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Gelneren [198K]
3 years ago
11

It has been found that women are excluded from social events with coworkers. One study at Company X, found that women only recei

ved 38% of the invites to events outside of the office compared to male colleagues.
In an effort to increase this invite rate, it was suggested that women implement 2 behavioral changes:

Go to work 10 minutes early. This doesn’t interfere with any childcare and can be used to bump into colleagues and form new connections.

Invite colleagues yourself to activities you feel comfortable with: whether it’s a rock climbing gym, wine tasting or a dinner hosted at your house with your family, this gives co-workers a chance to connect with you in a more personal, friendly way.

Women practiced these two strategies for 3 months and the researchers concluded that after this intervention women were found to receive 43% of the invites to events outside of the office compared to male colleagues.

The p-value in this case is the probability of getting data like that observed or more extreme (i.e., getting an observed invitation rate of 33% or higher) assuming that:

a.The sample was random.

b.The social strategies did increase the rate of invitations for female colleagues.

c.The social strategies did not increase the rate of invitations for female colleagues.

d.The sample size is large enough.
Mathematics
1 answer:
Natasha2012 [34]3 years ago
8 0

Answer:

Option C is correct.

The p-value in this case is the probability of getting data like that observed or more extreme (i.e., getting an observed invitation rate of 33% or higher) assuming that;

The social strategies did not increase the rate of invitations for female colleagues.

Step-by-step explanation:

The p-value is actually defined as the probability of obtaining results as extreme as the observed results in the statistical analysis provided that the null hypothesis is true.

The standing condition for the p-value definition is the assumption that the null hypothesis is true, then it is the probability of getting the extreme result (usually not due to just random chance), that is usually observed, being tested for and contained in the alternative hypothesis.

Normally, in hypothesis testing, especially one comparing two sets of data, the null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the directions of the test. It usually maintains that, with random chance responsible for the outcome or results of any experimental study/hypothesis testing, its statement is true.

The alternative hypothesis usually confirms the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the directions of the test. It usually maintains that significant factors other than random chance, affect the outcome or results of the experimental study/hypothesis testing and result in its own statement.

Since this test aimed to test whether the rate of invitation of women to outside work events after using the social strategies, the null hypothesis for the question would be that there is no significant difference in the rate of invitation of the women after using the social strategies while the alternative hypothesis would be that the social strategies worked and the rate invitation of women to outside of work events increased.

So, the null hypothesis that we need to assume was true in the p-value definition is that; the social strategies did not increase the rate of invitations for female colleagues.

This is option C!

Hope this Helps!!!

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Step-by-step explanation:

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The Z-score reflects how far the measure deviates from the mean. After determining the Z-score, we examine the z-score table to determine the p-value associated with this z-score. This p-value represents the likelihood that the measure's value is less than X, or the percentile of X. Subtracting 1 from the p-value yields the likelihood that the measure's value is larger than X.

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This means, ц  = 2000 and б = 100.

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This means that n = 20, s=\frac{100}{\sqrt{120} } =22.361

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This is the 100-95th percentile, or X when Z has a p-value of 0.05, or X when Z = -1.645. So Z=\frac{X-u}{a}

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Z=\frac{X-u}{a} \\-1.645=\frac{X-2000}{22.361} \\X-2000=-1.645*22.361

X =1963.2

<h3>Answer:</h3>

The sample mean that will cut off the top 95% of all size 20 samples obtained from the population is 1963.2 pounds.

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