Question

A microwave manufacturing company has just switched to a new automated production system. Unfortunately, the new machinery has been frequently failing and requiring repairs and service. The company has been able to provide its customers with a completion time of 6 days or less. To analyze whether the completion time has increased, the production manager took a sample of 36 jobs and found that the sample mean completion time was 6.5 days with a sample standard deviation of 1.5 days. At significance levels of .05 and .10, test whether the completion time has increased. Indicate which test you are performing; show the hypotheses, the test statistic and the critical values and mention whether one-tailed or two-tailed.

Answer 1

Answer:

Null hypothesis:$\mu \leq 6$  

Alternative hypothesis:$\mu > 6$  

$t=\frac{6.5-6}{\frac{1.5}{\sqrt{36}}}=2$    

$df=n-1=36-1=35$

$t_{crit}=1.690$ with the excel code:"=T.INV(0.95,35)"

$t_{crit}=1.306$ with the excel code:"=T.INV(0.90,35)"

$p_v =P(t_{(35)}>2)=0.0267$  

If we compare the p value and the significance level given $\alpha=0.05,0.1$ we see that $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the true mean it's significantly higher than 6 at 5% and 10% of significance.  

Step-by-step explanation:

Data given and notation  

$\bar X=6.5$ represent the mean time for the sample  

$s=1.5$ represent the sample standard deviation for the sample  

$n=36$ sample size  

$\mu_o =6$ represent the value that we want to test

$\alpha=0.05,0.1$ 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 higher than 6 days, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 6$  

Alternative hypothesis:$\mu > 6$  

If we analyze the size for the sample is > 30 but 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:  

$t=\frac{6.5-6}{\frac{1.5}{\sqrt{36}}}=2$    

Critical value and P-value

The first step is calculate the degrees of freedom, on this case:  

$df=n-1=36-1=35$

In order to calculate the critical value we need to find a quantile on the t distribution with 35 degrees of freedom that accumulates $\alpha$ on the right. Using the significance level of 0.05 we got:

$t_{crit}=1.690$ with the excel code:"=T.INV(0.95,35)"

And using the significance of 0.1 we got

$t_{crit}=1.306$ with the excel code:"=T.INV(0.90,35)"

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

$p_v =P(t_{(35)}>2)=0.0267$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.05,0.1$ we see that $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the true mean it's significantly higher than 6 at 5% and 10% of significance.