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Effectus [21]
3 years ago
7

A textile fiber manufacturer is investigating a new drapery yarn, which the company claims has a mean thread elongation of 12 ki

lograms with a standard deviation of 0.5 kilograms. The company wishes to test the hypothesis LaTeX: H_0=\mu=12H0=μ=12 against LaTeX: H_1=\mu>12H1=μ>12 using a random sample of specimens. Calculate the P-value if the observed statistic is. Round your final answer to five decimal places
Mathematics
1 answer:
shepuryov [24]3 years ago
5 0

Answer

given,

mean = 12 Kg

standard deviation = 0.5 Kg

assume the observed statistic is = 11.1

 now, \bar{X}=11.1 , \mu = 12 , \sigma = 0.5

assuming the number of sample = 4

n = 4

Hypothesis test:

H₀ : μ≥ 12

Ha : μ < 12

now,

significant level α = 0.05

z* = \dfrac{\bar{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}

z* = \dfrac{11.1-12}{\dfrac{0.5}{\sqrt{4}}}

     z* = -3.60

Test statistics, Z* = -3.60

P-value

P(Z<-3.60) = 0.002 (from z- table)

P- value = 0.002

now,

reject the value of H₀ when P-value < α

    0.002 < 0.05

since, it is less P-value < α , we have to reject the null hypothesis

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Step-by-step explanation:

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A random sample of 225 orders will be used to estimate the proportion of first-time-customers.

This means that n = 225

Mean and standard deviation:

\mu = p = 0.37

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.37*0.63}{225}} = 0.0322

What is the probability that the sample proportion is between 0.26 and 0.38?

This is the pvalue of Z when X = 0.38 subtracted by the pvalue of Z when X = 0.26.

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Z = \frac{X - \mu}{s}

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