A comparison between a major sporting goods chain and a specialty runners\' stor

Question

A comparison between a major sporting goods chain and a specialty runners' store was done to find differences in prices on running shoes. A sample of 17 different shoes was priced (in dollars) at both stores. It is found that the mean of differences prices is 8.8 with standard deviation 5.2.

1.) Choose the appropriate hypotheses. A) The hypotheses are Ho: md > or equal to 0 vs. Ha: md < 0. B) The hypotheses are Ho: md = 0 vs. Ha: md does not equal 0. C) The hypotheses are Ho: md > or equal to 0 vs Ha: md = 0. D) The hypotheses are Ho: md = 0 vs. > 0. E) None of the above

2.) The test statistic is A) -1.7570 B) 1.6923 C) 6.9780 D) -1.9600 E) None of the above

3.) The degrees of freedom are A) 17 B) 18 C) 15 D) 16 E) None of the above

4.) The critical value at 5% significance level is A) ±2.120 B) ±2,131 C) ±2.110 D) ±2.160 E) None of the above

5.) We can conclude that A) Discount Store is not different than Specialty Store based on 5% significance level. B) Discount Store is not different than Specialty Store based on 2.5% significance level. C) Discount Store is different than Specialty Store based on 5% significance level. D) Discount Store is different than Specialty Store based on 2.5% significance level. E) None of the above

Explanation / Answer

Solution:-

1) State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: d = 0

Alternative hypothesis: d 0

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a matched-pairs t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard deviation of the differences (s), the standard error (SE) of the mean difference, the degrees of freedom (DF), and the t statistic test statistic (t).

s = sqrt [ ((di - d)2 / (n - 1) ]

s = 5.2

SE = s / sqrt(n)

S.E = 1.26

DF = n - 1 = 17 -1

3) D.F = 16

t = [ (x1 - x2) - D ] / SE

2) t = 6.978

4) tcritical = + 2.12

where di is the observed difference for pair i, d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 16 degrees of freedom is more extreme than 6.978; that is, less than - 6.978 or greater than 6.978.

Thus, the P-value = less than 0.0001

Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we have to reject the null hypothesis.

5) (C) We can conclude that Discount Store is different than Specialty Store based on 5% significance level.

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