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levacccp [35]
3 years ago
15

A 90 % confidence interval for the average salary of all CEOs in the electronics industry was constructed using the results of a

random survey of 45 CEOs. the interval was ($133, 306, $150, 733). To make useful inferences from the data, it is desired to reduce the width of the confidence interval. Which of the following will result in a reduced interval width?
A) Increase the sample size and increase the confidence level.
B) Decrease the sample size and increase the confidence level.
C) Decrease the sample size and decrease the confidence level.
D) Increase the sample size and decrease the confidence level.
Mathematics
1 answer:
alisha [4.7K]3 years ago
8 0

Answer: D) Increase the sample size and decrease the confidence level.

Step-by-step explanation:

A reduced interval width means that the data is more accurate. This can only be achieved if the sample size is increased because a larger sample size is able to capture more of the characteristics of the variables being tested.

A smaller confidence interval will also lead to a reduced interval width because it means that the chances of the prediction being correct have increased.

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3 years ago
Find the value of x ANSWER IS NOT
gulaghasi [49]

Answer:

x = 11√2

Step-by-step explanation:

We can notice the triangle given is a special triangle, a 45°- 45°- 90° triangle. The hypotenuse of a 45°- 45°- 90° is √2 times the length of a side. We can set up the equation(using x as the side):

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This simplifies to:

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8 0
3 years ago
A company surveyed 2400 men where 1248 of the men identified themselves as the primary grocery shopper in their household. ​a) E
polet [3.4K]

Answer:

a) With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

b) The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

c) \alpha =1-0.98=0.02

Step-by-step explanation:

If np' and n(1-p') are higher than 5, a confidence interval for the proportion is calculated as:

p'-z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }\leq  p\leq p'+z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }

Where p' is the proportion of the sample, n is the size of the sample, p is the proportion of the population and z_{\alpha/2} is the z-value that let a probability of \alpha/2 on the right tail.

Then, a 98% confidence interval for the percentage of all males who identify themselves as the primary grocery shopper can be calculated replacing p' by 0.52, n by 2400, \alpha by 0.02 and z_{\alpha/2} by 2.33

Where p' and \alpha are calculated as:

p' = \frac{1248}{2400}=0.52\\\alpha =1-0.98=0.02

So, replacing the values we get:

0.52-2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\leq  p\leq 0.52+2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\\0.52-0.0238\leq p\leq 0.52+0.0238\\0.4962\leq p\leq 0.5438

With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

Finally, the level of significance is the probability to reject the null hypothesis given that the null hypothesis is true. It is also the complement of the level of confidence. So, if we create a 98% confidence interval, the level of confidence 1-\alpha is equal to 98%

It means the the level of significance \alpha is:

\alpha =1-0.98=0.02

4 0
3 years ago
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