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Bas_tet [7]
3 years ago
5

Last year, 50% of MNM. Inc. employees were female. It is believed that there has been a reduction in the percentage of females i

n the company. This year, in a random sample of 400 employees. 180 were female. a. What are the null and the alternative hypotheses?b. At 95% confidence using the critical value approach, determine if there has been a significant reduction in the proportion of females.
Mathematics
1 answer:
Serhud [2]3 years ago
3 0

Answer:

We conclude that there has been a significant reduction in the proportion of females.

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 400

p = 50% = 0.5

Alpha, α = 0.05

Number of women, x = 118

First, we design the null and the alternate hypothesis  

H_{0}: p = 0.50\\H_A: p < 0.50

This is a one-tailed test.  

Formula:

\hat{p} = \dfrac{x}{n} = \dfrac{118}{400} = 0.45

z = \dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}

Putting the values, we get,

z = \displaystyle\frac{0.45-0.50}{\sqrt{\frac{0.50(1-0.50)}{400}}} = -2

Now, we calculate the critical value.

Now, z_{critical} \text{ at 0.05 level of significance } = -1.645

Since the calculated z-statistic is less than the critical value, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

Thus, there has been a significant reduction in the proportion of females.

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Step-by-step explanation:

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5 0
3 years ago
Do male and female servers work the same number of hours? A sample of 25 female servers worked an average of 26 hours per week,
Vera_Pavlovna [14]

Answer:

Step-by-step explanation:

Given that

Group   Group One     Group Two  

Mean       26.00 23.00

SD                 2.00 4.00

SEM         0.40 1.21

N                     25     11    

where group I represents female servers and group II male servers.

We have to calculate confidence interval for 90% for difference in means

The mean of Group One minus Group Two equals 3.00

df = 34  

 standard error of difference = 0.993

t critical = 2.034 for 90% df 34

Hence confid. interval at 90%

=Mean diff ±2.034 * std error of diff

= (0.98, 5.02)

5 0
3 years ago
Need help on #7 help if u know how to solve thanks!
mash [69]
7a) 302.5
7b) 302500
Hope this helps! :)
8 0
3 years ago
Read 2 more answers
How i can prove the proofs?
Marianna [84]
You can prove the proofs by showing your work to prove the proofs.
3 0
3 years ago
The angle 118 degrees is also equal to
LekaFEV [45]
I don't understand your question
8 0
3 years ago
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