 65(a) Find a 98% confidence interval for the population mean. (b) Test H0 mean=75 versus Ha: mean < 75. Use alpha = 0.02.(c) Do the results of part (a) support the conclusion in part (b)?

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Denote the observations `x_i`, `i = 1,2,3,...,10`

Then the mean `bar(x)` is given by

`bar(x) = sum_1^10 x_i/10 = 70.5`

and the standard deviation is given by

`sd(x) = sqrt(V(x)) = sqrt(sum_1^10 (x_i-bar(x))^2/9) = sqrt(24.944) = 4.99`

a) A 98% confidence interval for the population mean `mu` is given by

`hat(u) pm z_0.99 (hat(sigma)/sqrt(n))`

where `sigma` is the population standard deviation and `z_0.99` is the 99th percentile of the standard Normal distribution. Estimating `mu` with `bar(x)` and `sigma` with `sd(x)`, a 98% CI is given by

`bar(x) pm 2.33 (hat(sigma)/sqrt(10)) = 70.5 pm ((2.33)(4.99))/sqrt(10) = [66.8,74.2]`

b) To test

`H_0: mu = 75`  vs. `H_A: mu < 75`

the test statistic is given by

`(hat(mu)-75)/(hat(sigma)/sqrt(n)) = (70.5-75)/(4.99/sqrt(10)) = -2.849`

Testing at the 98% level (`alpha=0.02`) and remembering that this is a one-sided test, we compare the test statistic with `z_0.02 = -2.054`

At the 98% level then we reject the null that `mu=75` given the alternative `mu <75` .

c) Although the hypothesis test in b) is one-sided and the confidence interval in a) is two-sided, the two do necessarily give the same conclusion that `mu` appears to be less than 75. The null is rejected at the 98% level and the 98% CI does not contain the value 75.

a) 98% CI is [66.8,74.2]

b) The null is rejected at the 98% level

c) The results of part a) do agree with the conclusion of part b)

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