Question

The manufacturer of an airport baggage scanning machine claims it can handle an average of530 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 507 and a standard deviation of 47 indicate that the manufacturer’s claim is overstated? Choose the appropriate hypothesis.a. H1: μ < 530. Reject H1 if tcalc > -1.753b. H0: μ < 530. Reject H0 if tcalc > -1.753c. H1: μ ≥ 530. Reject H1 if tcalc < -1.753d. H0: μ ≥ 530. Reject H0 if tcalc < -1.753

Answer 1

Answer:

b. H0: μ < 530. Reject H0 if tcalc > -1.753

Step-by-step explanation:

1) Data given and notation  

$\bar X=507$ represent the sample mean

$s=47$ represent the sample standard deviation  

$n=16$ 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)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is less than 530 (left tailed tes), the system of hypothesis would be:  

Null hypothesis:$\mu \geq 530$  

Alternative hypothesis:$\mu < 530$  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (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".  

Calculate the statistic  

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

$t=\frac{507-530}{\frac{47}{\sqrt{16}}}=-1.957$

Critical value

Since we are conducting a left tailed test we need to first find the degrees of freedom for the statistic given by:

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

Now we need to look on the t distribution with 15 degrees of freedom that accumulates 0.05 of the area on the left area. And on this case the critical value would be $t_{\alpha}=-1/753$.

And we can use the following excel code to find it: "=T.INV(0.05,15)"  

So then the correct rejection zone for H0 would be: Reject H0 if $t_{calc}$

P-value  

Since is a one side left tailed test the p value would be:  

$p_v =P(t_{(15)}$  

Conclusion  

If we compare the p value and the significance level for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the mean is significantly lower than 530 at 5% of significance.