Question

Write a hypothesis test problem using one of the two options below. For whichever option you choose, gather appropriate data and post your problem (without a solution) in the discussion topic. You may use the same data that you gathered in your Module Three Discussion Topic post. Allow time for your classmates to post their solutions, and then respond to your own post with a solution for others to check their work.

Option 1:

Think about a population mean that you may be interested in and propose a hypothesis test problem for this parameter. Gather appropriate data and post your problem, Later, respond to your own post with your own solution.

For example, you may believe that the population mean number of times that adults go out for dinner each week is less than 1.5. Your data could be that you spoke with 7 people and found that they went out 2, 0, 1, 5, 0, 2, and 3 times last week. You then would choose to test this hypothesis at the .05 (or another) significance level. Assume a random sample.

Answer 1

Solution

Conceived Problem

Five months back, I had collected data on the time the students in my class spent doing homework. The analysis had shown that on an average my friends spent 1.25 hours per day on working out the homework.

Informal discussions recently led me to feel that this time has gone down.

And I want to test my ‘feeling’

Data

I spoke with 16 of my friends and recorded the time they spent on doing homework the day before.

Here is the data in minutes:

55, 67, 75, 50, 90, 100, 45, 65, 50, 55, 80, 50, 60, 40, 35, 30.

Do these figures confirm my guess?

Solution

Let X = the time spent on doing homework (in minutes).

Let µ and σ be respectively the mean and standard deviation of X

Claim: The time spent on doing homework has come down

Hypotheses:

Null H₀: µ = µ₀ = 75   Vs Alternative H_A: µ < 75

Test statistic:

t = (√n)(Xbar - µ₀)/s = - 2.1315,

where

n = sample size;

Xbar = sample average;

s = sample standard deviation.

Summary of Excel Calculations is given below:

n	16
µ0	75
Xbar	59.1875
s	19.47381
tcal	- 3.24795
α assumed	0.05
tcrit	- 2.13145
p-value	0.002703

Distribution, significance level, α , Critical Value and p-value

Under H₀, t ~ t_{n – 1}

Critical value = lower α% point of t_{n – 1}.

p-value = P(t_{n – 1} < tcal)

Using Excel Function, Statistical TDIST and TINV, tcrit and p-value are found to be as shown in the above table.

Decision:

Since tcal < tcrit, or equivalently since p-value < α. H₀ is rejected.

Conclusion:

There is sufficient evidence to support the claim and hence we conclude that

The time spent on doing homework has come down Answer

DONE