Arecent study stated that if a person chewed gum, the average number of sticks of gum he or she chewed daily was 8. researcher believes the number of sticks is greater. he selected a random sample of 36 gum chewers and found the mean number of sticks of gum chewed per day was 9. the standard deviation of the population is 1. at α= 0.05 level of significance test the claim that the number of sticks of gum a person chews per day is actually greater than 8.

The average number of sticks of gum a person chewed daily was greater than 8.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 8

Sample mean, = 9

Sample size, n = 36

Alpha, α = 0.05

Population standard deviation, σ = 1

First, we design the null and the alternate hypothesis

We use one-tailed z test to perform this hypothesis.

Formula:

Putting all the values, we have

Now,

Since,  

We fail to accept the null hypothesis and reject the null hypothesis. We accept the alternate hypothesis.

Thus, the average number of sticks of gum a person chewed daily was greater than 8.

this is a direct variation proportion. y = kx

y is the total amount due, k is the constant of variation, and x is the number of texts sent.

plug in the 3.20 and the 16 to solve for the "k". this is the constant of proportionality. it is the value that will not change.

3.20 = k(16). solving you get $.20 per text.

the equation would then be y = .20x

this is a good question! i’ll do some research and what i find, i’ll contact you asap when i have something! ✊