Question

"Dalmatian Darnation," by Kathy Sparling

A greedy dog breeder named Spreckles

Bred puppies with numerous freckles

The Dalmatians he sought

Possessed spot upon spot

The more spots, he thought, the more shekels.

His competitors did not agree

That freckles would increase the fee.

They said, “Spots are quite nice

But they don't affect price;

One should breed for improved pedigree.”

The breeders decided to prove

This strategy was a wrong move.

Breeding only for spots

Would wreak havoc, they thought.

His theory they want to disprove.

They proposed a contest to Spreckles

Comparing dog prices to freckles.

In records they looked up

One hundred one pups:

Dalmatians that fetched the most shekels.

They asked Mr. Spreckles to name

An average spot count he'd claim

To bring in big bucks.

Said Spreckles, “Well, shucks,

It's for one hundred one that I aim.”

Said an amateur statistician

Who wanted to help with this mission.

“Twenty-one for the sample

Standard deviation's ample:

They examined one hundred and one

Dalmatians that fetched a good sum.

They counted each spot,

Mark, freckle and dot

And tallied up every one.

Instead of one hundred one spots

They averaged ninety six dots

Can they muzzle Spreckles’

Obsession with freckles

Based on all the dog data they've got?

Short answer

The alternate hypothesis states that the mean dots on Dalmatian are less than$101$remains true.

Step-by-step solution

Given information

The number of pups taken for sample standard deviation is$21$. Out of$101$spots on Dalmatian, they got an average of$96$dots.

Explanation

Hypothesis:

The null hypothesis claims that the Dalmatian mean dots are at least $101$, while the alternate is the Dalmatian mean dots are less than $101$

$$\begin{gathered} H_0:\mu \ge 101 \\ {H}_a:\mu <101 \end{gathered}$$

The student's t-distribution is,

$$\begin{gathered} t_n-1=t_{101}-1 \\ =t_{100} \end{gathered}$$

Thez value is calculated as

localid="1649765554909" $$\begin{gathered} z=\frac{\overline{X}-\mu }{\sigma /\sqrt{n}} \\ \text{wheresisthestandarddeviation,andnisthesamplesize.} \\ \overline{X}=96 \\ n=101 \\ z=\frac{96-100}{21/\sqrt{101}} \\ =\frac{-5}{2.089} \\ =-2.393 \end{gathered}$$