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Alenkinab [10]
3 years ago
9

A random 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.
Mathematics
1 answer:
Lina20 [59]3 years ago
6 0

Answer:

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

Step-by-step explanation:

 Group   Group One     Group Two  

Mean 41.300 54.800

SD 6.800 6.000

SEM 2.050 1.604

N 11      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

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Marianna [84]

Answer:

11.44% probability that exactly 12 members of the sample received a pneumococcal vaccination.

Step-by-step explanation:

For each adult, there are only two possible outcomes. Either they received a pneumococcal vaccination, or they did not. The probability of an adult receiving a pneumococcal vaccination is independent of other adults. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

70% of U.S. adults aged 65 and over have ever received a pneumococcal vaccination.

This means that p = 0.7

20 adults

This means that n = 20

Determine the probability that exactly 12 members of the sample received a pneumococcal vaccination.

This is P(X = 12).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 12) = C_{20,12}.(0.7)^{12}.(0.3)^{8} = 0.1144

11.44% probability that exactly 12 members of the sample received a pneumococcal vaccination.

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