Question

2. In a pilot study with 20 subjects evaluating the use of a new drug to lower resting heart rates (HR) of patients, the following data was recorded:

Subject #	Resting HR
001	72
002	88
003	71
004	87
005	64
006	77
007	79
008	59
009	77
010	68
011	78
012	89
013	91
014	81
015	85
016	75
017	69
018	75
019	77
020	81

Given that the average resting HR of the general population for this study is 72.  perform the appropriate t test. Use the two-tailed test. Explain why the textbook tells us to start with the two-tailed test. What is the value of t? Using an alpha of 0.05, is the t statistic significant? Why? What are the confidence limits for a 95% confidence interval here and what do they mean for this patient group

Now that you have answered question 6 with the two-tailed test, answer it with the one-tailed test.

How does this change your answers to our conventional questions:   What is the value of t? What is the p value? Why did this change? Using an alpha of 0.05, is the t statistic significant? Why?

Answer 1

$\dpi{100} \bar x$ = (72 + 88 + 71 + 87 + 64 + 77 + 79 + 59 + 77 + 68 + 78 + 89 + 91 + 81 + 85 + 75 + 69 + 75 + 77 + 81)/20 = 77.15

s = sqrt(((72 - 77.15)^2 + (88 - 77.15)^2 + (71 - 77.15)^2 +( 87 - 77.15)^2 + (64 - 77.15)^2 + (77 - 77.15)^2 + (79 - 77.15)^2 + (59 - 77.15)^2 + (77 - 77.15)^2 + (68 - 77.15)^2 + (78 - 77.15)^2 + (89 - 77.15)^2 + (91 - 77.15)^2 + (81 - 77.15)^2 + (85 - 77.15)^2 + (75 - 77.15)^2 + (69 - 77.15)^2 + (75 - 77.15)^2 + (77 - 77.15)^2 + (81 - 77.15)^2)/19) = 8.5

For two-tailed test

H₀: $\dpi{100} \mu$ = 72

H₁: $\dpi{100} \mu \neq$ 72

The test statistic t = ($\dpi{100} \bar x - \mu$)/(s/$\dpi{100} \sqrt n$)

                             = (77.15 - 72)/(8.5/$\dpi{100} \sqrt 20$)

                             = 2.71

df = 20 - 1 = 19

P-value = 2 * P(T > 2.71)

              = 2 * (1 - P(T < 2.71))

              = 2 * (1 - 0.9931)

              = 2 * 0.0069

              = 0.0138

At $\alpha$ = 0.05, since the P-value is less than $\alpha$(0.0138 < 0.05), so we should reject the null hypothesis.

For one-tailed test

H₀: $\dpi{100} \mu$ = 72

H₁: $\mu$ < 72

The test statistic t = ($\dpi{100} \bar x - \mu$)/(s/$\dpi{100} \sqrt n$)

                             = (77.15 - 72)/(8.5/$\dpi{100} \sqrt 20$)

                             = 2.71

df = 20 - 1 = 19

P-value = P(T > 2.71)

              = 1 - P(T < 2.71)

              = 1 - 0.9931

              = 0.0069

At $\alpha$ = 0.05, since the P-value is less than $\alpha$(0.0069 < 0.05), so we should reject the null hypothesis.