Question

The manufacturer of an airport baggage scanning machine claims it can handle an average of 530 bags per hour. (a-1) At α = .05 in a left-tailed test, would a sample of 16 randomly chosen hours with a mean of 510 and a standard deviation of 50 indicate that the manufacturer’s claim is overstated? Choose the appropriate hypothesis.

Answer 1

Answer:

Null hypothesis:$\mu \geq 250$        

Alternative hypothesis:$\mu < 250$  

$p_v =P(t_{15}$    

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and the the actual mean is not significantly lower than 530 at 5% of significance.      

Step-by-step explanation:

1) Data given and notation        

$\bar X=510$ represent the mean for the sample    

$s=50$ represent the standard deviation for the sample  

$n=16$ sample size        

$\mu_o =530$ represent the value that we want to test  

$\alpha$ 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 (variable of interest)    

2) State the null and alternative hypotheses.        

We need to conduct a hypothesis in order to determine if the claim thet the handle on average is 530 bags per hour:      

Null hypothesis:$\mu \geq 530$        

Alternative hypothesis:$\mu < 530$        

We don't know the population deviation and the sample size is lees than 30, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:        

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

3) Calculate the statistic        

We can replace in formula (1) the info given like this:        

$t=\frac{510-530}{\frac{50}{\sqrt{16}}}=-1.60$        

4) Calculate the critical value

We need to begin calculating the degrees of freedom

$df=n-1=16-1=15$

The critical value for this case would be :

$P(t_{15}$

The value of a that satisfy this on the normal standard distribution is a=-1.75 and would be the critical value on this case zc=-1.75

5) Calculate the P-value        

Since is a one-side upper test the p value would be:        

$p_v =P(t_{15}$    

6) Conclusion        

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and the the actual mean is not significantly lower than 530 at 5% of significance.