Question

A manufacturer of detergent claims that the contents of boxes sold weigh on average at least 20 ounces. The distribution of weight is known to be​ normal, with a standard deviation of 0.4 ounces. A random sample of 25 boxes yielded a sample mean weight of 19.87 ounces. Test at the 1​% significance level the null hypothesis that the population mean weight is at least 20 ounces.

Answer 1

Answer:

$z=\frac{19.87-20}{\frac{0.4}{\sqrt{25}}}=-1.625$    

The p value for this case would be given by:

$p_v =P(z$  

Since the p value is higher than the significance level provided we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly less than 20 ounces.

Step-by-step explanation:

Information provided

$\bar X=19.87$ represent the sample mean

$\sigma=0.4$ represent the population deviation

$n=25$ sample size  

$\mu_o =68$ represent the value that we want to test

$\alpha=0.01$ represent the significance level

z would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to test if the true mean is at least 20 ounces, the system of hypothesis would be:  

Null hypothesis:$\mu \geq 20$  

Alternative hypothesis:$\mu < 20$  

The statistic is given by:

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

Replacing the info given we got:

$z=\frac{19.87-20}{\frac{0.4}{\sqrt{25}}}=-1.625$    

The p value for this case would be given by:

$p_v =P(z$  

Since the p value is higher than the significance level provided we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly less than 20 ounces.