Respuesta :
Answer:
It would be less than 0.05.
Step-by-step explanation:
The hypothesis test was done to claim that the weight of participants has changed.
The null and alternative hypothesis could be written as:
[tex]H_0: \mu=0\\\\H_a: \mu\neq0[/tex]
being μ the population mean change in weight.
With that information, we can tell that is a two tailed test. Sample means that fall in any of the tails, with a z-statistic over 1.96 or under -1.96 (at a significance level of 0.05), will be evidence to reject the null hypothesis.
The information we have about the sample is the 95% confidence interval calculated from the sample information.
This confidence interval does not include the value μ=0.
Then, there is 2.5% of probabiltity that the population mean is under 2.177, the lower bound of the interval, which includes the value μ=0.
With this information, we can conclude that the P-value have to be under 0.05.
From the confidence interval given, we have that the correct option is:
It would be less than 0.05.
At the null hypothesis, we test that there are no changes, that is, the means are equal, so:
[tex]H_0: \mu_A = \mu_B[/tex]
[tex]H_0: \mu_A - \mu_B = 0[/tex]
At the alternative hypothesis, we test if there was a change, that is, if the means are different, so:
[tex]H_1: \mu_A - \mu_B \neq 0[/tex].
The confidence interval for the difference of means is: (2.177, 4.977).
- It does not contain 0, which means that they are different.
- Statistically significant results, which indicate difference, have low p-values, that is, lower than 0.05, thus the p-value would be less than 0.05.
A similar problem is given at https://brainly.com/question/22968227
