Question

A bag of chips states that it contains 24 oz of chips. A sample of 40 bags finds a mean of 23.7 and a standard deviation of 0.2. With α=0.05, is there evidence that the bags are underfilled?
What is the null hypothesis?


What is the alternative hypothesis?


What is the value of the test statistic?


What is the p-value for the rejection region?


What conclusion do you draw?


What does this mean in terms of the problem situation?

Answer 1

Answer:

a) Null hypothesis:$\mu \geq 24$      

b) Alternative hypothesis:$\mu < 40$      

c) $t=\frac{23.7-24}{\frac{0.2}{\sqrt{40}}}=-9.487$      

d) $df=n-1=40-1=39$  

$p_v =P(t_{39}$  

e) Since the p value is almost 0 we have enough evidence to reject the null hypothesis at the significance level of 0.05

f) Since we reject the null hypothesis we have enough evidence to conclude the true mean is significantly less than 24 ounces

Step-by-step explanation:

Information provided

$\bar X=23.7$ represent the sample mean for the weigth in ounces

$s=0.2$ represent the standard deviation

$n=40$ sample size      

$\mu_o =24$ represent the value to verify

$\alpha=0.05$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

Part a

For this case the null hypothesis is:

Null hypothesis:$\mu \geq 24$      

Part b

And the alternative would be given by the claim:

Alternative hypothesis:$\mu < 40$      

Part c

We don't know the population deviation so then the statistic would be given by:  

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

And replacing the data given we got:

$t=\frac{23.7-24}{\frac{0.2}{\sqrt{40}}}=-9.487$      

Part d

The degrees of freedom are given by:

$df=n-1=40-1=39$  

And the p value for this case would be:

$p_v =P(t_{39}$  

Part e

Since the p value is almost 0 we have enough evidence to reject the null hypothesis at the significance level of 0.05

Part f

Since we reject the null hypothesis we have enough evidence to conclude the true mean is significantly less than 24 ounces