Question

Let X1 be an observation from an exponential distribution with the p.d.f.

 

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Test the null hypothesis that the mean of the distribution is equation against the alternative hypothesis that is equation. The null hypothesis is accepted if and only if the observed value of the random variable is less than 3. Find the probabilities of type-I and type-II errors.
b) The mean and standard deviation of 20 items is found to be 10 and 2 respectively. At the time of checking it was found that one item having value 8 was incorrect. Calculate the mean and standard deviation if the wrong item is omitted.

09 Jan 2026
Answer :
Word Count : 429
For part (a), we have a single observation (X_1) from an exponential distribution with pdf [ f(x) = \frac{1}{\theta} e^{-x/\theta}, \quad x>0 ] The hypotheses are: [ H_0: \theta = 2, \quad H_1: \theta = 5 ] The decision rule is: accept (H_0) if (X_1 < 3). Type-I error ((\alpha)): rejecting (H_0) when it is true. This occurs when (X_1 \ge 3) under (H_0). The probability is [ \alpha = P(X_1 \ge 3 \mid \theta = 2) = 1 - P(X_1 < 3 \mid \theta=2) = 1 - F_{X_1}(3) ] The CDF of an exponential distribution is [ F(x) = 1 - e^{-x/\theta} ] So, [ \alpha ______ ____ _______ __________ ___ ______ ____ _______ ________ _______.
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