Question

The output voltage for an electric circuit is specified to be 127 volts. A sample of 40 independent readings on the voltage for this circuit gave a sample mean 125.8 volts and standard deviation 2 volts. Test the hypothesis that the average output voltage is 127 volts against the alternative that it is less than 127 volts. Use a test with level 0.05. State the null and alternative hypotheses.

Answer 1

`t=(125.8-127)/((2)/(√(40)))=-3.795`

`p_v =P(t_((39))<-3.795)=0.00025`

If we compare the p value and the significance level given
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, we can conclude that the true mean is significantly les than 127 at 5% of signficance.

Explanation:

Data given and notation

`\bar X=125.8` represent the sample mean

`s=2` represent the sample standard deviation

`n=40` sample size

`\mu_o =127` represent the value that we want to test

`\alpha=0.05` 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 ture mean is less than 127 or no, the system of hypothesis would be:

Null hypothesis:
`\mu = 127`

Alternative hypothesis:
`\mu < 127`

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=(\bar X-\mu_o)/((s)/(√(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=(125.8-127)/((2)/(√(40)))=-3.795`

P-value

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

`df=n-1=40-1=39`

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

`p_v =P(t_((39))<-3.795)=0.00025`

Conclusion

If we compare the p value and the significance level given
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, we can conclude that the true mean is significantly les than 127 at 5% of signficance.