Question

1 The data in table 1 below are measurements from a group of 10 normal males and 11 males with left-heart disease taken at autopsy at a particular hospital. Measurements were made on several variables at that time, and the table presents the measurements on total heart weight (THW) and total body weight (BW). Assume that the diagnosis of left-heart disease is made independently of these variables. BW(kg) + Table 1 Left-heart disease males Observation THW (g) BW(kg) number 452 55 755 72 331 56 499 Normal males Observation THW (g) number 312 364 349 286 312 52 58 301 450 78 303 66 348 469 375 310 592 424 52 | 400 49 10 1277 10 11 73 1.1 1.3 Using the data in table 1 construct a 95% confidence interval for the mean of body weight for normal males. = 54.8, 65.7 Using the data in table 1 construct a 90% confidence interval for the variance of total heart weight for left-heart disease males. ==9732.2, 45,277.4 Using the data in table 1 construct a 99% confidence interval for the variance of total heart weight for normal males. = 642.27, 8757.88 Are the variances of total heart weight (THW) in the two groups (Left-heart disease males and normal males) same at a significant alpha 0.05? = NOT EQUAL Based on the result from 1.4, which t-test (equal or unequal variances) is used to test if the means of total heart weight (THW) the same or not at a significant alpha 0.05 by two-sided test? Implement the test and report a p-value (two-sided)? 1.4

ONLY NEED HELP WITH 1.5! PLEASE SHOW WORK

Answer 1

GIVEN:

The data on total heart weight from a group of 10 normal males and 11 males with left heart disease:

THW of Left heart disease males	THW of Normal males
452	312
755	364
331	349
499	286
301	312
450	277
469	303
375	348
310	400
592	277
424

The given problem is to test whether the means of total heart weight between normal and left heart diseased males is same or not. Thus we use "Two sample t test with unequal variance" to test this claim.

HYPOTHESIS:

(That is, there is no significant difference in mean total heart weight between normal and left heart diseased males)

HO: M1 – M2 = 0

(That is, there is significant difference in mean total heart weight between normal and left heart diseased males)

0 7 11 – 17:17

LEVEL OF SIGNIFICANCE:
g= 0,05

TEST STATISTIC:

t = [(-11-22)]/(si/nı) + (s/n2)

which follows t distribution with degrees of freedom given by,

df = [(st/n1) +($}/n2)]/[(s/nı)?/(n1 – 1)] + [(8/12)/(n2 – 1))

CALCULATION:

I have used excel function to calculate mean and standard deviation of two samples.

Mean: "=AVERAGE(select array of data values)"

Standard deviation: "=STDEV(select array of data values)"

Sample size of left heart diseased males

(ni) = 11

Sample size of normal males

(n2) = 10

Sample mean heart weight of left heart diseased males

(21) = 450.73

Sample mean heart weight of normal males

(T2) = 322.8

Sample standard deviation of left heart diseased males

(si) = 133.496

Sample standard deviation of normal males

(S2) = 41.033

DEGREES OF FREEDOM:

df = [(st/n1) + ($}/n2)]/[[{s} /nı)?/(n1 – 1)] + [(8/12)?/(n2 – 1)]]

  

= [(133.4962/11)+(41.033/10)] /[[(133.4962/11)/(11-1)]+[(41.033/10)/(10– 1)]]

=3198654.14632/262474.81+3149.855

df = 12.042 12

TEST STATISTIC:

t = [(-11-22)]/(si/nı) + (s/n2)

= [450.73 – 322.8]/ (133.4962/11) + (41.0332/10)

= 127.93/42.2904

t = 3,03

P VALUE:

The two tailed p value for test statistic with 12 degrees of freedom is,

t = 3,03

2 * Plt > 3.031 = 2 * Plt >3.03]

= 2*(1 – Pſt <3.031)

=2+1-0.9948)

=?*0,0052

= 0.0105

The p value is .

0.0105

DECISION RULE:

Reject H, if pualue <a

CONCLUSION:

Since the calculated p value (0.0105) is less than the significance level , we reject the null hypothesis and conclude that there is significant difference in mean total heart weight between normal and left heart diseased males.

g= 0,05