Question

A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 409 gram setting. It is believed that the machine is underfilling the bags. A 42 bag sample had a mean of 404 grams. Assume the population standard deviation is known to be 24. A level of significance of 0.01 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

Answer:

$z=\frac{404-409}{\frac{24}{\sqrt{42}}}=-1.35$  

The p value for this case is given by:

$p_v =P(z$  

For this case the p value is higher than the significance level given so we have enough evidence to FAIL to reject the null hypothesis and we can't conclude that the true mean is significantly less than 409

Step-by-step explanation:

Information given

$\bar X=404$ represent the sample mean

$\sigma=24$ represent the population standard deviation

$n=42$ sample size  

$\mu_o =409$ represent the value to verify

$\alpha=0.01$ represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

$p_v$ represent the p value

Hypothesis to test

We want to verify if the true mean is less than 409, the system of hypothesis would be:  

Null hypothesis:$\mu \geq 409$  

Alternative hypothesis:$\mu < 409$  

The statistic for this case is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)  

Replacing the info we got:

$z=\frac{404-409}{\frac{24}{\sqrt{42}}}=-1.35$  

The p value for this case is given by:

$p_v =P(z$  

For this case the p value is higher than the significance level given so we have enough evidence to FAIL to reject the null hypothesis and we can't conclude that the true mean is significantly less than 409