Someone claims that the breaking strength of their climbing rope is 2,000 psi, with a standard deviation of 10 psi. We think the actual amount is lower than that and want to run the test at an alpha level of 5%. What would our sample size need to be if we want to reject the null hypothesis if the sample mean is at or below 1,997.2956?

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 = mean breaking strength of their climbing rope

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.

3,6 and 7.

step-by-step explanation:

a) unit place of 37 is 7

so the unit place of cube of 37 will be

7³= 343=3

the answer is 3.

b) unit place of 56 is 6, so the unit place of cube of 56 will be

6³= 216=6

the answer is 6.

c) unit place of 133 is 3, so the unit place of cube of 133 will be

3³= 27= 7.

the answer is 7.