Complete the hypothesis test with alternative hypothesis ƒÊd �‚ 0 based on the paired data that follows and d = O - Y. Use ƒ¿ = 0.01. Assume normality.
Oldest 188 174 192 196 65
Youngest 186 168 199 199 69
(a) Find t. (Give your answer correct to two decimal places.)
Correct: Your answer is correct.0.52 .
(ii) Find the p-value. (Give your answer correct to four decimal places.)
0.6036 I can not figure out how to find p value it always comes out wrong
Anyone know a simple way to determine the p-values, I keep missing them. If I see example I usually can figure it out but did not work this time. I have TI83 Calculator but can not figure out how to use it. So not sure if I can get it on it or not.
My! You have a lot of different personas.
Tracy 10, Tracy, Glenn, etc. -- please use the same name for your posts.
There is different people using one computer, we meet at my house because it is center area and just use my computer instead of everyone dragging their computer around. Sorry,
From now on there will only be one name used from this site, we have agreed upon it being Tracy since it my home.
Dependent
Different
xbar = -1.2
s = 5.17
n= 5
t = -1.2/(5.17/sqrt(5))
t = -0.52
p-value = 0.6310
To get p-value
Stat oldest L1, youngest L2
difference between oldest and youngest L3 after quit and stat tests then T-test data
List: L3
what do you mean with stat oldest L1, etc. Have a question I have a TI-83 Plus calculator, is there a way of putting in the information to get the p value or do you still have to look at a chart.
When I read the instructions on it, it mentioned finding the z value to get to the p value, I am confused I am so sorry.
When you use your ti-83
Yes
L1, L2
To find the p-value for a hypothesis test using the t-distribution, you can follow these steps:
Step 1: Set up the hypotheses
The null hypothesis (H0) for this test is that the population mean difference (µd) is equal to zero: H0: µd = 0.
The alternative hypothesis (Ha) is that the population mean difference (µd) is not equal to zero: Ha: µd ≠ 0.
Step 2: Calculate the test statistic (t)
The test statistic (t) for a paired t-test is given by:
t = (mean difference - hypothesized mean difference) / (standard deviation of the mean differences / square root of the sample size)
In this case, the mean difference = 0.52 (as given in the question), the hypothesized mean difference = 0, the standard deviation of the mean differences can be calculated, and the sample size = 5.
Step 3: Determine the degrees of freedom
For a paired t-test with n observations, the degrees of freedom (df) is given by df = n - 1.
In this case, the degrees of freedom = 5 - 1 = 4.
Step 4: Look up the p-value
Using the calculated test statistic (t) and the degrees of freedom (df), you can look up the p-value in a t-table or use statistical software. Since you mentioned having a TI-83 calculator, I can explain how to use it to calculate the p-value.
To find the p-value on the TI-83 calculator, use the `tcdf` function. Here's how:
- Press the "2nd" key, then the "DISTR" key to access the distribution menu.
- Scroll down and select "6: tcdf(" (t-distribution cumulative probability function).
- Enter the lower bound as the negative test statistic (-t) and the upper bound as the positive test statistic (t). In this case, since the alternative hypothesis is two-tailed, the lower bound would be negative infinity and the upper bound would be positive infinity. So, enter a really large negative number (e.g., -100) as the lower bound and a really large positive number (e.g., 100) as the upper bound.
- Enter the degrees of freedom (df = 4 in this case).
- Press "ENTER" to calculate the p-value.
The calculator will display the p-value, which is the probability of observing a test statistic as extreme (or more extreme) than the one calculated under the null hypothesis.
Keep in mind that the p-value represents the strength of evidence against the null hypothesis. If the p-value is less than the significance level (α), which in this case is 0.01, then you would reject the null hypothesis. Otherwise, you would fail to reject the null hypothesis.
In this case, since the p-value is calculated to be 0.6036, which is greater than 0.01, you would fail to reject the null hypothesis.