Question

a manufacturer of cookies and crackers does a small survery of the age at sale of one of its brands. A tandom sample of 225 retail markets in a particular region is chosen. In each store, the number of days since manufacture of the frontmost box of crackers is determined by date code on the box. The samples mean is 56.7 and the standard deviation is 30. Can we infer at the 10% significance level that the true mean age is less than 60?

Answer 1

Answer:

$t=\frac{56.7-60}{\frac{30}{\sqrt{225}}}=-1.65$    

$df=n-1=225-1=224$  

The p value would be:

$p_v =P(t_{(224)}$  

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we cna conclude that the true mean is less than 60 at 10% of significance.

Step-by-step explanation:

Information given

$\bar X=56.7$ represent the sample mean of interest

$s=20$ represent the sample standard deviation

$n=225$ sample size  

$\mu_o =60$ represent the value to check

$\alpha=0.1$ represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test

System of hypothesis

we want to check if the true mean is lower than 60, so thn the system of hypothesis are:

Null hypothesis:$\mu \geq 60$  

Alternative hypothesis:$\mu < 60$  

The statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$  (1)  

Replcing into the formula we got:

$t=\frac{56.7-60}{\frac{30}{\sqrt{225}}}=-1.65$    

P value

The degrees of freedom are given by:

$df=n-1=225-1=224$  

The p value would be:

$p_v =P(t_{(224)}$  

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is less than 60 at 10% of significance.