A company produces millions of 1-pound packages of bacon every week. Company specifications allow for no more than 3 percent of the 1-pound packages to be underweight. To investigate compliance with the specifications, the company's quality control manager selected a random sample of 1,000 packages produced in one week and found 40 packages, or 4 percent, to be underweight. Assuming all conditions for inference are met, do the data provide convincing statistical evidence at the significance level of α= 0.05 that more than 3 percent of all the packages produced ir one week are underweight?

a. Yes, because the sample estimate of 0.04 is greater than the company specification of 0.03.
b. Yes, because the p-value of 0.032 is less than the significance level of 0.05.
c. Yes, because the p-value of 0.064 is greater than the significance level of 0.05.
d. No, because the p-value of 0.032 is less than the significance level of 0.05.
e. No, because the p-value of 0.064 is greater than the significance level of 0.05.

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jtellezd jtellezd · Answer 1 · 2021-05-20T04:37:14+02:00

Answer:

We accept H₀ we don´t have enough evidence to support that more than 3 % of the packages are undeweight

Step-by-step explanation:

Sample size n = 1000

x ( defective packages found) = 40

p = 40 /1000 p = 0,04 p = 4 % then q = 0,96

np = 0,04 * 1000 and nq / 0.96 * 1000 both greater than 5

Then we can use the approximation of binomial distribution to normal distribution.

α is significance level α = 0,05

From z- table we find z(c) for α = 0,05 z(c) = 1,64

Test Hypothesis

Null Hypothesis H₀ p = 0,03

Alternative Hypothesis Hₐ p > 0,03

We can see from alternative Hypothesis that the test is a one tail test to the right of the bell shape curve of the normal distribution

To calculate z(c) = ( 0,04 - 0,03 ) / √ ( p*q)/n

z(c) = 0,01 * 31,62 / √ 0,04*0,96

z(c) = 0,3162 / 0,1959

z(c) = 1,6140 z(c) ≈ 1,61

Comparing z(s) and z(c)

z(c) > z(s) 1.64 > 1.61

Then z(s) is in the acceptance region for H₀ we accept H₀