Question

1. At CES 2019, L’Oreal won the award for best innovation in wearable technology for their product, My Skin Track pH. The scientific and medical communities have long known the link between skin pH and common skin concerns, yet there has never been a consumer-friendly way to measure it. My Skin Track pH is the first-ever, wearable sensor and companion app to measure personal skin pH levels and create customized product regimens to better care for skin. Using microfluidics technology, the sensor captures trace amounts of sweat to provide an accurate skin pH reading. Conventional methods for reading pH levels include swabbing the skin for sweat and using a bulky digital pH meter. Researchers know that both methods provide accurate data,but were not sure which would provide results faster. A sample of seven subjects were brought in and each tested both methods. The data is shown below.

Subject ID	My SkinTrack pH (sec)	Conventional method (sec)
AX120	151	110.6
JT635	156	127.9
IP894	141.5	122.6
SB260	115.7	134.9
EJ789	140.3	139.9
AS696	200.2	102.0
NC510	179.5	121.5

Using a significance level of 0.05, does one method provide results that are significantly faster?

Answer 1

Solution:-

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

Null hypothesis: u_d = 0

Alternative hypothesis: u_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 = 38.5913

SE = s / sqrt(n)

S.E = 14.586

DF = n - 1 = 7 -1

D.F = 6

t = [ (x₁ - x₂) - D ] / SE

t = 2.202

where d_i 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 6 degrees of freedom is more extreme than 2.202; that is, less than - 2.202 or greater than 2.202.

Thus, the P-value = 0.07

Interpret results. Since the P-value (0.07) is greater than the significance level (0.05), we have to reject the null hypothesis.

Do not reject H₀. hence one method does one method does not provide results that are significantly faster.