Question

A doctor is interested in knowing whether systolic blood pressure differs among sexes. The doctor obtains a sample of patients and measures their blood pressures during their regular check-up. Below is the systolic blood pressure for the sample. What can the doctor conclude with α = 0.01?

male	female
137 105 119 115 127 108 131 124 111 136	106 119 98 105 100 132 109 113 98 105

a) What is the appropriate test statistic?
---Select--- na z-test One-Sample t-test Independent-Samples t-test Related-Samples t-test

b)
Condition 1:
---Select--- female systolic blood pressure sex male doctor
Condition 2:
---Select--- female systolic blood pressure sex male doctor

c) Compute the appropriate test statistic(s) to make a decision about H₀.
(Hint: Make sure to write down the null and alternative hypotheses to help solve the problem.)
critical value =  ; test statistic =  
Decision:  ---Select--- Reject H0 Fail to reject H0

d) If appropriate, compute the CI. If not appropriate, input "na" for both spaces below.
[  ,  ]

e) Compute the corresponding effect size(s) and indicate magnitude(s).
If not appropriate, input and/or select "na" below.
d =  ;  ---Select--- na trivial effect small effect medium effect large effect
r² =  ;  ---Select--- na trivial effect small effect medium effect large effect

f) Make an interpretation based on the results.

Males have significantly higher systolic blood pressure than females.Females have significantly higher systolic blood pressure than males.    Males and females do not have significantly different systolic blood pressure.

Answer 1

Solution:-

a) Independent-Samples t-test.

b) Condition 1: Male systolic blood pressure.
Condition 2: Female systolic blood pressure.

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

Null hypothesis: u₁ = u₂
Alternative hypothesis: u₁$\neq$ u₂

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.01. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

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

Male		Female
Mean	121.3	Mean	108.5
Standard Error	3.63639	Standard Error	3.344149
Median	121.5	Median	105.5
Mode	-	Mode	98
Standard Deviation	11.49928	Standard Deviation	10.57513
Sample Variance	132.2333	Sample Variance	111.8333
Kurtosis	-1.44157	Kurtosis	1.723437
Skewness	0.011476	Skewness	1.314842
Range	32	Range	34
Minimum	105	Minimum	98
Maximum	137	Maximum	132
Sum	1213	Sum	1085
Count	10	Count	10

SE = sqrt[(s₁²/n₁) + (s₂²/n₂)]
SE = 4.9403
DF = 18

c)

t_Critical = + 2.879
t = [ (x₁ - x₂) - d ] / SE

t = 2.59

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is the size of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 18 degrees of freedom is more extreme than -2.59; that is, less than - 2.59 or greater than 2.59.

Thus, the P-value = 0.018

Interpret results. Since the P-value (0.018) is greater than the significance level (0.10), hence we failed to reject null hypothesis.

Fail to reject H₀.

d) 99% confidence interval for the difference in the means is C.I = ( 3.52, 22.08)

C.I (1 2) ± ta/2 n2 ni

C.I = 12.8 + 1.879*4.9403

C.I = 12.8 + 9.2828

C.I = ( 3.52, 22.08)