State whether each of the following changes would make a confidence interval wider or narrower. (assume that nothing else changes.) a. changing from a 95% confidence level to a 99% confidence level. b. changing from a sample size of 15 to a sample size of 350. c. changing from a standard deviation of 15 pounds to a standard deviation of 20 pounds.

Step-by-step explanation:

The formula for determining confidence interval is expressed as

Confidence interval

= mean ± z × s/ √n

Where

z is the value of the z score

s = standard deviation

n = sample size

a) The 95% confidence level has a z value of 1.96

The 99% confidence level has a z value of 2.58

Since 99% confidence level z value is greater than 95% confidence level z value, if we input it into the formula, it will result to a higher confidence interval. So changing from a 95% confidence level to a 99% confidence level would make a confidence interval wider.

b) The √15 is smaller than the √350. This means that if we make use of the formula, √350 will give a lower confidence interval than that of √15. Therefore, the confidence interval would be narrower changing from a sample size of 15 to a sample size of 350.

c) Applying the formula, a standard deviation of 15 pounds would result to a lower confidence interval than a standard deviation of 20 pounds. Therefore, the confidence interval would be wider changing from a standard deviation of 15 pounds to a standard deviation of 20 pounds.

the answer is vanilla

step-by-step explanation:

45.36 feet per second.

step-by-step explanation:

to find the most accurate determination for the number of feet per second tyrone traveled on his way to work let us convert miles to feet and minutes to seconds.

1 mile = 5280 feet.

1 minute = 60 seconds.

now we will divide 158400 by 3492 to find the most accurate determination for the number of feet per second tyrone traveled on his way to work.

therefore, tyrone traveled on his way to work at the rate of 45.36 feet per second.