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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 48 47 67 79 73 66
N 52 55 68 87 78 70
(a) Find t. (Give your answer correct to two decimal places.)
(ii) Find the p-value. (Give your answer correct to four decimal places.)
(b) State the appropriate conclusion.
Reject the null hypothesis, there is significant evidence that ìd < 0.
Reject the null hypothesis, there is not significant evidence that ìd < 0.
Fail to reject the null hypothesis, there is significant evidence that ìd<0.
Fail to reject the null hypothesis, there is not significant evidence that ìd < 0.
t = -4.56
p-value = .0.0030
Reject the null hypothesis
To complete the hypothesis test with the alternative hypothesis ìd < 0 based on the paired data, we need to follow these steps:
Step 1: State the null and alternative hypotheses
The null hypothesis (H0) states that there is no significant difference between the means of the paired data, ìd = 0.
The alternative hypothesis (Ha) states that there is a significant difference between the means of the paired data, ìd < 0.
Step 2: Set the level of significance (álpha)
In this case, the level of significance is given as á = 0.02.
Step 3: Calculate the test statistic (t)
The formula to calculate the test statistic (t) in a paired t-test is:
t = (x̄d - ìd) / (s / √n)
where x̄d is the sample mean of the differences, ìd is the population mean of the differences under the null hypothesis, s is the sample standard deviation of the differences, and n is the number of pairs of data.
Given the data:
M: 48 47 67 79 73 66
N: 52 55 68 87 78 70
We can calculate the differences (d) as follows:
d = M - N: -4 -8 -1 -8 -5 -4
The sample mean of the differences (x̄d) is:
x̄d = sum(d) / n = (-4 -8 -1 -8 -5 -4) / 6 ≈ -5.00
To calculate s, we need to calculate the sample variance (s^2) of the differences:
s^2 = [sum(d^2) - (sum(d)^2 / n)] / (n - 1)
= [((-4)^2 + (-8)^2 + (-1)^2 + (-8)^2 + (-5)^2 + (-4)^2) - ((-4 -8 -1 -8 -5 -4)^2 / 6)] / (6 - 1)
= 48
Hence, s = √s^2 = √48 ≈ 6.93
Now, we can calculate the test statistic (t):
t = (x̄d - ìd) / (s / √n)
= (-5 - 0) / (6.93 / √6)
≈ -5 / 2.83
≈ -1.77
Step 4: Calculate the p-value
To find the p-value corresponding to the test statistic (t), we can use a t-distribution table or a statistical calculator. The p-value is the probability of obtaining a test statistic as extreme as the observed value, assuming the null hypothesis is true.
The degrees of freedom for a paired t-test is n - 1. In this case, n = 6, so the degrees of freedom is 6 - 1 = 5.
Looking up the t-distribution table or using a calculator for a one-tailed test with 5 degrees of freedom, we find that the p-value is approximately 0.0543.
Step 5: Make the decision
Based on our level of significance (á = 0.02) and comparing it to the p-value we obtained (0.0543), we can make the decision.
If the p-value is less than or equal to the level of significance (p-value ≤ á), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
In this case, the p-value (0.0543) is not less than or equal to the level of significance (0.02), so we fail to reject the null hypothesis.
Therefore, the appropriate conclusion is:
Fail to reject the null hypothesis, there is not significant evidence that ìd < 0.