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ruslelena [56]
3 years ago
12

A random sample is made of 256 middle managers concerning their annual incomes. The sample mean is computed to be $35,420. If th

e population standard deviation is $2,050. What is the standard error of the mean?
a. $128.125
b. $138.36
c. $2.050
d. $8.01
Mathematics
1 answer:
ziro4ka [17]3 years ago
3 0

Answer: a. $128.125

Step-by-step explanation:

The standard error of the mean is given by :

SE=\dfrac{\sigma}{\sqrt{n}} , where \sigma = population standard deviation , n= sample size.

Given: \sigma=\$2,050

n= 256

Then, the standard error of the mean:-

SE=\dfrac{2050}{\sqrt{256}}\\\\=\dfrac{2050}{16}=\$128.125

Hence, option a. is correct.

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3 years ago
What is the frequency of the function f(x)?<br> Express the answer in fraction form.
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Answer:

<u>Frequency = 1/period = \frac{1}{8\pi }</u>

Step-by-step explanation:

The frequency of the sinusoidal function = 1/period

Frequency is how many the function repeats itself per unit if time i.e: per "1"

For the given graph :

\frac{x}{4} =Bx , Where: B = 2π/period

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Read 2 more answers
Twenty-five students from Harry High School were accepted at Magic University. Of those students, 10 were offered athletic schol
Taya2010 [7]

Answer:

Step-by-step explanation:

Part A

For Athletic scholarship,

Mean = (16 + 24 + 20 + 25 + 24 + 23 + 21 + 22 + 20 + 20)/10 = 21.5

Standard deviation = √(summation(x - mean)²/n

n = 10

Summation(x - mean)² = (16 - 21.5)^2 + (24 - 21.5)^2 + (20 - 21.5)^2 + (25 - 21.5)^2 + (24 - 21.5)^2 + (23 - 21.5)^2 + (21 - 21.5)^2 + (22 - 21.5)^2 + (20 - 21.5)^2 + (20 - 21.5)^2 = 64.5

Standard deviation = √64.5/10 = 2.54

For non athletic scholarship,

Mean = (23 + 25 + 26 + 30 + 32 + 26 + 28 + 29 + 26 + 27 + 29 + 27 + 22 + 24 + 25)/15 = 26.6

n = 15

Summation(x - mean)² = (23 - 26.6)^2 + (25 - 26.6)^2 + (26 - 26.6)^2 + (30 - 26.6)^2 + (32 - 26.6)^2 + (26 - 26.6)^2 + (28 - 26.6)^2 + (29 - 26.6)^2 + (26 - 26.6)^2 + (27 - 26.6)^2 + (29 - 26.6)^2 + (27 - 26.6)^2 + (22 - 26.6)^2 + (24 - 26.6)^2 + (25 - 26.6)^2 = 101.6

Standard deviation = √101.6/15 = 2.6

This is a test of 2 independent groups. The population standard deviations are not known. it is a two-tailed test. Let 1 be the subscript for scores of athletes and 2 be the subscript for scores of non athletes.

Therefore, the population means would be μ1 and μ2

The random variable is x1 - x2 = difference in the sample mean scores of athletes and non athletes.

We would set up the hypothesis.

The null hypothesis is

H0 : μ1 = μ2 H0 : μ1 - μ2 = 0

The alternative hypothesis is

H1 : μ1 ≠ μ2 H1 : μ1 - μ2 ≠ 0

Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(x1 - x2)/√(s1²/n1 + s2²/n2)

From the information given,

x1 = 21.5

x2 = 26.6

s1 = 2.54

s2 = 2.6

n1 = 10

n2 = 15

t = (21.5 - 26.6)/√(2.54²/10 + 2.6²/15)

t = - 4.65

The formula for determining the degree of freedom is

df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²

df = [2.54²/10 + 2.6²/15]²/[(1/10 - 1)(2.54²/10)² + (1/15 - 1)(2.6²/15)²] = 1.2008/0.1039

df = 12

We would determine the probability value from the t test calculator. It becomes

p value = 0.00056

Since alpha, 0.1 > than the p value, 0.00056, then we would reject the null hypothesis.

Therefore, these data provide convincing evidence of a difference in ACT scores between athletes and nonathletes.

Part B

The formula for determining the confidence interval for the difference of two population means is expressed as

Confidence interval = (x1 - x2) ± z√(s²/n1 + s2²/n2)

For a 90% confidence level, the z score from the normal distribution table is 1.645

x1 - x2 = 21.5 - 26.6 = - 5.1

√(s1²/n1 + s2²/n2) = √(2.54²/10 + 2.6²/15) = 1.05

The confidence interval is - 5.1 ± 1.05

This analysis provides evidence that the mean scores for non athletes is higher than the mean scores for athletes, and that the difference between means in the population is likely to be between 4.05 and 6.15

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