For a given confidence level, t ? df is larger than z ? . Explain how t ∗ df being slightly larger than z ∗ affects the width of the confidence interval.

The width of confidence interval is a function of the margin of error. If the critical value of t(t*) is slightly larger than the critical value of z(z*), then the width of the confidence interval will be larger.

The margin of error is the product of the critical value and the standard error. Therefore, given the same standard error value, the value of the margin of error will increases based on the value of the critical value.

Since, t* is slightly larger than z*, then the confidence interval, t will be wider.

Learn more : link

Answer is below

Step-by-step explanation:

The width of the CI is directly proportional to critical value. When t* is greater than z value, the t value would then cause the margin of error to be larger and this will in turn cause the width of the confidence interval to be larger.

Greater t*df than z* gives us a bigger margin of error. This would in turn give bigger width of confidence interval. t distribution has greater width confidence interval compared to z distribution.

connie < felicia < stew

step-by-step explanation:

stew:

we need to find the slope

m = (y2-y1)/ (x2-x1)

    = (2650-795)/(10-3)

    = 1855/7

265 words per minute

connie:

we need to find the slope

m = (y2-y1)/ (x2-x1)

we have two points (0,0) and (4,800)

m = (800-0)/(4-0)

  m = 800/4

    200 words per minute

felicia:

260 words per minute

200< 260 < 265

connie < felicia < stew