IGNOU MTE 11 2026 - Let X1 be an observation from an exponential distribution with th

Question

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

$$f(x) = \frac{1}{\theta}e^{-x/\theta}; x > 0$ $
Test the null hypothesis that the mean of the distribution is $\theta = 2$ against the alternative hypothesis that is $\theta = 5$ . 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 = 1 - (1 - e^{-3/2}) = e^{-3/2} ] Numerically: [ e^{-3/2} = e^{-1.5} \approx 0.2231 ] So, (\alpha \approx 0.223). Type-II error ((\beta)): accepting (H_0) when (H_1) is true. This occurs when (X_1 < 3) under (\theta = 5). [ \beta =

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