Statistics Questions

QNo.1 Answer As claimed by the company let us consider the null hypothesis to be, H0 : \mu \ge 19.8μ≥19.8 against the alternate hypothesis H1 : \mu < 19.8μ<19.8 where \muμ = population mean weight of the durian crackers. But since this is a left tailed test the rejection region is (- \infin∞, Z (1 - \alpha)(1−α)] Let us assume the level of significance = 0.05 So the value of Z (1 - \alpha)(1−α) from the standard normal distribution table we get, Z (1 - 0.05)(1−0.05) = -1.6449 Hence the rejection region becomes (- \infin∞, -1.6449] QNo.2 Answer. ANSWER 1) Let us consider the null hypothesis H0 : \mu = 300μ=300 against the claim of the researcher H1 : \mu < 300μ<300 . Test Criteria : Reject the null hypothesis H0 if Zcalc < Z\alphaα Given that, Sample quantity N = 50 Sample mean U = 296 Standard deviation \sigma = 5σ=5 Considering the population mean \mu = 300μ=300 We can find Zcalc = \frac{U-\mu}{\frac{\sigma}{\sqrt N}}NσU−μ Zcalc = \frac{296-300}{\frac{5}{\sqrt 50}} =...

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Step by step solution attached. Q1) we use t-test to conduct the hypothesis. Since it is left tailed test, we ne...

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