What is the following equation written in standard form?

The manufacturer of an airport baggage scanning machine claims it can handle an average of530 bags per hour.(a-1) At α = .05 in

-Dominant- [34]

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.

6 0

3 years ago

PTC is a substance that has a strong bitter taste for some people and is tasteless for others. The ability to taste PTC is inher

lana66690 [7]

Answer:

a) $n=\frac{0.76(1-0.76)}{(\frac{0.01}{1.64})^2}=4905.83$

And rounded up we have that n=4906

b) For this case since we don't have prior info we need to use as estimatro for the proportion $\hat p =0.5$

$n=\frac{0.5(1-0.5)}{(\frac{0.01}{1.64})^2}=6724$

And rounded up we have that n=6724

Step-by-step explanation:

We need to remember that the confidence interval for the true proportion is given by :

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

Part a

The estimated proportion for this case is $\hat p =0.76$

Our interval is at 90% of confidence, and the significance level is given by $\alpha=1-0.90=0.1$ and $\alpha/2 =0.05$ . The critical values for this case are: