Question

Jennifer’s manager Dr. Jonathan Steinberg wonders whether Healthy Life members needed more chiropractic help in 2019 than in 2018, on average. Jennifer selected a random sample of those who were treated by chiropractic doctors in both years. Data provided

Jennifer’s manager Dr. Jonathan Steinberg

wonders whether Healthy Life members needed

more chiropractic help in 2019 than in 2018, on

average. Jennifer selected a random sample of

those who were treated by chiropractic doctors in

both years. Data provided. Please help Jennifer

Nguyen to check whether annual expenses and

number of visits increased, on average. For both

tests use Data Analysis t-Test: Paired Two

Sample for Means and 2% significance level. Jennifer Nguyen and her manager know that in both cases differences are normally distributed, so here again there is no need to build histograms.


Number of visitsAnnual expenses
Patient #2018201920182019
11011$340.00$374.00
21514$560.00$476.00
32425$816.00$850.00
474$238.00$136.00
51110$374.00$340.00
61922$646.00$748.00
71815$612.00$510.00
843$136.00$162.00
92524$850.00$816.00
103135$1,054.00$1,190.00
1177$298.00$238.00
1289$272.00$366.00
131715$578.00$510.00
141012$340.00$408.00
151415$476.00$510.00
162220$748.00$680.00
1765$204.00$170.00
182124$774.00$816.00
191515$510.00$570.00
201212$408.00$468.00
2134$140.00$120.00
221110$450.00$420.00
2320$90.00$0.00
2424$80.00$160.00
2576$300.00$280.00


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Answer #1

The hypothesis being tested is:

H0: µd = 0

Ha: µd < 0

12.840 mean 2018
12.840 mean 2019
0.000 mean difference (2018 - 2019)
1.871 std. dev.
0.374 std. error
25 n
24 df
0.000 t
.5000 p-value (one-tailed, lower)

The p-value is 0.5000.

Since the p-value (0.5000) is greater than the significance level (0.02), we fail to reject the null hypothesis.

Therefore, we cannot conclude that the number of visits increased, on average.

The hypothesis being tested is:

H0: µd = 0

Ha: µd < 0

451.760 mean 2018
452.720 mean 2019
-0.960 mean difference (2018 - 2019)
69.470 std. dev.
13.894 std. error
25 n
24 df
-0.069 t
.4727 p-value (one-tailed, lower)

The p-value is 0.4727.

Since the p-value (0.4727) is greater than the significance level (0.02), we fail to reject the null hypothesis.

Therefore, we cannot conclude that annual expenses increased, on average.

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