How will decreasing the level of confidence without changing the sample size affect the width of a confidence interval for a population mean? Assume that the population standard deviation is unknown and the population distribution is approximately normal

Without changing the sample size the change of CI may affect the results

Step-by-step explanation:

According to Central Limit theorem the sampling distribution is given by

Z= x`- u/ σ/√n

Z has in the limit a standard normal distribution and

x`= u ± zσ/√n

From the above

x`- z∝(σ/√n) ≤ u ≤ x`+ z∝(σ/√n)

This formula is used for the confidence interval  with normal population and unknown standard deviation.

But if the different values of Z∝ are used the results will be different.

If the CI of 99% or 95% or 90% is used the values of acceptance and rejection regions change and therefore the results will change.

The value of Z∝ for

∝= 0.1 is ± 1.645

∝= 0.05 is ± 1.96

∝= 0.01 is ± 2.58

Suppose we get the calculated Z value equal 2.59 but we decrease the CI from 0.05 to 0.01 the acceptance region would become rejection region  and the level of confidence will change.

step-by-step explanation:

a.   since the profit made in 2012 is equal to the profit made in 2014, the profit is           symmetric about the line y = 120,000.

step-by-step explanation:

given that megan owns a clothing boutique.

the profits of 2010, 11, are recorded and shown as follows:

years since     2010 0     1     2     3       4       5

profits made in 1000's   39       84     111     120   111     84

we find that profit gradually increases form 2010 to 2013 upto 120,   then starts decreasing.

the profits for 2012 and 2014 are the same while that of 2011 and 2015 are also the same.

this shows that there is symmetry about the line profit = 120000

hence answer is:

a.   since the profit made in 2012 is equal to the profit made in 2014, the profit is           symmetric about the line y = 120,000.

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