Mathematics: A manufacturer of a new nicotine nasal spray claims that their product has a 30% success rate for smoking cessation. In a clinical study involving 150 smokers, 93 of them quit smoking. Test the hypothesis that the success rate claimed by the manufacturer is valid at the 5% level of significance.

it s between c or dStep-by-step explanation:...

A manufacturer of a new nicotine nasal spray claims

lindsayb2000

22.04.2021

we reject null and conclude that this manufacturers claim is false

Step-by-step explanation:

p = 30% = 0.30

p^ = 93/150 = 0.62

we state the hypothesis

H0: p = 0.30

h1: p not equal to 0.30

we find the z test stattistics

$z=\frac{p^--p}{\sqrt{\frac{p(1-p)}{n} } }$

$z=\frac{0.62-0.30}{\sqrt{\frac{0.30(1-0.30)}{150} } }$

$z=\frac{0.32}{\sqrt{\frac{0.30*0.70}{150} } }$

$z=\frac{0.32}{0.03741}$

z = 8.5538

at alpha = 0.05

z-critical = Z₀.₀₅/₂ = Z₀.₀₂₅

= 1.96

we compare z critical with the test statistic

z statistic > z critical so we have to reject H₀ and conclude that the manufacturers claim is not valid at 0.05 level of significance.

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