number 4 Sample : Pe-rining weigs mple Poet-training weights of 18 people depent
ID: 3311254 • Letter: N
Question
number 4
Sample : Pe-rining weigs mple Poet-training weights of 18 people depentent t) Independent Classity the two give samples as independent or dependent Sample . The scores of 22 stadente who took the ACT Samle 2: The scores of 22 diflerent students whe took the SAT A) dependent B) independent SHORT ANSWER. Write the word or phrase that best completes each statement or answers the q 3) Find the standardized test statistic to test the claim that 2, Assume the two samples 3) are random and independent Population statistics: 1" 1.5 and 2-1.9 Simple statistics: xt = 24, n1-50 and x2 = 22,n2 = 60 4) A local bank claims that the waiting time for its customers to be served is the lowest in the 4) area. A competitor bank checks the waiting times at both banks Assume the two samples are random and independent. Usea 0.05 and a confidence interval to test the local bank's claim. Local Bank n1 = 45 x1-5.3 minutesxz-5.6 minutes 071-1.1 minutes 02-10 minute Competitor Bank n2- 50 5) Find the critical values, to-to test the claim that 2. Two samples are random, independent, and come from populstions that are normal. The sample statistics are glven below. Assume that . 2. Use a:005. 2 n2-30 x1 17 51 1.5 x2 15 5) = 1.9Explanation / Answer
Solution:-
4)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: Local> Competitor
Alternative hypothesis: Local < Competitor
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 of freedom (DF), and the t statistic test statistic (t).
SE = sqrt[(s12/n1) + (s22/n2)]
SE = 0.217
DF = 93
t = [ (x1 - x2) - d ] / SE
t = - 1.38
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size 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.
The observed difference in sample means produced a t statistic of - 1.38. We use the t Distribution Calculator to find P(t < - 1.38).
Therefore, the P-value in this analysis is 0.171
Interpret results. Since the P-value (0.171) is greater than the significance level (0.05), we cannot reject the null hypothesis.
From the above test we do not have sufficient evidence in the favor of the claim that local bank waiting time for its costumer is lowest.
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