In there a relationship between confidence intervals and two-tailed hypothesis tests? The answer is yes. Let c be the level of confidence used to cons

Significance tests
In there a relationship between confidence intervals and two-tailed hypothesis tests? The answer is yes. Let c be the level of confidence used to construct a confidence interval from sample data. Let * be the level of significance for a two-tailed hypothesis test. The following statement applies to hypothesis tests of the mean:
For a two-tailed hypothesis test with level of significance a and null hypothesis $$H_{0} : \mu = k$$ we reject Ho whenever k falls outside the $$c = 1 — \alpha$$ confidence interval for $$\mu$$ based on the sample data. When A falls within the $$c = 1 — \alpha$$ confidence interval. we do reject $$H_{0}$$.
For a one-tailed hypothesis test with level of significance Ho : $$\mu = k$$ and null hypothesiswe reject Ho whenever A falls outsidethe $$c = 1 — 2\alpha$$ confidence interval for p based on the sample data. When A falls within the $$c = 1 — 2\alpha$$ confidence interval, we do not reject $$H_{0}$$.
A corresponding relationship between confidence intervals and two-tailed hypothesis tests is also valid for other parameters, such as p, $$\mu1 — \mu_2,\ and\ p_{1}, - p_{2}$$.
(a) Consider the hypotheses $$H_{0} : \mu_{1} — \mu_{2} = O\ and\ H_{1} : \mu_{1} — \mu_{2} \neq$$ Suppose a 95% confidence interval for $$\mu_{1} — \mu_{2}$$ contains only positive numbers. Should you reject the null hypothesis when $$\alpha = 0.05$$? Why or why not?

2021-02-01
The level of significance, $$\alpha = 0.05$$
The null hypothesis:
$$H_{0}: \mu_{1} - \mu_{2} =0$$
The alternative hypothesis:
$$H_{0}: \mu_{1} - \mu_{2} \neq0$$
Here, from above hypothesis D = 0 and we know that for a two-tailed hypothesis test with level of significance $$\alpha$$, we reject $$H_{0}$$ whenever D falls outside the $$c= 1-\alpha$$ confidence interval for $$\mu$$ based on the sample data. If a 95% confidence interval for $$\mu_{1} — \mu_{2}$$ contains only positive numbers then we have to reject $$H_{0}$$ at the level of significance $$\alpha = 0.05$$.
Since, the confidence interval does not contain D = 0 and hence it falls outside the 95% confidence interval.