Question

A sample of 64 observations is selected from a normal population. The sample mean is 215, and the population standard deviation is 15. Conduct the following test of hypothesis using the .03 significance level. What is the p-value?H0 : μ ≥ 220H1 : μ < 220

Answer 1

`z=(215-220)/((15)/(√(64)))=-2.67`

`p_v =P(z<-2.67)=0.0038`

If we compare the p value and the significance level given
`\alpha=0.03` we see that
`p_v<\alpha` then we have enough evidence to conclude that the true mean is significantly lower than 220 at 3% of significance.

Explanation:

Data given

`\bar X=215` represent the sample mean

`\sigma=15` represent the population standard deviation

`n=64` sample size

`\mu_o =220` represent the value that we want to test

`\alpha=0.03` represent the significance level for the hypothesis test.

z 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 lower than 220, the system of hypothesis would be:

Null hypothesis:
`\mu \geq 220`

Alternative hypothesis:
`\mu < 220`

The statistic is given by:

`z=(\bar X-\mu_o)/((\sigma)/(√(n)))` (1)

Calculate the statistic

We can replace in formula (1) the info given like this:

`z=(215-220)/((15)/(√(64)))=-2.67`

P-value

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

`p_v =P(z<-2.67)=0.0038`

Conclusion

If we compare the p value and the significance level given
`\alpha=0.03` we see that
`p_v<\alpha` then we have enough evidence to conclude that the true mean is significantly lower than 220 at 3% of significance.