Question #242021

Write the null and alternative hypothesis of given problem:


    1. It is claimed that the average weight of a bag of biscuit is 250 grams with a standard deviation of 20.5 grams. Would you agree to this claim if a random sample of 50 bags of biscuits showed an average weight of 240 grams, using a 5% level of significance.



Expert's answer

Solution:

Null hypothesis, H0:μ=250H_0:\mu=250

Alternative hypothesis, H1:μ≠250H_1:\mu\ne250

Assume that the population standard deviation is σ=20.5\sigma=20.5 . The population standard deviation is known and the sample size n is large (n≥30)(n \geq 30) . Hence, the test statistic is z=xˉ−μσ/nz=\frac{\bar{x}-\mu}{\sigma / \sqrt{n}} , and the value is z=xˉ−μσ/n=240−25520.5/50=−5.174z=\frac{\bar{x}-\mu}{\sigma / \sqrt{n}}=\frac{240-255}{20.5 / \sqrt{50}}=-5.174 . The test is two tailed (because H1 contains inequality ≠\neq ) and the critical values are z−0.025=−1.96z_{-0.025}=-1.96 and z0.025=1.96z_{0.025}=1.96 . Since z<−z0.025z<-z_{0.025} , i.e. the z lies to the left of the critical value, we reject the null hypothesis.




We would not agree with a claim, the true average weight of a bag of biscuits is not 250 grams. Answer: the true average weight of a bag of biscuits is not 250 grams.


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