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Statistics Questions

  • From Mathematics, Statistics
  • Due on 26 Apr, 2021 12:05:00
  • Asked On 25 Apr, 2021 08:14:29
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1. A company sells packets of cheese crackers that they claim on average, contain at least 20g of cheese crackers. They admit that their claim is wrong (and will refund any money), if a sample of 40 cheese crackers have a mean less than 19.8g. If the standard deviation is 0.5g, determine the critical value that specifies the rejection region. 

2. A researcher wishes to test the claim of a particular cereal manufacturer that the mean weight of cereal in the boxes is less than 300g. A sample of 50 boxes yields a sample mean weight of 296g. Assume that the population standard deviation is 5g.

a) Can we conclude that the claim is true? Test at α = 0.05.

b) Obtain a 95% confidence interval for μ. 

3. The time taken to assemble a car in a certain plant has a normal distribution with mean of 25.4 hours and a standard deviation of 4.1 hours. Calculate the probability that a car can be assembled at this plant in the following period of time:

a) More than 28.8 hours 

b) Between 18.6 and 27.5 hours 

c) Between 25.0 and 34.0 hours

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[Solved] Statistics Questions Answers

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  • Submitted On 25 Apr, 2021 09:11:23
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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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[Solved] Step solution for hypothesis test and CI

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  • Submitted On 25 Apr, 2021 09:14:40
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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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