A random sample of 65 bags of white cheddar popcorn​ weighed, on​ average, 5.74 ounces with a standard deviation of 0.26 ounce. Test the hypothesis that muequals5.8 ounces against the alternative​ hypothesis, muless than5.8 ​ounces, at the 0.10 level of significance.

Respuesta :

Answer:

We failed to accept null hypothesis

Step-by-step explanation:

x= 5.74

Standard deviation = [tex]0.26[/tex]

[tex]H_0:\mu = 5.8\\H_a:\mu < 5.8[/tex]

n = 65

We will use z test

Formula : [tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

Substitute the values :

[tex]z=\frac{5.74-5.8}{\frac{0.26}{\sqrt{65}}}[/tex]

[tex]z=-1.86[/tex]

Refer the z table for p value

p value = 0.0314

α = 0.10

p value < α

So, we failed to accept null hypothesis

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