Answer:
The null hypothesis μ=4.0 could not be rejected.
Step-by-step explanation:
In this problem we have to perform a hypothesis test of the mean, with s.d. of the population unknown.
The sample is: [3.999, 4.037, 4.116, 4.063, 3.969, 3.955, 4.091]
This sample has a mean of 4.033 and a standard deviation of 0.061.
The null and alternative hypothesis are:
[tex]H_0: \mu = 4.0\\\\H_1: \mu \neq 4.0[/tex]
We assume a significance level of 0.05.
The test statistic for this test is:
[tex]t=\frac{M-\mu}{\sigma}=\frac{4.033-4.0}{0.061}=0.536[/tex]
We have a sample size n=7, so the degrees of freedom are 7-1=6.
For a two-tailed test, t=0.536 and df=6, the P-value can be look up in a t-table.
The P-value is 0.61. Is greater than the significance level, so the effect is not significant and the null hypothesis can't be rejected.
