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evablogger [386]
2 years ago
14

A sample of 75 information systems managers had an average hourly income of $40.75 with a standard deviation of $7.00. The 95% c

onfidence interval for the average hourly wage (in $) of all information system managers is
a. 37.54 to 43.96.
b. 38.61 to 42.89.
c. 39.14 to 42.36.
d. 39.40 to 42.10.
Mathematics
1 answer:
IrinaVladis [17]2 years ago
6 0

Answer: c. 39.14 to 42.36

Step-by-step explanation:

We want to determine a 95% confidence interval for the average hourly wage (in $) of all information system managers

Number of sample, n = 75

Mean, u = $40.75

Standard deviation, s = $7.00

For a confidence level of 95%, the corresponding z value is 1.96. This is determined from the normal distribution table.

We will apply the formula

Confidence interval

= mean +/- z ×standard deviation/√n

It becomes

40.75 +/- 1.96 × 7/√75

= 40.75 ± 1.96 × 0.82

= 40.75 + 1.6072

The lower end of the confidence interval is 40.75 - 1.6072 =39.14

The upper end of the confidence interval is 40.75 + 1.6072 =42.36

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Using four or more complete sentences, contrast race and ethnicity and provide an example of each.
Brums [2.3K]

Answer:

Race and Ethnicity are considered to be social constructs that enable humans to be grouped or categorized. They are sometimes used interchangeably but most argue that they are different and give them different definitions.

Race is generally used to refer to people that have the same physical characteristics such as skin color or the same texture of hair for instance Black and White/Caucasian.

Ethnicity on the other hand relates to people sharing a common culture, language, nation, and/ or tribal heritage for instance, the English, the Irish and the Igbo.

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2 years ago
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Concentric circles are circles with the same center but different radii. Which equations represent concentric circles along with
Harrizon [31]
<span>(x – 8)² + (y – 9)² = 3
(x – 8)² + (y – 9)² = 14
</span>are the correct answers
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2 years ago
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Suppose you are an expert on the fashion industry and wish to gather information to compare the amount earned per month by model
Ann [662]

Answer:

(1) The degrees of freedom for unequal variance test is (14, 11).

(2) The decision rule for the 0.01 significance level is;

  • If the value of our test statistics is less than the critical values of F at 0.01 level of significance, then we have insufficient evidence to reject our null hypothesis.      
  • If the value of our test statistics is more than the critical values of F at 0.01 level of significance, then we have sufficient evidence to reject our null hypothesis.  

(3) The value of the test statistic is 0.3796.

Step-by-step explanation:

We are given that you are an expert on the fashion industry and wish to gather information to compare the amount earned per month by models featuring Liz Claiborne's attire with those of Calvin Klein.

The following is the amount ($000) earned per month by a sample of 15 Claiborne models;

$3.5, $5.1, $5.2, $3.6, $5.0, $3.4, $5.3, $6.5, $4.8, $6.3, $5.8, $4.5, $6.3, $4.9, $4.2 .

The following is the amount ($000) earned by a sample of 12 Klein models;

$4.1, $2.5, $1.2, $3.5, $5.1, $2.3, $6.1, $1.2, $1.5, $1.3, $1.8, $2.1.

(1) As we know that for the unequal variance test, we use F-test. The degrees of freedom for the F-test is given by;

\text{F}_(_n__1-1, n_2-1_)

Here, n_1 = sample of 15 Claiborne models

         n_2 = sample of 12 Klein models

So, the degrees of freedom = (n_1-1, n_2-1) = (15 - 1, 12 - 1) = (14, 11)

(2) The decision rule for 0.01 significance level is given by;

  • If the value of our test statistics is less than the critical values of F at 0.01 level of significance, then we have insufficient evidence to reject our null hypothesis.      
  • If the value of our test statistics is more than the critical values of F at 0.01 level of significance, then we have sufficient evidence to reject our null hypothesis.  

(3) The test statistics that will be used here is F-test which is given by;

                          T.S. = \frac{s_1^{2} }{s_2^{2} } \times \frac{\sigma_2^{2} }{\sigma_1^{2} }  ~ \text{F}_(_n__1-1, n_2-1_)

where, s_1^{2} = sample variance of the Claiborne models data = \frac{\sum (X_i-\bar X)^{2} }{n_1-1} = 1.007

s_2^{2} = sample variance of the Klein models data = \frac{\sum (X_i-\bar X)^{2} }{n_2-1} = 2.653    

So, the test statistics =  \frac{1.007}{2.653 } \times 1  ~ \text{F}_(_1_4,_1_1_)

                                   = 0.3796

Hence, the value of the test statistic is 0.3796.

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2 years ago
A sample of size 60 from one population of weights had a sample average of 10.4 lb. and a sample standard deviation of 2.7 lb. A
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Answer:

z=  0.278

Step-by-step explanation:

Given data

n1= 60 ; n2 = 100

mean 1= x1`= 10.4;     mean 2= x2`= 9.7

standard deviation  1= s1= 2.7 pounds ;  standard deviation 2= s2 = 1.9 lb

We formulate our null and alternate hypothesis as

H0 = x`1- x`2 = 0 and H1 = x`1- x`2 ≠ 0 ( two sided)

We set level of significance α= 0.05

the test statistic to be used under H0 is

z = x1`- x2`/ √ s₁²/n₁ + s₂²/n₂

the critical region is z > ± 1.96

Computations

z= 10.4- 9.7/ √(2.7)²/60+( 1.9)²/ 100

z= 10.4- 9.7/ √ 7.29/60 + 3.61/100

z= 0.7/√ 0.1215+ 0.0361

z=0.7 /√0.1576

z= 0.7 (0.396988)

z= 0.2778= 0.278

Since the calculated value of z does not fall in the critical region so we accept the null hypothesis H0 = x`1- x`2 = 0  at 5 % significance level. In other words we conclude that the difference between mean scores is insignificant or merely due to chance.

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2 years ago
Write an inequality that represents the situation: a large box of golf balls has more than 12 balls. Describe how your inequalit
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12>x

X = the number of golf balls, inequality states that the number of golf balls is over 12
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