Question

A certain pen has been designed so that true average writing lifetime under controlled conditions (involving the use of awriting machine) is at least 10 hours. A random sample of 18 pens is selected, the writing lifetime of each is determined,and a normal probability plot of the resulting data supports the use of a one-sample t test.a. What hypotheses should be tested if the investigators believe a priori that the design specification has been satisfied?b. What conclusion is appropriate if the hypotheses of part (a) are tested,t = —23, and a = .05?c. What conclusion is appropriate if the hypotheses of part (a) are tested, t = —1.8, and u = .01?d. What should be concluded if the hypotheses of part (a) are tested and t = —3.6 ?

Answer 1

Answer:

a) Null hypothesis:$\mu \geq 10$  

Alternative hypothesis:$\mu < 10$  

b) $p_v =P(t_{17}<-2.3)=0.017$

And on this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis.

c) $p_v =P(t_{17}<-1.8)=0.0448$

And on this case since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis.

d) $p_v =P(t_{17}<-3.6)=0.0011$

And on this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis.

Step-by-step explanation:

1) Data given and notation  

$\bar X$ represent the sample mean

$s$ represent the sample deviation

$n=18$ sample size  

$\mu_o =10$ 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 true mean is at least 10 hours, the system of hypothesis would be:  

Part a

Null hypothesis:$\mu \geq 10$  

Alternative hypothesis:$\mu < 10$  

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)  

Part b

For this case we have t=-2.3 , $\alpha=0.05$

First we need to find the degrees of freedom $df=n-1=18-1=17$

Now since we are conducting a left tailed test the p value is given by:

$p_v =P(t_{17}<-2.3)=0.017$

And on this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis.

Part c

For this case we have t=-1.8 , $\alpha=0.01$

First we need to find the degrees of freedom $df=n-1=18-1=17$

Now since we are conducting a left tailed test the p value is given by:

$p_v =P(t_{17}<-1.8)=0.0448$

And on this case since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis.

Part d

For this case we have t=-3.6 , $\alpha=0.05$

First we need to find the degrees of freedom $df=n-1=18-1=17$

Now since we are conducting a left tailed test the p value is given by:

$p_v =P(t_{17}<-3.6)=0.0011$

And on this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis.