Question 101 pts Suppose that you had consumer group wanted to test to see if weight of participants in a weight loss program changed (up or down). They computed a 95% confidence interval of the result (2.177, 4.977). What do we know about the p-value for the test? Group of answer choices It would be greater than 0.05. Can not be determined. It would be 0.05. It would be less than 0.05.

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:

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.

total amount of money that he should pay if he has not used his duty free exemption is $98

step-by-step explanation:

given a united states citizen returning to the states declares the following items at the customs office: 3 shirts at $8.50 each 2 dresses at $27.50 each 1 pair of gold cuff links at $17.50 per pair. we have to find how much he pay

cost of 3 shirts if one cost at $8.50 is

cost of 2 dresses if one cost at $27.50 is  

cost of 1 pair of gold cuff at $17.50

so, total amount of money that he should pay if he has not used his duty free exemption is   $25.5+$55+$17.50=$98

$30.32

step-by-step explanation:

65.00*.12=2.80

65.00-2.80=57.20

57.20*1.06=60.632

60.632/2=30.316