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murzikaleks [220]
3 years ago
11

A university wants to compare out-of-state applicants' mean SAT math scores (?1) to in-state applicants' mean SAT math scores (?

2). The university looks at 35 in-state applicants and 35 out-of-state applicants. The mean SAT math score for in-state applicants was 540, with a standard deviation of 20. The mean SAT math score for out-of-state applicants was 555, with a standard deviation of 25. It is reasonable to assume the corresponding population standard deviations are equal.
At the 5% significance level, can the university conclude that the mean SAT math scores for in-state students and out-of-state students differ?
a. No, because the confidence interval contains zero
b. Yes, because the confidence interval contains zero
c. No, because the confidence interval does not contain zero
d. Yes, because the confidence interval does not contain zero
Mathematics
1 answer:
nordsb [41]3 years ago
6 0

Answer:

d. Yes, because the confidence interval does not contain zero.

Step-by-step explanation:

We are given that the university looks at 35 in-state applicants and 35 out-of-state applicants. The mean SAT math score for in-state applicants was 540, with a standard deviation of 20.

The mean SAT math score for out-of-state applicants was 555, with a standard deviation of 25.

Firstly, the Pivotal quantity for 95% confidence interval for the difference between the population means is given by;

                P.Q. =  \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }  ~ t__n__1-_n__2-2

where, \bar X_1 = sample mean SAT math score for in-state applicants = 540

\bar X_2 = sample mean SAT math score for out-of-state applicants = 555

s_1 = sample standard deviation for in-state applicants = 20

s_2 = sample standard deviation for out-of-state applicants = 25

n_1 = sample of in-state applicants = 35

n_2 = sample of out-of-state applicants = 35

Also, s_p=\sqrt{\frac{(n_1-1)s_1^{2} +(n_2-1)s_2^{2} }{n_1+n_2-2} } = \sqrt{\frac{(35-1)\times 20^{2} +(35-1)\times 25^{2} }{35+35-2} }  = 22.64

<em>Here for constructing 95% confidence interval we have used Two-sample t test statistics.</em>

So, 95% confidence interval for the difference between population means (\mu_1-\mu_2) is ;

P(-1.997 < t_6_8 < 1.997) = 0.95  {As the critical value of t at 68 degree

                                         of freedom are -1.997 & 1.997 with P = 2.5%}  

P(-1.997 < \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } < 1.997) = 0.95

P( -1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } < {(\bar X_1-\bar X_2)-(\mu_1-\mu_2)} < 1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ) = 0.95

P( (\bar X_1-\bar X_2)-1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } < (\mu_1-\mu_2) < (\bar X_1-\bar X_2)+1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ) = 0.95

<u>95% confidence interval for</u> (\mu_1-\mu_2) =

[ (\bar X_1-\bar X_2)-1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } , (\bar X_1-\bar X_2)+1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ]

=[(540-555)-1.997 \times {22.64 \times \sqrt{\frac{1}{35} +\frac{1}{35} } },(540-555)+1.997 \times {22.64 \times \sqrt{\frac{1}{35} +\frac{1}{35} } }]

= [-25.81 , -4.19]

Therefore, 95% confidence interval for the difference between population means SAT math score for in-state and out-of-state applicants is [-25.81 , -4.19].

This means that the mean SAT math scores for in-state students and out-of-state students differ because the confidence interval does not contain zero.

So, option d is correct as Yes, because the confidence interval does not contain zero.

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Murljashka [212]

Answer:

b and a

Step-by-step explanation:

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Answer:

The probability of success for this case would be:

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Let X the random variable of interest, on this case we now that:  

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The probability mass function for the Binomial distribution is given as:  

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Where (nCx) means combinatory and it's given by this formula:  

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Step-by-step explanation:

Adduming the following info: In a list of 15 households, 9 own homes and 6 do not own homes. Five households are randomly selected from these 15 households. Find the probability that the number of households in these 5 own homes is exactly 3.

Round your answer to four decimal places

P (exactly 3)=

Previous concepts  

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".  

Solution to the problem

The probability of success for this case would be:

p =\frac{9}{15}= 0.6 representing the proportion of homes that are own homes

Let X the random variable of interest, on this case we now that:  

X \sim Binom(n=15, p=0.6)  

The probability mass function for the Binomial distribution is given as:  

P(X)=(nCx)(p)^x (1-p)^{n-x}  

Where (nCx) means combinatory and it's given by this formula:  

nCx=\frac{n!}{(n-x)! x!}  

And we want this probability:

P(X=3)

And uing the probability mass function we got:

P(X=3)= 15C3 (0.6)^3 (1-0.6)^{15-3}= 0.00165

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