A company that packages peanuts states that at a maximum 6% of the peanut shells contain...

A company that packages peanuts states that at a maximum 6% of the peanut shells contain no nuts. At random, 300 peanuts were selected and 21 of them were empty.

With a significance level of 1%, can the statement made by the company be accepted? What is the P-Value?

Solutions

Expert Solution

Solution:

Point estimate = sample proportion = = x / n = 0.07

This a right (One) tailed test.

The null and alternative hypothesis is,

Ho: p = 0.06

Ha: p 0.06

Test statistics

z = ( - ) / *(1-) / n

= ( 0.07 - 0.06) / (0.06*0.94) /300

= 0.729

P-value = P(Z > z )

= 1 - P(Z < 0.729 )

= 1 - 0.7670

= 0.2330

orchestra answered 1 year ago

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