Could someone work a problem like this one, for an example.
Complete the hypothesis test with alternative hypothesis ìd < 0 based on the paired data that follow and d = M - N. Use á = 0.02. Assume normality.
M 49 67 45 57 57 48
N 57 68 47 60 64 55
(a) Find t. (Give your answer correct to two decimal places.)
(ii) Find the p-value. (Give your answer correct to four decimal places.)
M 49 67 45 57 57 48
N 57 68 47 60 64 55
d 8 1 2 3 7 7
dbar = (8+ 1+ 2 +3 +7 +7)/6 = 28/6
sd = sqrt ((8-28/6)^2 + (1-28/6)^2 + (2-28/6)^2 +(3-28/6)^2 +(7-28/6)^2 + (7-28/6)^2 )/5))
Sd = sqrt(132/3/5)) = 3.011
t = (dbar-0)/sd
t = 4.667/3.011 = 1.55
To get critical value look t table
Degree of freedom D = 5
ta/2 = 0.02
Critical value is 3.37
I have worked it out and found the t value as 3.80 and that came out right but the p value is not coming out right.
t = 3.80
P-value = 0.0063
Thank you, I keep missing my p-values, and I am not sure why. I look them up on the chart, I have done everything the example says to but mine always come up wrong.
To complete the hypothesis test, we need to follow several steps. Let's go through each step and find the answers to the given problem:
Step 1: State the null and alternative hypotheses:
Null hypothesis (H0): d = 0 (There is no significant difference between the paired data)
Alternative hypothesis (Ha): d < 0 (There is a significant difference and the first set of data is lower than the second set)
Step 2: Calculate the difference (d) between the pairs of data:
To find d, subtract the second set of data (N) from the first set (M). We have:
M: 49 67 45 57 57 48
N: 57 68 47 60 64 55
The differences (d) are:
d: -8 -1 -2 -3 -7 -7
Step 3: Calculate the mean (µd) and standard deviation (σd) of the differences:
Mean (µd) = Σd / n (where n is the number of pairs)
µd = (-8 -1 -2 -3 -7 -7) / 6
µd = -5.33 (rounded to two decimal places)
Standard Deviation (σd) = sqrt(Σ(d - µd)^2 / (n - 1))
σd = sqrt(((-8 - (-5.33))^2 + (-1 - (-5.33))^2 + ... + (-7 - (-5.33))^2) / (6-1))
σd = sqrt((34.4489 + 16.9921 + ... + 2.8489) / 5)
σd = sqrt(73.0674 / 5)
σd = sqrt(14.6135)
σd = 3.82 (rounded to two decimal places)
Step 4: Calculate the t-statistic (t):
t = (µd - δ0) / (σd / √n)
δ0 is the hypothesized difference under the null hypothesis, which is 0 in this case.
t = (-5.33 - 0) / (3.82 / √6)
t = -5.33 / (3.82 / 2.45)
t = -5.33 / 1.56
t = -3.42 (rounded to two decimal places)
Step 5: Find the p-value:
The p-value is the probability of observing a test statistic (t) as extreme as the one calculated if the null hypothesis is true. Since the alternative hypothesis is one-tailed (d < 0), we need to find the probability to the left of t on the t-distribution with (n-1) degrees of freedom.
Using a t-table or statistical software, we find that the p-value for t = -3.42 with 5 degrees of freedom is approximately 0.0127 (rounded to four decimal places).
So, the p-value is 0.0127.
In conclusion, the t-value is -3.42 (rounded to two decimal places) and the p-value is 0.0127 (rounded to four decimal places).