Question

In: Statistics and Probability

Jennifer’s manager Dr. Jonathan Steinberg wonders whether Healthy Life members needed more chiropractic help in 2019...

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 visits Annual expenses
Patient # 2018 2019 2018 2019
1 10 11 $340.00 $374.00
2 15 14 $560.00 $476.00
3 24 25 $816.00 $850.00
4 7 4 $238.00 $136.00
5 11 10 $374.00 $340.00
6 19 22 $646.00 $748.00
7 18 15 $612.00 $510.00
8 4 3 $136.00 $162.00
9 25 24 $850.00 $816.00
10 31 35 $1,054.00 $1,190.00
11 7 7 $298.00 $238.00
12 8 9 $272.00 $366.00
13 17 15 $578.00 $510.00
14 10 12 $340.00 $408.00
15 14 15 $476.00 $510.00
16 22 20 $748.00 $680.00
17 6 5 $204.00 $170.00
18 21 24 $774.00 $816.00
19 15 15 $510.00 $570.00
20 12 12 $408.00 $468.00
21 3 4 $140.00 $120.00
22 11 10 $450.00 $420.00
23 2 0 $90.00 $0.00
24 2 4 $80.00 $160.00
25 7 6 $300.00 $280.00

Solutions

Expert Solution

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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The manager of a health insurance company in Toronto wonders whether the members needed more chiropractic help in 2018 than in 2017
Hypothesis test (2- sample)The manager of a health insurance company in Toronto wonders whether the members needed more chiropractic help in 2018 than in 2017, on average. He found a random sample of those who were treated by chiropractic doctors in both years. Data provided in TERMTESTDATA file. Perform a Hypothesis test to check whether annual expenses increased, on average. For the test use Data Analysis t-Test: Paired Two Sample for Means and 1% significance level.Annual expenses2017201834037456047681685023813637434064674861251013616285081610541190298238272366578510340408476510748680204170774816510570408468
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