design engineers (long question in photos) 190124 (9 complete) Design engineers
ID: 3269517 • Letter: D
Question
design engineers (long question in photos) 190124 (9 complete) Design engineers want to know whether you may be more likely to purchase a vice product (for example, a candy bar) when your arm is flexed (as when carrying a shopping basket) than when your arm is extended (as when pushing a shopping cart) To test this theory, the researchers recruited 20 consumers and had each push their hand against a table while they were asked a series of shopping questions. Half of the consumers were told to put their arm in a flex position (similar to a shopping basket), and the other half were told to put their arm in an extended position (similar to a shopping cart). Participants were offered several choices between a vice and a virtue (for example, a movie ticket vs. a shopping coupon, pay later with a larger amount vs. pay now) and a choice score (on a scale of 0 to 100) was determined for each. (Higher scores indicate a greater preference for vice options.) The average choice score for consumers with a flexed arm was 56, while the average for consumers with an extended arm was 45. Suppose the standard deviations of the choice scores for the flexed arm and extended arm conditions are 11 and 5, respectively. Does this information support the researchers theory? Answer the question by conducting a hypothesis test. Use =0 05. Specify the null and alternative hypotheses. Let the first group be the consumers assigned to the flex position, and let the second group be the consumers assigned to the extended position H, Compute the test statistic The test statistic is Round to two decimal places as needed.) Enter your answer in each of the answer boxes javascript doExercise(12) 8:29 PMExplanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: Flexed< Extended
Alternative hypothesis: Flexed > Extended
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 = 3.821
DF = 10 + 10 - 2
D.F = 18
t = [ (x1 - x2) - d ] / SE
t = 2.88
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 2.88. We use the t Distribution Calculator to find P(t > 2.88) = 0.0049
Therefore, the P-value in this analysis is 0.00498
Interpret results. Since the P-value (0.00498) is less than the significance level (0.05), we have to reject the null hypothesis.
There is sufficient evidence in the favor of the claim that difference in the means is different than that given by the null hypothesis.
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