Question

Conduct the appropriate hypothesis test and compute the test statistic. A company that produces fishing line undergoes random testing to see if their fishing line holds up to the advertised specifications. Currently they are producing 30-pound test line and 20 randomly selected pieces are selected to test the strength. The 20 pieces broke with an average force of 29.1 pounds and a sample standard deviation of 2 pounds. Assuming that the strength of the fishing line is normally distributed, perform the appropriate hypothesis test at a 0.05 significance level in order to determine whether there is sufficient sample evidence to conclude the fishing line breaks with an average force of less than 30 pounds.

a. No, because the test statistic is -2.01.
b. No, because the test statistic is -2.52
c. Yes, because the test statistic is -2.52
d. Cannot be determined Yes, because the test statistic is -2.01

Answer 1

Option D) Yes, because the test statistic is -2.01

Explanation:

We are given the following in the question:

Population mean, μ = 30 pound

Sample mean,
`\bar{x}` = 29.1 pounds

Sample size, n = 20

Alpha, α = 0.05

Sample standard deviation, s = 2 pounds

First, we design the null and the alternate hypothesis

`H_(0): \mu = 30\text{ pounds}\\H_A: \mu < 30\text{ pounds}`

We use one-tailed(left) t test to perform this hypothesis.

Formula:

`t_(stat) = \displaystyle\frac{\bar{x} - \mu}{(\sigma)/(√(n)) }`

Putting all the values, we have

`t_(stat) = \displaystyle(29.1 - 30)/((2)/(√(20)) ) = -2.012`

Now,

`t_(critical) \text{ at 0.05 level of significance, 19 degree of freedom } = -1.729`

Since,

`t_(stat) < t_(critical)`

We fail to accept the null hypothesis and reject it. We accept the alternate hypothesis. Thus, there were enough evidence to conclude that the fishing line breaks with an average force of less than 30 pounds.

Option D) Yes, because the test statistic is -2.01