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.

560

hope this : )

step-by-step explanation:

part a. the differences are decreasing, so we know the sequence is not arithmetic. the ratios are constant (=3/4), so we know the sequence is geometric.

part b. from part a, we know the common ratio is 3/4, so each term is 3/4 of the one before. the initial term is the number of gallons at h=1.

g[h] = (3/4)·g[h-1]

g[1] = 384 . . . . initial value required for a recursive definition

part c. the general formula for a geometric sequence is

a[n] = a[1]·r^(n-1)

we have a[1] = 384, r = 3/4, and the term number given by h. the formula that follows is

g[h] = 384·(3/4)^(h-1)

part d. filling in the value and doing the arithmetic, we have

g[12] = 384·(3/4)^11 ≈ 384·0.042235136 ≈ 16.2183 . . gallons

part e. each gallon is 4 quarts, so

16.2183 gal × 4 qt/gal = 64.8732 qt