Question

Air crew escape systems are powered by a solid propellant. The burning rate of this propellant is an important product characteristic. Specifications require that the mean burning rate must be 50 centimeters per second. The experimenter selects a random sample of n=40 and obtains an observed sample mean of the burning rate 51.3 centimeters per second and an observed sample standard deviation of the burning rate 2.0 centimeters per second. Based on a 5% type I error threshold, construct a hypothesis test and draw your conclusion.

Answer 1

Answer:

Conclusion: Reject the mean of 50 cm/sec and establish a new mean of 51.3 cm/sec

Step-by-step explanation:

In this case we can set our hypothesis the following way:

$\bf H_0$:The mean burning rate equals 50 cm/sec.

$\bf H_a$: The mean burning rate is greater than 50 cm/sec.

So this is a right-tailed hypothesis test.

For a 5% level of significance, the Z-score against with we are going to compare is $\bf Z_{0.05}=1.64$ (the area under the normal curve N(0;1) to the right of Z is less than 5%)

We compute the Z score corresponding to the sample with the formula

$\bf Z=\frac{\bar x-\mu}{s/\sqrt n}$

where  

$\bf \bar x$ = mean of the sample

$\bf \mu$ mean of the null hypothesis

s = standard deviation of the sample

n = size of the sample

Computing our Z, we get

$\bf Z=\frac{51.3-50}{2/\sqrt {40}}=4.1109$

Since our Z is greater than $\bf Z_{0.05}$, we reject the null hypothesis and we could also state a new mean equals to the alternative, that is to say, we could say that the mean is 51.3 instead of 50.