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Suppose that the 12 observations \({X_1},....,{X_{12}}\)form a random sample from the normal distribution with unknown mean \(\mu \)and unknown variance\({\sigma ^2}\). Describe how to carry out a \(t\)test of the following hypotheses at the level of significance\({\alpha _0} = 0.005\):

\(\begin{array}{l}{H_0}:\;\;\;\mu \ge 3 \\{H_1}:\;\;\;\mu < 3\end{array}\)

Short Answer

Expert verified

Reject the null if\(\bar X \le 3 - 0.078\sqrt {\prod\limits_{i = 1}^{12} {{{\left( {{X_i} - \bar X} \right)}^2}} } \)

Step by step solution

01

Step 1: To determine the \(t\) test of the following hypotheses at  \({\alpha _0} = 0.005\)

Here\(n = 12{\rm{ and }}{X_1}, \cdots ,{X_n}\~N\left( {\mu ,{\sigma ^2}} \right)\)where both\(\mu \)and\({\sigma ^2}\)are unknown.

We wish to test

\(\begin{array}{l}{H_0}:\mu \ge {\mu _0}\;\;\;\\{H_1}:\mu < {\mu _0}\end{array}\)

Here\({\mu _0} = 3\). The test statistic is

\(T = \frac{{\sqrt n \left( {\bar X - {\mu _0}} \right)}}{{\sqrt {\frac{1}{{n - 1}}} _{i = 1}^n{{\left( {{X_i} - \bar X} \right)}^2}}}\)

Which follows a\(t\)distribution with\(n - 1 = 11\)degrees of freedom. We reject the null for small values of\(T\), where the cutoff for being small is the test statistic being smaller than\(T_{n - 1}^{ - 1}\left( {{\alpha _0}} \right) = - 3.106\)

Thus, we reject the null if

\(\begin{array}{c}\bar X \le 3 - \frac{{3.106}}{{12}}\sqrt {\frac{1}{{11}}_{i = 1}^n{{\left( {{X_i} - \bar X} \right)}^2}} \\ = 3 - 0.078\sqrt {\underbrace {12}_{i = 1}{{\left( {{X_i} - \bar X} \right)}^2}} \end{array}\)
Reject the null if\(\bar X \le 3 - 0.078\sqrt {\prod\limits_{i = 1}^{12} {{{\left( {{X_i} - \bar X} \right)}^2}} } \)

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Most popular questions from this chapter

2. Suppose that a random variable X has the F distribution with three and eight degrees of freedom. Determine the value of c such that Pr(X>c) = 0.975

1. Consider again the situation described in Exercise 11 of Sec. 9.6. Test the null hypothesis that the variance of the fusion time for subjects who saw a picture of the object is no smaller than the variance for subjects who did see a picture. The alternative hypothesis is that the variance for subjects who saw a picture is smaller than the variance for subjects who did not see a picture. Use a level of significance of 0.05.

Suppose that a single observation X is taken from the normal distribution with unknown mean μ and known variance is 1. Suppose that it is known that the value of μ must be −5, 0, or 5, and it is desired to test the following hypotheses at the level of significance 0.05:

H0: μ = 0, H1: μ = −5 or μ = 5.

Suppose also that the test procedure to be used specifies rejecting H0 when |X| > c, where the constant c is chosen so that Pr(|X| >c|μ)=0.05.

a. Find the value of C, and show that if X = 2, then will be rejected.

b. Show that if X = 2, then the value of the likelihood function at μ = 0 is 12.2 times as large as its value at μ = 5 and is 5.9 × 109 times as large as its value at μ = −5.

Consider the situation described immediately before Eq. (9.1.12). Prove that the expression (9.1.12) equals the smallestα0such that we would reject H0 at level of significanceα0.

Consider again the conditions of Exercise 2, but suppose now that it is desired to test the following hypotheses:

H0: μ ≤ 0,

H1: μ > 0.

Suppose also that in the random sample of 10,000 observations, the sample means Xn is 0.03. At what level of significance is this result just significant?

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