Statistics

(b) A manufacturer claims that at most 5 5 5 percent of the time a given product will sustain fewer than 1000 1000 1000 hours of operation before requiring service. Twenty products were selected randomly from the production line and tested. It was found that three of them required service before 1000 1000 1000 hours of operation. Comment on the manufacturer's claim. [6] Statistical Hypothesis Testing: Claim Assessment To evaluate the manufacturer's claim, we employ the binomial distribution where the probability of success (product failure before 1000 hours) is p = 0.05 p = 0.05 p = 0.05 . We define the random variable $X$ as the number of products requiring service in a sample of size n = 20 n = 20 n = 20 . 1. Hypothesis Formulation Null Hypothesis ( $H_0$ ): p $p \leq 0.05$ (Claim is true) Alternative Hypothesis ( $H_1$ ): p > 0.05 p > 0.05 p > 0.05 2. Calculation of Probability We calculate the probability of observing 3 or more failures given p = 0.05 p = 0.05 p = 0.05 using the binomial formula: P ( X = k ) = ( n k ) p k ( 1 − p ) n − k P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} P ( X = k ) = ( k n ​ ) p k ( 1 − p ) n − k . The probability of observing 3 or more failures, P ( X $P(X \geq 3)$ , is 1 − P ( X < 3 ) 1 - P(X < 3) 1 − P ( X < 3 ) : $P(X < 3) = P(X=0) + P(X=1) + P(X=2)$ P ( X = 0 ) = ( 20 0 ) ( 0.05 ) 0 ( 0.95 ) 20 ≈ 0.3585 P(X=0) = \binom{20}{0} (0.05)^0 (0.95)^{20} \approx 0.3585 P ( X = 0 ) = ( 0 20 ​ ) ( 0.05 ) 0 ( 0.95 ) 20 ≈ 0.3585 P ( X = 1 ) = ( 20 1 ) ( 0.05 ) 1 ( 0.95 ) 19 ≈ 0.3774 P(X=1) = \binom{20}{1} (0.05)^1 (0.95)^{19} \approx 0.3774 P ( X = 1 ) = ( 1 20 ​ ) ( 0.05 ) 1 ( 0.95 ) 19 ≈ 0.3774 P ( X = 2 ) = ( 20 2 ) ( 0.05 ) 2 ( 0.95 ) 18 ≈ 0.1887 P(X=2) = \binom{20}{2} (0.05)^2 (0.95)^{18} \approx 0.1887 P ( X = 2 ) = ( 2 20 ​ ) ( 0.05 ) 2 ( 0.95 ) 18 ≈ 0.1887 P ( X < 3 ) = 0.3585 + 0.3774 + 0.1887 = 0.9246 P(X < 3) = 0.3585 + 0.3774 + 0.1887 = 0.9246 P ( X < 3 ) = 0.3585 + 0.3774 + 0.1887 = 0.9246 P ( X $P(X \geq 3) = 1 - 0.9246 = 0.0754$ 3. Conclusion Using a standard significance level of $\alpha = 0.05$ , we observe that P ( X $P(X \geq 3) = 0.0754 > 0.05$ . Since the p-value is greater than the significance level, we fail to reject $H_0$ . There is insufficient statistical evidence at the 5% level to refute the manufacturer's claim. 5 A pharmaceutical company ABC recently launched a new medicine to provide an early recovery to severe headache patients. The company has announced that their medicine named NewMed provides, on average, early recovery than the existing medicine ExMed. Following table shows the recovery times of 13 such patients: Medicine Recovery Times ExMed 12, 23, 22, 12, 13, 14 NewMed 12, 9, 11, 10, 9, 8, 10

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