Arandom sample of 11 nursing students from group 1 resulted in a mean score of 41.3 with a standard deviation of 6.8. a random sample of 14 nursing students from group 2 resulted in a mean score of 54.8 with a standard deviation of 6. can you conclude that the mean score for group 1 is significantly lower than the mean score for group 2? let μ1 represent the mean score for group 1 and μ2 represent the mean score for group 2. use a significance level of α=0.1 for the test. assume that the population variances are equal and that the two populations are normally distributed.

At 1% significance level,  this difference is considered to be extremely statistically significant.

Step-by-step explanation:

 Group   Group One     Group Two  

Mean41.30054.800

SD6.8006.000

SEM2.0501.604

N11      14      

H0: Mean of group I = Mean of group II

Ha: Mean of group I < mean of group II

(Left tailed test at 1% significance level)

  The mean of Group One minus Group Two equals -13.500

 standard error of difference = 2.563

 t = 5.2681

 df = 23

 p value= 0.00005

Since p < significance level, reject H0

a) px = P(X=1) = 1/2

b) py = P(Y=1) = 1/2

c) pz = P(Z=1) = 1/4

d) X and Y are independent events as they do not depend on each other to occur.

e) yes, pz = px py

f) yes, Z = XY

Step-by-step explanation:

The sample space for tossing both a penny and a nickel includes HH, HT, TH, TT

n(sample space) = 4

a) Let px denote the success probability for X. Find px.

X = 1 if the penny comes up heads, and X = 0

px = success probability for X and that is p(X=1)

p(X=1) = 2/4 = 1/2 (out of the four possible outcomes, only 2 have the penny come up heads; HH and HT)

b) py = P(Y=1) = 2/4 = 1/2 (out of the four possible outcomes, only 2 have the nickel come up heads; TH and HH)

c) pz = P(Z=1) = 1/4 (out of the four possible outcomes, only one has the penny and nickel come up heads; HH)

d) X and Y are independent events as they do not depend on each other to occur. Occurrence of a penny turning up heads, doesn't affect the probability of a nickel turning up heads.

Mathematically, for two independent events,

P(X n Y) = P(X) × P(Y) = (1/2) × (1/2) = 1/4 = P(Z)

e) pz = P(Z) = the probability of both penny and nickel turn up heads

And since we've established that X, probability of a penny head is independent of getting a nickel head, Y.

pz = px py = (1/2)(1/2) = 1/4 (proved)

f) To prove Z = XY

when X = 1, And Y = 1, that is, HH

XY = 1×1 = 1 and Z = 1 too since HH is its conditiin to be a 1. Hence Z = XY = 1 here.

when X = 1 and Y = 0, that is, HT

XY = 1×0 = 0 and Z = 0, since any deviation from both heads (HH) is a 0 for Z. Hence, Z = XY = 0 here.

when X = 0 and Y = 1, that is, TH

XY = 0×1 = 0 and Z = 0, since any deviation from both heads (HH) is a 0 for Z. Hence, Z = XY = 0 here too.

when X = 0 and Y = 0, that is, TT

XY = 0×0 = 0 and Z = 0, since any deviation from both heads (HH) is a 0 for Z. Hence, Z = XY = 0 here too.

Since Z = XY for all the cases, Z is indeed equal to XY.