Complete each table and describe the rule for the relationship. Meal time: 12:00,12:30,1:00,1:30
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Answer:
The sample size must be greater than 37 if we want to reject the null hypothesis.
Step-by-step explanation:
We are given that someone claims that the breaking strength of their climbing rope is 2,000 psi, with a standard deviation of 10 psi.
Also, we are given a level of significance of 5%.
Let = <u><em>mean breaking strength of their climbing rope</em></u>
SO, Null Hypothesis, :
= 2,000 psi {means that the mean breaking strength of their climbing rope is 2,000 psi}
Alternate Hypothesis, :
< 2,000 psi {means that the mean breaking strength of their climbing rope is lower than 2,000 psi}
Now, the test statistics that we will use here is One-sample z-test statistics as we know about population standard deviation;
T.S. = ~ N(0,1)
where, = ample mean strength = 1,997.2956 psi
= population standard devaition = 10 psi
n = sample size
Now, at the 5% level of significance, the z table gives a critical value of -1.645 for the left-tailed test.
So, to reject our null hypothesis our test statistics must be less than -1.645 as only then we have sufficient evidence to reject our null hypothesis.
SO, T.S. < -1.645 {then reject null hypothesis}
n > 36.99 ≈ 37.
SO, the sample size must be greater than 37 if we want to reject the null hypothesis.