The null and alternate hypotheses are A random sample of 18 items from the first
ID: 3052561 • Letter: T
Question
The null and alternate hypotheses are A random sample of 18 items from the first population showed a mean of 110 and a standard deviation of 10 A sample of 14 items for the second population showed a mean of 101 and a standard deviation of 8 Assume the sample populations do not have equal standard deviations a. Find the degrees of freedom for unequal variance test (Round down your answer to the nearest whole number.) Degrees of freedom b. State the decision rule for o 05 significance leve! (Round your answer to 3 decimal places.) Reject Ho if t c. Compute the value of the test statistic (Round your answer to 3 decimal places.) Value of the test statistic d. What is your decision regarding the nul hypothesis? Use the 005 significance level Null hypothesis (Click to select not rejected eBook & ResourduelectedExplanation / Answer
Solution:-
Mean = 110, S.D = 14
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: 1< 2
Alternative hypothesis: 1 > 2
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the mean difference between sample means is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees offreedom (DF), and the t statistic test statistic (t).
SE = sqrt[(s12/n1) + (s22/n2)]
SE = 3.182
a)
DF = 30
b)
tcritical = 1.697
Reject H0 if t > 1.697
c)
t = [ (x1 - x2) - d ] / SE
t = 2.83
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is thesize of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.
We use the t Distribution Calculator to find P(t > 2.83) = 0.0082.
d) Interpret results. Since the P-value (0.0082) is less than the significance level (0.05), we have to reject the null hypothesis.
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