Question

A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 433 gram setting. It is believed that the machine is underfilling the bags. A 42 bag sample had a mean of 425 grams. Assume the population variance is known to be 625. A level of significance of 0.05 will be used. Find the P-value of the test statistic. You may write the P-value as a range using interval notation, or as a decimal value rounded to four decimal places.

Answer 1

Solution:

Here, we have to use one sample z test for the population mean.

The null and alternative hypotheses are given as below:

H₀: µ =433  versus H_a: µ < 433 [Left Tailed Test]

This is a lower tailed test.

The test statistic formula is given as below:

Z = (Xbar - µ)/[σ/sqrt(n)]

From given data, we have

µ = 433

Xbar = 425

= 625

n = 42

α = 0.05

the test statistics

Z = (425– 433)/[25/sqrt(42)

Z = -2.0738

#test statistics=-2.0738

p-value=P(|Z|>2.07384)

P-value = 2*P(Z>2.07384)=2*0.0191=0.0381

#P-value=0.0381

So, we reject the null hypothesis